Advanced General Mathematics Worked Answers Enhanced Edition NO COPY WRITE INFRINGEMENT INTENDED

Chapter 1 – Matrices Exercise 1A Solutions 1 a Number of rows × number of columns = 2 × 2 b Number of rows × number of columns = 2 × 3 c Number of rows × number of columns = 1 × 4 d Number of rows × number of columns = 4 × 1 2 a There will be 5 rows and 5 columns to

match the seating. Every seat of both diagonals is occupied, and so the diagonals will all be ones, and the rest of the numbers, representing unoccupied seats, will all be 0.

⎣⎢⎢⎢⎢⎢⎡ 1

0001

01010

00100

01010

10001⎦⎥⎥⎥⎥⎥⎤

b If all seats are occupied, then every

number in the matrix will be 1.

⎣⎢⎢⎢⎢⎢⎡ 1

1111

11111

11111

11111

11111⎦⎥⎥⎥⎥⎥⎤

3 i = j for the leading diagonal only, so the

leading diagonal will be all ones, and the rest of the numbers 0.

⎣⎢⎢⎢⎢⎢⎡ 1

0000

01000

00100

00010

00001⎦⎥⎥⎥⎥⎥⎤

4 We can present this as a table with the girls on the top row, and the boys on the bottom row, in order of year level, i.e. years 7, 8, 9, 10, 11 and 12 going from left to right.

⎣⎢⎡ 200

110180117

13598

11089

5653

2833⎦⎥⎤

Alternatively, girls and boys could be the two columns, and year levels could run down from year 7 to 12, in order. This would give:

⎣⎢⎢⎢⎢⎢⎢⎢⎡ 200

1801351105628

11011798895333 ⎦

⎥⎥⎥⎥⎥⎥⎥⎤

5 a Matrices are equal only if they have the

same number of rows and columns, and all pairs of corresponding entries are equal.

The first two matrices have the same dimensions, but the top entries are not equal, so the matrices cannot be equal.

The last two matrices have the same dimensions and equal first (left) entries, so they will be equal if x = 4.

Thus, [0 x] = [0 4] if x = 4. b The first two matrices cannot be equal

because corresponding entries are not equal, nor can the second and third for the same reason.

The last matrix cannot equal any of the others because it has different dimensions. The only two that can be equal are the first and third.

⎣⎢⎡ 4

17–2⎦⎥⎤ =

⎣⎢⎡ x

17–2⎦⎥⎤ if x = 4

Cambridge Essential Advanced General Mathematics 3rd Edition Worked Solutions CD-ROM 1

c All three matrices have the same dimensions and all corresponding numerical entries are equal. They could all be equal.

⎣⎢⎡ 2

–1x

1043⎦⎥⎤ =

⎣⎢⎡ y

–10

1043⎦⎥⎤

=⎣⎢⎡ 2

–10

1043⎦⎥⎤ if x = 0, y = 2

6 a The entry corresponding to x is 2, and

the entry corresponding to y is 3, so x = 2 and y = 3. b The entry corresponding to x is 3, and

the entry corresponding to y is 2, so x = 3 and y = 2. c The entry corresponding to x is 4, and

the entry corresponding to y is –3, so x = 4 and y = –3. d The entry corresponding to x is 3, and

the entry corresponding to y is –2, so x = 3 and y = –2.

7 Let A, B, C and D be the columns and rows, in that order.

There are no roads from A to A, so the top left entry will be 0. There are 3 roads from A to B, so the next entry right will be 3. There is 1 road from A to C and no roads directly from A to D, so the next two entries right will be 1 and 0.

Continue to fill in the matrix.

⎣⎢⎢⎢⎡ 0

310

3021

1201

0110⎦⎥⎥⎥⎤

8 Write it as set out, with each row

representing players A, B, C, D and E respectively, and columns showing points, rebounds and assists respectively.

⎣⎢⎢⎢⎢⎢⎡ 21

84140

52181

531602 ⎦⎥⎥⎥⎥⎥⎤


Exercise 1B Solutions 1 Add the corresponding entries. X + Y =

⎣⎢⎡ 1 + 3

–2 + 0⎦⎥⎤ =

⎣⎢⎡ 4

–2⎦⎥⎤

Double each entry.

2X =⎣⎢⎡ 2 × 1

2 × –2⎦⎥⎤ =

⎣⎢⎡ 2

–4⎦⎥⎤

Multiply each entry in Y by 4 and add the corresponding entry for X.

4Y + X =⎣⎢⎡ 4 × 3 + 1

4 × 0 + –2⎦⎥⎤ =

⎣⎢⎡ 13

–2⎦⎥⎤

Subtract corresponding entries. X – Y =

⎣⎢⎡ 1 – 3

–2 – 0⎦⎥⎤ =

⎣⎢⎡ –2

–2⎦⎥⎤

Multiply each entry by –3.

–3A =⎣⎢⎡ –3 × 1

–3 × 2–3 × –1–3 × 3 ⎦

⎥⎤ =

⎣⎢⎡ –3

–63

–9⎦⎥⎤

Add B to the previous answer. –3A + B =

⎣⎢⎡ –3

–63

–9⎦⎥⎤ +

⎣⎢⎡ 4

–102⎦⎥⎤

=⎣⎢⎡ 1

–73

–7⎦⎥⎤

2 a Double each entry.

⎣⎢⎡ 3 × 2

4 × 26 × 22 × 2

2 × 21 × 2⎦

⎥⎤ =

⎣⎢⎡ 6

8124

42⎦⎥⎤

b

⎣⎢⎡ 1

402

03⎦⎥⎤ +

⎣⎢⎡ 2

804

06⎦⎥⎤ +

⎣⎢⎡ 2

611

04⎦⎥⎤

=⎣⎢⎡ 5

1817

013⎦⎥⎤

c The average will be the total divided by 3,

so divide each entry by 3.

⎣⎢⎡ 5 ÷ 3

18 ÷ 31 ÷ 37 ÷ 3

0 ÷ 313 ÷ 3⎦

⎥⎤ =

⎣⎢⎢⎢⎡ 5

3

6

1373

0

133 ⎦⎥⎥⎥⎤

3 Multiply each entry by the factor. 2A =

⎣⎢⎡ 2

0–24 ⎦⎥⎤

–3A =⎣⎢⎡ –3

03

–6⎦⎥⎤

–6A =⎣⎢⎡ –6

06

–12⎦⎥⎤

4 a As the matrices have the same

dimensions, corresponding terms can be added. They will simply be added in the opposite order.

Since the commutative law holds true for numbers, all corresponding entries in the added matrices terms will be equal, so the matrices will be equal.

b As the matrices have the same

dimensions, corresponding terms can be added. The first matrix will add the first two numbers, then the third, and the second matrix will add the second and third numbers first, then add the result to the first number.

Since the associative law holds true for numbers, all corresponding entries in the added matrices terms will be equal, so the matrices will be equal.

5 a Multiply each entry by 2. 2A =

⎣⎢⎡ 6

–44–4⎦⎥⎤

b Multiply each entry by 3. 3B =

⎣⎢⎡ 0

12–93 ⎦⎥⎤

c Add answers to a and b. 2A + 3B =

⎣⎢⎡ 6

–44

–4⎦⎥⎤ +

⎣⎢⎡ 0

12–93 ⎦⎥⎤

=⎣⎢⎡ 6

8–5–1⎦⎥⎤


d Subtract a from b. 3 B – 2A =

⎣⎢⎡ 0

12–93 ⎦⎥⎤ –

⎣⎢⎡ 6

–44

–4⎦⎥⎤

=⎣⎢⎡ –6

16–13

7 ⎦⎥⎤

6 a Add corresponding entries.

⎣⎢⎡ 1

003⎦⎥⎤ +

⎣⎢⎡ –1

210⎦⎥⎤ =

⎣⎢⎡ 0

213⎦⎥⎤

b Triple entries in Q, then add to

corresponding entries in P.

⎣⎢⎡ 1

003⎦⎥⎤ +

⎣⎢⎡ –3

630⎦⎥⎤ =

⎣⎢⎡ –2

633⎦⎥⎤

c Double entries in P, then subtract Q and add R.

⎣⎢⎡ 2

006⎦⎥⎤ –

⎣⎢⎡ –1

210⎦⎥⎤ +

⎣⎢⎡ 0

141⎦⎥⎤ =

⎣⎢⎡ 3

–137⎦⎥⎤

7 a If 2A – 3X = B, then 2A – B = 3X

3X = 2A – BX = 2

3 A – 1

3 B

= 23

⎣⎢⎡ 3

–114⎦⎥⎤ – 1

3 ⎣⎢⎡ 0

–2–1017 ⎦

⎥⎤

=

⎣⎢⎢⎢⎡ 2

3× 3 – 1

3× 0

23× –1 – 1

3× 2

23× 1 – 1

3× –10

23× 4 – 1

3× 17 ⎦

⎥⎥⎥⎤

=⎣⎢⎡ 2

04–3⎦⎥⎤

b If 3A + 2Y = 2B then 2Y = 2B – 3A Y = B – 1 1

2 A

=⎣⎢⎡ 0

–2–1017 ⎦

⎥⎤ – 1 1

2 ⎣⎢⎡ 3

–114⎦⎥⎤

=

⎣⎢⎢⎢⎡ 0 – 3

2× 3

–2 – 32× –1

–10 – 32× 1

17 – 32× 4 ⎦

⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡ – 9

2

– 12

– 232

11 ⎦⎥⎥⎥⎤

8 X + Y =

⎣⎢⎡ 150 + 160

100 + 10090 + 900 + 0

100 + 12075 + 50

50 + 400 + 0 ⎦

⎥⎤

=⎣⎢⎡ 310

200180

0220125

900 ⎦⎥⎤

The matrix represents the total production at two factories in two successive weeks.


Exercise 1C Solutions 1 AX =

⎣⎢⎡ 1

–1–23 ⎦⎥⎤ ⎣⎢⎡ 2

–1⎦⎥⎤

=⎣⎢⎡ 1 × 2 + –2 × –1

–1 × 2 + 3 × –1 ⎦⎥⎤

=⎣⎢⎡ 4

–5⎦⎥⎤

BX =⎣⎢⎡ 3

121⎦⎥⎤ ⎣⎢⎡ 2

–1⎦⎥⎤

=⎣⎢⎡ 3 × 2 + 2 × –1

1 × 2 + 1 × –1⎦⎥⎤

=⎣⎢⎡ 4

1⎦⎥⎤

AY =⎣⎢⎡ 1

–1–23 ⎦⎥⎤ ⎣⎢⎡ 1

3⎦⎥⎤

=⎣⎢⎡ 1 × 1 + –2 × 3

–1 × 1 + 3 × 3 ⎦⎥⎤

=⎣⎢⎡ –5

8 ⎦⎥⎤

IX =⎣⎢⎡ 1

001⎦⎥⎤ ⎣⎢⎡ 2

–1⎦⎥⎤

=⎣⎢⎡ 1 × 2 + 0 × –1

0 × 2 + 1 × –1⎦⎥⎤

=⎣⎢⎡ 2

–1⎦⎥⎤

AC =⎣⎢⎡ 1

–1–23 ⎦⎥⎤ ⎣⎢⎡ 2

111⎦⎥⎤

=⎣⎢⎡ 1 × 2 + –2 × 1

–1 × 2 + 3 × 11 × 1 + –2 × 1–1 × 1 + 3 × 1 ⎦

⎥⎤

=⎣⎢⎡ 0

1–12 ⎦⎥⎤

CA =⎣⎢⎡ 2

111⎦⎥⎤ ⎣⎢⎡ 1

–1–23 ⎦⎥⎤

=⎣⎢⎡ 2 × 1 + 1 × –1

1 × 1 + 1 × –12 × –2 + 1 × 31 × –2 + 1 × 3⎦

⎥⎤

=⎣⎢⎡ 1

0–11 ⎦⎥⎤

Use AC =⎣⎢⎡ 0

1–12 ⎦⎥⎤

(AC)X =⎣⎢⎡ 0

1–12 ⎦⎥⎤ ⎣⎢⎡ 2

–1⎦⎥⎤

=⎣⎢⎡ 0 × 2 + –1 × –1

1 × 2 + 2 × –1 ⎦⎥⎤

=⎣⎢⎡ 1

0⎦⎥⎤

Use BX =⎣⎢⎡ 4

1⎦⎥⎤

C(BX) =⎣⎢⎡ 2

111⎦⎥⎤ ⎣⎢⎡ 4

1⎦⎥⎤

=⎣⎢⎡ 2 × 4 + 1 × 1

1 × 4 + 1 × 1⎦⎥⎤

=⎣⎢⎡ 9

5⎦⎥⎤

AI =⎣⎢⎡ 1

–1–23 ⎦⎥⎤ ⎣⎢⎡ 1

001⎦⎥⎤

=⎣⎢⎡ 1 × 1 + –2 × 0

–1 × 1 + 3 × 01 × 0 + –2 × 1–1 × 0 + 3 × 1 ⎦

⎥⎤

=⎣⎢⎡ 1

–1–23 ⎦⎥⎤

IB =⎣⎢⎡ 1

001⎦⎥⎤ ⎣⎢⎡ 3

121⎦⎥⎤

=⎣⎢⎡ 1 × 3 + 0 × 1

0 × 3 + 1 × 11 × 2 + 0 × 10 × 2 + 1 × 1⎦

⎥⎤

=⎣⎢⎡ 3

121⎦⎥⎤

AB =⎣⎢⎡ 1

–1–23 ⎦⎥⎤ ⎣⎢⎡ 3

121⎦⎥⎤

=⎣⎢⎡ 1 × 3 + –2 × 1

–1 × 3 + 3 × 11 × 2 + –2 × 1–1 × 2 + 3 × 1 ⎦

⎥⎤

=⎣⎢⎡ 1

001⎦⎥⎤

BA =⎣⎢⎡ 3

121⎦⎥⎤ ⎣⎢⎡ 1

–1–23 ⎦⎥⎤

=⎣⎢⎡ 3 × 1 + 2 × –1

1 × 1 + 1 × –13 × –2 + 2 × 31 × –2 + 1 × 3⎦

⎥⎤

=⎣⎢⎡ 1

001⎦⎥⎤


A2 = AA =⎣⎢⎡ 1

–1–23 ⎦⎥⎤ ⎣⎢⎡ 1

–1–23 ⎦⎥⎤

=⎣⎢⎡ 1 × 1 + –2 × –1

–1 × 1 + 3 × –11 × –2 + –2 × 3–1 × –2 + 3 × 3 ⎦

⎥⎤

=⎣⎢⎡ 3

–4–811⎦⎥⎤

B2 = BB =⎣⎢⎡ 3

121⎦⎥⎤ ⎣⎢⎡ 3

121⎦⎥⎤

=⎣⎢⎡ 3 × 3 + 2 × 1

1 × 3 + 1 × 13 × 2 + 2 × 11 × 2 + 1 × 1⎦

⎥⎤

=⎣⎢⎡ 11

483⎦⎥⎤

Use CA =⎣⎢⎡1

0–11 ⎦⎥⎤

A(CA) =⎣⎢⎡ 1

–1–23 ⎦⎥⎤ ⎣⎢⎡1

0–11 ⎦⎥⎤

=⎣⎢⎡1 × 1 + –2× 0

–1× 1 + 3 × 01 × –1 + –2× 1–1× –1 + 3 × 1 ⎦

⎥⎤

=⎣⎢⎡ 1

–1–34 ⎦⎥⎤

Use A2 =⎣⎢⎡ 3

–4–811⎦⎥⎤

A2C =⎣⎢⎡ 3

–4–811⎦⎥⎤ ⎣⎢⎡2

111⎦⎥⎤

=⎣⎢⎡ 3 × 2 + –8 × 1

–4 × 2 + 11 × 13 × 1 + –8 × 1–4 × 1 + 11 × 1⎦

⎥⎤

=⎣⎢⎡–2

3–57 ⎦⎥⎤

2 a A product is defined only if the number

of columns in the first matrix equals the number of rows of the second. A has 2 columns and Y has 2 rows, so AY is defined. Y has 1 column and A has 2 rows, so YA is not defined. X has 1 column and Y has 2 rows, so XY is not defined. X has 1 column and 2 rows, so is not defined.

2X

C has 2 columns and I has 2 rows, so CI is defined. X has 1 column and I has 2 rows, so XI is not defined.

b AB =⎣⎢⎡ 2

000⎦⎥⎤ ⎣⎢⎡ 0

–302⎦⎥⎤

=⎣⎢⎡ 2 × 0 + 0 × –3

0 × 0 + 0 × –32 × 0 + 0 × 20 × 0 + 0 × 2⎦

⎥⎤

=⎣⎢⎡ 0

000⎦⎥⎤

3 No, because Q.2 part b shows that AB

can equal O, and A ≠ O, B ≠ O. 4 LX = [2 – 1]

⎣⎢⎡ 2

–3⎦⎥⎤

= [2 × 2 + –1 × –3] = [7]

XL =⎣⎢⎡ 2

–3⎦⎥⎤ [2 – 1]

=⎣⎢⎡ 2 × 2

–3 × 22 × –1–3 × –1⎦

⎥⎤

=⎣⎢⎡ 4

–6–23 ⎦⎥⎤

5 A product is defined only if the number

of columns in the first matrix equals the number of rows of the second.

This can only happen if m = n, in which case both products will be defined.

6

⎣⎢⎡ a

cbd⎦⎥⎤⎣⎢⎡ d

– c– ba ⎦⎥⎤

=⎣⎢⎡ a × d + b × – c

c × d + d × – ca × – b + b × ac × – b + d × a ⎦

⎥⎤

=⎣⎢⎡ ad – bc

00

ad – bc⎦⎥⎤ =

⎣⎢⎡ 1

001⎦⎥⎤

For the equations to be equal, all corresponding entries must be equal, therefore ad – bc = 1.

When written in reverse order, we get

⎣⎢⎡ d

– c– ba ⎦⎥⎤⎣⎢⎡ a

cbd⎦⎥⎤

=⎣⎢⎡ d × a + – b × c

– c × a + a × cd × b + – b × d– c × b + a × d ⎦

⎥⎤

=⎣⎢⎡ ad – bc

00

ad – bc⎦⎥⎤ =

⎣⎢⎡ 1

001⎦⎥⎤

Since ad – bc = 1.


7 We can use any values of a, b, c and d as long as ad – bc = 1.

(B + C)A =⎣⎢⎡ –1

034⎦⎥⎤ ⎣⎢⎡ 1

423⎦⎥⎤

=⎣⎢⎡ –1 × 1 + 3 × 4

0 × 1 + 4 × 4–1 × 2 + 3 × 30 × 2 + 4 × 3 ⎦

⎥⎤

=⎣⎢⎡ 11

167

12⎦⎥⎤

For example, a = 5, d = 2, b = 3, c = 3 satisfy ad – bc = 1 and give AB =

⎣⎢⎡ 5

332⎦⎥⎤⎣⎢⎡ 2

–3–35 ⎦⎥⎤ =

⎣⎢⎡ 1

001⎦⎥⎤

BA =⎣⎢⎡ 2

–3–35 ⎦⎥⎤⎣⎢⎡ 5

332⎦⎥⎤ =

⎣⎢⎡ 1

001⎦⎥⎤

9 ⎣⎢⎡ 5

2.5012

3.00⎦⎥⎤⎣⎢⎡ 1

2⎦⎥⎤ =

⎣⎢⎡ 5 × 1 + 12 × 2

2.50 × 1 + 3.00 × 2⎦⎥⎤

=⎣⎢⎡ 29

850⎦⎥⎤

Other values could be chosen. 8 One possible answer. A =

⎣⎢⎡ 1

423⎦⎥⎤ , B =

⎣⎢⎡ 0

213⎦⎥⎤ , C =

⎣⎢⎡ –1

–221⎦⎥⎤

1 × 5 min plus 2 × 12 min means 29 min for one milkshake and two banana splits. The total cost is $8.50.

A + B =⎣⎢⎡1 + 0

4 + 22 + 13 + 3⎦

⎥⎤ =

⎣⎢⎡1

636⎦⎥⎤

B + C =⎣⎢⎡0 + –1

2 + –21 + 23 + 1⎦

⎥⎤ =

⎣⎢⎡–1

034⎦⎥⎤

A(B + C) =⎣⎢⎡1

423⎦⎥⎤ ⎣⎢⎡–1

034⎦⎥⎤

=⎣⎢⎡1 × –1 + 2 × 0

4 × –1 + 3 × 01 × 3 + 2 × 44 × 3 + 3 × 4⎦

⎥⎤

=⎣⎢⎡–1

–41124⎦⎥⎤

AB =⎣⎢⎡1

423⎦⎥⎤ ⎣⎢⎡0

213⎦⎥⎤

=⎣⎢⎡1 × 0 + 2 × 2

4 × 0 + 3 × 21 × 1 + 2 × 34 × 1 + 3 × 4⎦

⎥⎤

=⎣⎢⎡4

6713⎦⎥⎤

AC =⎣⎢⎡1

423⎦⎥⎤ ⎣⎢⎡–1

–221⎦⎥⎤

=⎣⎢⎡1 × –1 + 2 × –2

4 × –1 + 3 × –21 × 2 + 2 × 14 × 2 + 3 × 1⎦

⎥⎤

=⎣⎢⎡ –5

–10411⎦⎥⎤

AB + AC =⎣⎢⎡4

6713⎦⎥⎤ +

⎣⎢⎡ –5

–10411⎦⎥⎤

=⎣⎢⎡ 4 + –5

6 + –107 + 4

13 + 11⎦⎥⎤

=⎣⎢⎡–1

–41124⎦⎥⎤

⎣⎢⎡ 5

2.5012

3.00⎦⎥⎤ ⎣⎢⎡ 1

221

01⎦⎥⎤

=⎣⎢⎡ 5 × 1 + 12 × 2

2.5 × 1 + 3 × 25 × 2 + 12 × 12.5 × 2 + 3 × 1

5 × 0 + 12 × 12.5 × 0 + 3 × 1⎦

⎥⎤

=⎣⎢⎡ 29

8.5022

8.0012

3.00⎦⎥⎤

The matrix shows that John spent 29 min and $8.50, one friend spent 22 min and $8.00 (2 milkshakes and 1 banana split) while the other friend spent 12 min and $3.00 (no milkshakes and 1 banana split).

10

⎣⎢⎢⎢⎢⎢⎡ 0

1110

00011

11010

11011⎦⎥⎥⎥⎥⎥⎤

⎣⎢⎢⎢⎡ 2.00

3.002.503.50⎦

⎥⎥⎥⎤

=

⎣⎢⎢⎢⎢⎢⎡ 0 × 2.00 + 0 × 3.00 + 1 × 2.50 + 1 × 3.50

1 × 2.00 + 0 × 3.00 + 1 × 2.50 + 1 × 3.501 × 2.00 + 0 × 3.00 + 0 × 2.50 + 0 × 3.501 × 2.00 + 1 × 3.00 + 1 × 2.50 + 1 × 3.500 × 2.00 + 1 × 3.00 + 0 × 2.50 + 1 × 3.50⎦

⎥⎥⎥⎥⎥⎤

=

⎣⎢⎢⎢⎢⎢⎡ 6.00

8.002.0011.006.50 ⎦

⎥⎥⎥⎥⎥⎤

This shows the total amount spent on magazines by each student. A spent $6.00, B spent $8.00, C spent $2.00, D spent $11.00 and E spent $6.50.


11

a SC =⎣⎢⎡ s11

s21

s12

s22

s13

s23⎦⎥⎤

⎣⎢⎢⎢⎡ c1

c2

c3⎦⎥⎥⎥⎤

=⎣⎢⎡ s11 c1 + s12 c2 + s13 c3

s21 c1 + s22 c2 + s23 c3⎦⎥⎤

b SC represents the income from car sales for each showroom.

c SC =⎣⎢⎡ s11

s21

s12

s22

s13

s23⎦⎥⎤

⎣⎢⎢⎢⎡ c1

c2

c3

u1

u2

u3⎦⎥⎥⎥⎤

=⎣⎢⎡ s11 c1 + s12 c2 + s13 c3

s21 c1 + s22 c2 + s23 c3

s11 u1 + s12 u2 + s13 u3

s21 u1 + s22 u2 + s23 u3⎦⎥⎤

SC now represents the income from each showroom for both new and used car sales. d CV gives the profit on each new car and each used car for the three models.


Exercise 1D Solutions 1 a det(A) = 2 × 2 – 1 × 3 = 1 b A–1 = 1

1 ⎣⎢⎡ 2

–3–12 ⎦⎥⎤ =

⎣⎢⎡ 2

–3–12 ⎦⎥⎤

c det(B) = –2 × 2 – –2 × 3 = 2 d B–1 = 1

2 ⎣⎢⎡ –2

–322⎦⎥⎤

The answer could also be given as

B–1 =

⎣⎢⎢⎢⎡ –1

–1 12

1

1⎦⎥⎥⎥⎤

2 a Determinant = 3 × –1 – –1 × 4 = 1 A–1 = 1

1 ⎣⎢⎡ –1

413⎦⎥⎤ =

⎣⎢⎡ –1

413⎦⎥⎤

b Determinant = 3 × 4 – 1 × –2 = 14 A–1 = 1

14 ⎣⎢⎡ 4

2–13 ⎦⎥⎤

=

⎣⎢⎢⎢⎡ 2

717

– 114

314 ⎦

⎥⎥⎥⎤

c Determinant = 1 × k – 0 × 0 = k

A–1 = 1k ⎣⎢⎡ k

001⎦⎥⎤ =

⎣⎢⎢⎢⎡1

0

01k ⎦⎥⎥⎥⎤

d Determinant = cos θ × cos θ – –sin θ × sin θ = 1 since cos2 θ + sin2 θ = 1

A–1 = 11

⎣⎢⎡ cos θ

– sin θsin θcos θ⎦

⎥⎤

=⎣⎢⎡ cos θ

– sin θsin θcos θ⎦

⎥⎤

3 det(A) = 2 × –1 – 1 × 0 = –2 A–1 = 1

–2 ⎣⎢⎡ –1

0–12 ⎦⎥⎤

=

⎣⎢⎢⎢⎡120

12

–1⎦⎥⎥⎥⎤

det(B) = 1 × 1 – 0 × 3 = 1

B–1 = 11

⎣⎢⎡ 1

–301⎦⎥⎤

=⎣⎢⎡ 1

–301⎦⎥⎤

AB =⎣⎢⎡ 2

01

–1⎦⎥⎤ ⎣⎢⎡ 1

301⎦⎥⎤

=⎣⎢⎡ 2 × 1 + 1 × 3

0 × 1 + –1 × 32 × 0 + 1 × 1

0 × 0 + –1 × 1⎦⎥⎤

=⎣⎢⎡ 5

–31

–1⎦⎥⎤

det(AB) = 5 × –1 – 1 × –3 = –2 (AB)–1 = 1

–2 ⎣⎢⎡ –1

3–15 ⎦⎥⎤

=

⎣⎢⎢⎢⎡ 1

2

– 32

12

– 52⎦⎥⎥⎥⎤

A–1B–1 =

⎣⎢⎢⎢⎡120

12

–1⎦⎥⎥⎥⎤ ⎣⎢⎡ 1

–301⎦⎥⎤

=

⎣⎢⎢⎢⎡ 1

2× 1 + 1

2× –3

0 × 1 + –1 × –3

12× 0 + 1

2× 1

0 × 0 + –1 × 1⎦⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡–1

3

12

–1⎦⎥⎥⎥⎤


B–1A–1 =⎣⎢⎡ 1

–301⎦⎥⎤

⎣⎢⎢⎢⎡120

12–1⎦⎥

⎥⎥⎤

=

⎣⎢⎢⎢⎡ 1 × 1

2+ 0 × 0

–3 × 12

+ 1 × 0

1 × 12

+ 0 × –1

–3 × 12

+ 1 × –1⎦⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡ 1

2

– 32

12

– 52⎦⎥⎥⎥⎤

(AB)–1 = B–1A–1

4 a det(A) = 4 × 1 – 3 × 2 = –2 A–1 = 1

–2 ⎣⎢⎡ 1

–2–34 ⎦⎥⎤

=

⎣⎢⎢⎢⎡ – 1

21

32

–2⎦⎥⎥⎥⎤

b If AX = , multiply both sides

from the left by ⎣⎢⎡ 3

146⎦⎥⎤

1.−A A–1AX = A–1

⎣⎢⎡3

146⎦⎥⎤

∴ IX = X

=

⎣⎢⎢⎢⎡ – 1

21

32–2⎦⎥⎥⎥⎤ ⎣⎢⎡3

146⎦⎥⎤

=

⎣⎢⎢⎢⎡ – 1

2× 3 + 3

2× 1

1 × 3 + –2× 1

– 12× 4 + 3

2× 6

1 × 4 + –2× 6 ⎦⎥⎥⎥⎤

=⎣⎢⎡0

17–8⎦⎥⎤

c If YA = ⎣⎢⎡ 3

146⎦⎥⎤ , multiply both sides

from the right by 1.−A YAA–1 =

⎣⎢⎡ 3

146⎦⎥⎤A–1

∴ YI = Y

=⎣⎢⎡ 3

146⎦⎥⎤

⎣⎢⎢⎢⎡ – 1

21

32–2⎦

⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡ 3 × – 1

2+ 4 × 1

1 × – 12

+ 6 × 1

3 × 32

+ 4 × –2

1 × 32

+ 6 × –2⎦⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡ 5

2112

– 72

– 212 ⎦⎥⎥⎥⎤

5 a If AX + B = C then AX = C – B ∴ AX =

⎣⎢⎡ 3

246⎦⎥⎤ –

⎣⎢⎡ 4

2–12 ⎦⎥⎤

=⎣⎢⎡ –1

054⎦⎥⎤

det(A) = 3 × 6 – 2 × 1 = 16 A–1 = 1

16 ⎣⎢⎡ 6

–1–23 ⎦⎥⎤

=

⎣⎢⎢⎢⎡ 3

8

– 116

– 18

316 ⎦⎥⎥⎥⎤

If AX = ⎣⎢⎡ –1

054⎦⎥⎤ , multiply both sides

from the left by 1.−A

A–1AX = A–1

⎣⎢⎡ –1

054⎦⎥⎤


∴ IX = X

=

⎣⎢⎢⎢⎡ 3

8

– 116

– 18

316 ⎦⎥⎥⎥⎤

⎣⎢⎡–1

054⎦⎥⎤

=

⎣⎢⎢⎢⎡ 3

8× –1+ – 1

8× 0

– 116

× –1+ 316× 0

38× 5 + – 1

8× 4

– 116

× 5 + 316× 4⎦⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡ – 3

8116

118716⎦⎥⎥⎥⎤

b If YA + B = C then YA = C – B ∴ YA =

⎣⎢⎡ 3

246⎦⎥⎤ –

⎣⎢⎡ 4

2–12 ⎦⎥⎤

=⎣⎢⎡ –1

054⎦⎥⎤

From part a, A–1 =

⎣⎢⎢⎢⎡ 3

8

– 116

– 18

316 ⎦⎥⎥⎥⎤

If YA = , multiply both sides

from the right by ⎣⎢⎡ –1

054⎦⎥⎤

1.−A YAA–1 =

⎣⎢⎡–1

054⎦⎥⎤A–1

YI = Y

=⎣⎢⎡–1

054⎦⎥⎤

⎣⎢⎢⎢⎡ 3

8

– 116

–18

316⎦⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡–1×3

8+ 5× – 1

16

0×38

+ 4× – 116

–1× –18

+ 5× 316

0× –18

+ 4× 316⎦⎥⎥⎥⎤

X=

⎣⎢⎢⎢⎡ –11

16

–14

171634 ⎦⎥⎥⎥⎤

6 A must be ⎣⎢⎡ a11

00

a22⎦⎥⎤ .

det(A) = a11 × a22 – 0 × 0 = a11a22 det(A) ≠ 0 since a11 ≠ 0 and a22 ≠ 0 and

the product of two non-zero numbers cannot be zero.

∴A is regular.

A–1 = 1a11 a22

⎣⎢⎡ a22

00

a11⎦⎥⎤

=

⎣⎢⎢⎢⎡ 1

a11

0

0

1a22⎦⎥⎥⎥⎤

7 If A is regular, it will have an inverse, 1.−A Multiply both sides of the equation AB = 0

from the left by 1.−A A–1AB = A–1 0

∴ IB = 0B = 0

8 Let A be any matrix

⎣⎢⎡a

cbd⎦⎥⎤ .

If the determinant is n, then the inverse of A is given by 1

n ⎣⎢⎡ d

– c– ba ⎦⎥⎤ .

∴ ⎣⎢⎡ a

cbd⎦⎥⎤ = 1

n ⎣⎢⎡ d

– c– ba ⎦⎥⎤

a = dn and d = a

n

Substituting for d, a = a ÷ nn

= an2

This gives 2 1n = , or n = ± 1. If n = 1, a = d and –b = b, which gives b = 0 and similarly c = 0. det(A) = 2 1ad a= =

This leads to two matrices, and ⎣⎢⎡ 1

001⎦⎥⎤

⎣⎢⎡ –1

00–1⎦⎥⎤ .


If n = –1, a = –d; there are no restrictions on b and c but the determinant = ad – bc = –1. ∴ a2 + bc = 1 (since a = – d)

If b = 0, a = ±1, giving , which

can be written ⎣⎢⎡±1

c0

±1⎦⎥⎤

⎣⎢⎡ 1

k0–1⎦⎥⎤ or

⎣⎢⎡ –1

k01⎦⎥⎤ .

If b ≠ 0, = 1 gives 2a bc+

c = 1 – a2

b , giving

⎣⎢⎢⎢⎡ a

1 – a2

b

b

– a⎦⎥⎥⎥⎤

, which

includes the cases ⎣⎢⎡ 1

0k

–1⎦⎥⎤ and

⎣⎢⎡ –1

0k1⎦⎥⎤

when a = ± 1.


Exercise 1E Solutions 1 First find the inverse of A. det(A) = 3 × –1 – –1 × 4 = 1 A–1 = 1

1 ⎣⎢⎡ –1

–413⎦⎥⎤ =

⎣⎢⎡ –1

–413⎦⎥⎤

a If AX = K then A–1AX = A–1K ∴ IX = X = A–1K

X =⎣⎢⎡ –1

–413⎦⎥⎤ ⎣⎢⎡ –1

2 ⎦⎥⎤

=⎣⎢⎡ –1 × –1 + 1 × 2

–4 × –1 + 3 × 2⎦⎥⎤

=⎣⎢⎡ 3

10⎦⎥⎤

b If AX = K then A–1AX = A–1K

∴ IX = X = A–1K X =

⎣⎢⎡ –1

–413⎦⎥⎤ ⎣⎢⎡ –2

3 ⎦⎥⎤

=⎣⎢⎡ –1 × –2 + 1 × 3

–4 × –2 + 3 × 3⎦⎥⎤

=⎣⎢⎡ 5

17⎦⎥⎤

2 First find the inverse of A. det(A) = 3 × 4 – 1 × –2 = 14

A–1 = 114

⎣⎢⎡ 4

2–13 ⎦⎥⎤ =

⎣⎢⎢⎢⎡ 2

717

– 114

314 ⎦

⎥⎥⎥⎤

a If AX = K then A–1AX = A–1K

∴ IX = X = A–1K

X =

⎣⎢⎢⎢⎡ 2

717

– 114

314 ⎦

⎥⎥⎥⎤

⎣⎢⎡ 0

1⎦⎥⎤

=

⎣⎢⎢⎢⎡ 2

7× 0 + – 1

14× 1

17× 0 + 3

14× 1 ⎦

⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡ – 1

143

14 ⎦⎥⎥⎥⎤

b If AX = K then A–1AX = A–1K

∴ IX = X = A–1K

X =

⎣⎢⎢⎢⎡ 2

717

– 114

314 ⎦

⎥⎥⎥⎤

⎣⎢⎡ 2

0⎦⎥⎤

=

⎣⎢⎢⎢⎡ 2

7× 2 + – 1

14× 0

17× 2 + 3

14× 0 ⎦

⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡ 4

727⎦⎥⎥⎥⎤

3 a

⎣⎢⎡ –2

341⎦⎥⎤⎣⎢⎡ x

y⎦⎥⎤ =

⎣⎢⎡ 6

1⎦⎥⎤

Determinant = –2 × 1 – 4 × 3 = –14 Inverse = 1

–14 ⎣⎢⎡ 1

–3–4–2⎦⎥⎤

=

⎣⎢⎢⎢⎡ – 1

14314

2717⎦⎥⎥⎥⎤

⎣⎢⎡ x

y⎦⎥⎤ =

⎣⎢⎢⎢⎡ – 1

143

14

2717⎦⎥⎥⎥⎤

⎣⎢⎡ 6

1⎦⎥⎤

=⎣⎢⎢⎢⎡ – 1

14× 6 + 2

7× 1

314

× 6 + 17× 1 ⎦

⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡ – 1

7107 ⎦⎥⎥⎥⎤

x = – 17, y = 10

7


b ⎣⎢⎡ –1

–124⎦⎥⎤⎣⎢⎡ x

y⎦⎥⎤ =

⎣⎢⎡ –1

2 ⎦⎥⎤

Determinant = –1 × 4 – 2 × –1 = –2

Inverse = 1–2

⎣⎢⎡ 4

1–2–1⎦⎥⎤

=

⎣⎢⎢⎢⎡ –2

– 12

112⎦⎥⎥⎥⎤

⎣⎢⎡ x

y⎦⎥⎤ =

⎣⎢⎢⎢⎡ –2

– 12

112⎦⎥⎥⎥⎤

⎣⎢⎡ –1

2 ⎦⎥⎤

=

⎣⎢⎢⎢⎡ –2 × –1 + 1 × 2

– 12× –1 + 1

2× 2

⎦⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡4

32⎦⎥⎥⎥⎤

x = 4, y = 32 or 1.5

c Convert the second equation to –x + y = 4, giving

⎣⎢⎡ 2

–151⎦⎥⎤⎣⎢⎡ x

y⎦⎥⎤ =

⎣⎢⎡ –10

4 ⎦⎥⎤

Determinant = 2 × 1 – 5 × –1 = 7

Inverse = 17 ⎣⎢⎡ 1

1–52 ⎦⎥⎤

⎣⎢⎡ x

y⎦⎥⎤ = 1

7 ⎣⎢⎡ 1

1–52 ⎦⎥⎤⎣⎢⎡ –10

4 ⎦⎥⎤

= 17⎣⎢⎡ 1 × –10 + –5 × 4

1 × –10 + 2 × 4 ⎦⎥⎤

= 17 ⎣⎢⎡ –30

–2 ⎦⎥⎤

x = – 307

, y = – 27

d Re-write the second equation as –3.5x + 4.6y = 11.4, giving

⎣⎢⎡ 1.3

–3.52.74.6⎦

⎥⎤⎣⎢⎡ x

y⎦⎥⎤ =

⎣⎢⎡ –1.2

11.4⎦⎥⎤

Determinant = 1.3 × 4.6 – 2.7 × –3.5 = 15.43

Inverse = 115.43

⎣⎢⎡ 4.6

3.5–2.71.3 ⎦

⎥⎤

⎣⎢⎡ x

y⎦⎥⎤ = 1

15.43 ⎣⎢⎡ 4.6

3.5–2.71.3 ⎦

⎥⎤ ⎣⎢⎡ –1.2

11.4⎦⎥⎤

= 115.43

⎣⎢⎡ 4.6 × –1.2 + –2.7 × 11.4

3.5 × –1.2 + 1.3 × 11.4 ⎦⎥⎤

= 115.43

⎣⎢⎡ –36.3

10.62⎦⎥⎤

x ≈ – 2.35, y ≈ 0.69

4 Solve the simultaneous equations 2x – 3y = 7 3x + y = 5

⎣⎢⎡ 2

3–31 ⎦⎥⎤⎣⎢⎡ x

y⎦⎥⎤ =

⎣⎢⎡ 7

5⎦⎥⎤

Determinant = 2 × 1 – –3 × 3 = 11 Inverse = 1

11 ⎣⎢⎡ 1

–332⎦⎥⎤

⎣⎢⎡ x

y⎦⎥⎤ = 1

11 ⎣⎢⎡ 1

–332⎦⎥⎤⎣⎢⎡ 7

5⎦⎥⎤

= 111

⎣⎢⎡ 1 × 7 + 3 × 5

–3 × 7 + 2 × 5⎦⎥⎤

= 111

⎣⎢⎡ 22

–11⎦⎥⎤

x = 2, y = –1

The point of intersection is (2, –1).

5 If x is the number of books they are buying and y is the number of CDs they are buying, then the following equations apply. 4x + 4y = 120

5x + 3y = 114

⎣⎢⎡ 4

543⎦⎥⎤⎣⎢⎡ x

y⎦⎥⎤ =

⎣⎢⎡ 120

114⎦⎥⎤

Determinant = 4 × 3 – 4 × 5 = –8 Inverse = 1

–8 ⎣⎢⎡ 3

–5–44 ⎦⎥⎤ = 1

8 ⎣⎢⎡ –3

54

–4⎦⎥⎤

⎣⎢⎡ x

y⎦⎥⎤ = 1

8 ⎣⎢⎡ –3

54

–4⎦⎥⎤⎣⎢⎡ 120

114⎦⎥⎤

= 18 ⎣⎢⎡ –3 × 120 + 4 × 114

5 × 120 + –4 × 114⎦⎥⎤

= 18 ⎣⎢⎡ 96

144⎦⎥⎤

x = 12, y = 18

One book costs $12, a CD costs $18.


7 Enter the 4 × 4 matrix A and the 4 × 1 matrix B into the graphics calculator.

6 a

⎣⎢⎡ 2

4–3–6⎦⎥⎤⎣⎢⎡ x

y⎦⎥⎤ =

⎣⎢⎡ 3

6⎦⎥⎤

Fill in missing coefficients with zeros, so that r + s = 1 becomes 0p + 0q + 1r + 1s = 1 b det(A) = 2 × –6 – –3 × 4 = 0, so the

A =

⎣⎢⎢⎢⎡ 1

020

10–1–1

–1120

–1101 ⎦⎥⎥⎥⎤

; B =

⎣⎢⎢⎢⎡ 5

1–20 ⎦⎥⎥⎥⎤

A–1B =

⎣⎢⎢⎢⎡ 2

4–12 ⎦⎥⎥⎥⎤

matrix is singular. c Yes. For example x = 0, y = –1 is an obvious solution. d You should notice that the second

equation is simply the first with both sides multiplied by 2. There is an infinite number of solutions to these equations, just as there is an infinite number of ordered pairs that make 2x – 3y = 3 a true equation.

p = 2, q = 4, r = –1, s = 2


Solutions to Multiple-choice Questions 1 The dimension is number of rows by

number of columns, i.e. 4 × 2. B 2 The matrices cannot be added as they

have different dimensions. E 3 D – C =

⎣⎢⎡ 1

2–33

1–1⎦⎥⎤ –

⎣⎢⎡ 2

1–3

01

–2⎦⎥⎤

=⎣⎢⎡ 1 – 2

2 – 1–3 – –3

3 – 01 – 1

–1 – –2⎦⎥⎤

= C ⎣⎢⎡ –1

103

01⎦⎥⎤

4 Multiply every entry by –1. – M = –

⎣⎢⎡ –4

–20

–6⎦⎥⎤

= E ⎣⎢⎡ 4

206⎦⎥⎤

5 2M – 2N = 2 ×

⎣⎢⎡ 0

–321⎦⎥⎤ – 2 ×

⎣⎢⎡ 0

340⎦⎥⎤

=⎣⎢⎡ 0

–642⎦⎥⎤ –

⎣⎢⎡ 0

680⎦⎥⎤

C =⎣⎢⎡ 0

–12–42⎦⎥⎤

6 A + B will have the same dimension as A and B, i.e. m × n. A

7 The number of columns of Q is not the

same as the number of rows of P, so they cannot be multiplied. E

8 Determinant = 2 × 1 – 2 × –1 = 4 A 9 Determinant = 1 × –2 – –1 × 1 = –1 Inverse = 1

–1 ⎣⎢⎡ –2

–111⎦⎥⎤

=⎣⎢⎡ 2

1–1–1⎦⎥⎤ E

10 NM =

⎣⎢⎡ 0

321⎦⎥⎤⎣⎢⎡ 0

–3–2

1⎦⎥⎤

=⎣⎢⎡ 0 × 0 + 2 × –3

3 × 0 + 1 × –30 × –2 + 2 × 13 × –2 + 1 × 1⎦

⎥⎤

=⎣⎢⎡ –6

–32

–5⎦⎥⎤ D


Solutions to Short-answer Questions 1 a A + B =

⎣⎢⎡ 1

203⎦⎥⎤ +

⎣⎢⎡ –1

001⎦⎥⎤

=⎣⎢⎡ 0

204⎦⎥⎤

A – B =⎣⎢⎡ 1

203⎦⎥⎤ –

⎣⎢⎡ –1

001⎦⎥⎤

=⎣⎢⎡ 2

202⎦⎥⎤

(A + B)(A – B) =⎣⎢⎡ 0

204⎦⎥⎤⎣⎢⎡ 2

202⎦⎥⎤

=⎣⎢⎡ 0

1208⎦⎥⎤

b A2 = AA =

⎣⎢⎡ 1

203⎦⎥⎤⎣⎢⎡ 1

203⎦⎥⎤

=⎣⎢⎡ 1

809⎦⎥⎤

B2 = BB =⎣⎢⎡ –1

001⎦⎥⎤⎣⎢⎡ –1

001⎦⎥⎤

=⎣⎢⎡ 1

001⎦⎥⎤

A2 – B2 =⎣⎢⎡ 1

809⎦⎥⎤ –

⎣⎢⎡ 1

001⎦⎥⎤

=⎣⎢⎡ 0

808⎦⎥⎤

2 Find the inverse of

⎣⎢⎡3

648⎦⎥⎤ .

Determinant = 3 × 8 – 4 × 6 = 0 This is a singular matrix.

If A then this corresponds to the

simultaneous equations:

=⎣⎢⎡ xy⎦⎥⎤ ,

3x + 4y = 8 6x + 8y = 16

The second equation is equivalent to the first, as it is obtained by multiplying both sides of the first by 2.

Thus if x = a, 3a + 4y = 8

4y = 8 – 3ay = 2 – 3a

4

The matrices may be expressed as

⎣⎢⎢⎢⎡ a

2 – 3a4 ⎦⎥⎥⎥⎤.

3 a For a product to exist, the number of

columns of the first matrix must equal the number of rows of the second.

This is true only for AC, CD and BE, so these products exist.

b DA = [2 4]

⎣⎢⎡ 1

32

–1⎦⎥⎤

=⎣⎢⎡ 2 × 1 + 4 × 3

2 × 2 + 4 × –1⎦⎥⎤

=⎣⎢⎡ 14

0⎦⎥⎤

det(A) = 1 × –1 – 2 × 3 = –7 A–1 = 1

–7 ⎣⎢⎡ –1

–3–21⎦⎥⎤

= 17 ⎣⎢⎡ 1

32

–1⎦⎥⎤ or

⎣⎢⎢⎢⎡ 1

737

27

– 17⎦⎥⎥⎥⎤

4 AB =⎣⎢⎡ 1

–5–21

12⎦⎥⎤

⎣⎢⎢⎢⎡113

–4–6–8⎦⎥⎥⎥⎤

=⎣⎢⎡1 × 1 + –2× 1 + 1 × 3

–5× 1 + 1 × 1 + 2 × 31 × –4 + –2× –6 + 1 × –8–5× –4 + 1 × –6 + 2 × –8⎦

⎥⎤

=⎣⎢⎡2

20

–2⎦⎥⎤

det(C) = 1 × 4 – 2 × 3 = –2 C–1 = 1

–2⎣⎢⎡ 4

–3–21 ⎦⎥⎤

=

⎣⎢⎢⎢⎡–2

32

1

– 12⎦⎥⎥⎥⎤


5 Find the inverse of ⎣⎢⎡1

324⎦⎥⎤ .

Determinant = 1 × 4 – 2 × 3 = –2 Inverse = 1

–2 ⎣⎢⎡ 4

–3–2

1⎦⎥⎤

= 12 ⎣⎢⎡ –4

32

–1⎦⎥⎤

Multiply by the inverse on the right: A =

⎣⎢⎡ 5

126

14⎦⎥⎤ × 1

2 ⎣⎢⎡ –4

32

–1⎦⎥⎤

=⎣⎢⎡ –1

–325⎦⎥⎤

6 A2 =

⎣⎢⎢⎢⎡2

00

002

020⎦⎥⎥⎥⎤

⎣⎢⎢⎢⎡2

00

002

020⎦⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡4

00

040

004⎦⎥⎥⎥⎤

A–1 =

⎣⎢⎢⎢⎢⎢⎡1

2

0

0

0

0

12

0

12

0⎦⎥⎥⎥⎥⎥⎤

7 The determinant must be zero. 1 × x – 2 × 4 = 0

x – 8 = 0x = 8

8 a i MM =

⎣⎢⎡ 2

1–13⎦⎥⎤⎣⎢⎡ 2

1–1

3⎦⎥⎤

=⎣⎢⎡ 3

5–5

8⎦⎥⎤

ii MMM = MM(M) =

⎣⎢⎡ 3

5–5

8⎦⎥⎤⎣⎢⎡ 2

1–1

3⎦⎥⎤

=⎣⎢⎡ 1

18–18

19⎦⎥⎤

iii Determinant = 2 × 3 – –1 × 1 = 7

M–1 = 1

7 ⎣⎢⎡ 3

–112⎦⎥⎤

b M–1M

⎣⎢⎡ x

y⎦⎥⎤ = M–1

⎣⎢⎡ 3

5⎦⎥⎤

⎣⎢⎡ x

y⎦⎥⎤ = 1

7 ⎣⎢⎡ 3

–112⎦⎥⎤ ⎣⎢⎡ 3

5⎦⎥⎤

= 17 ⎣⎢⎡ 14

7⎦⎥⎤

=⎣⎢⎡ 2

1⎦⎥⎤

x = 2, y = 1


Chapter 2 – Algebra I Exercise 2A Solutions 1 a Add indices: x3 × x4 = x3 + 4 = x7

b Add indices: a5 × a–3 = a5 + –3 = a2 c Add indices: x2 × x–1 × x2 = x2 + –1 + 2 = x3 d Subtract indices:

y3

y7 = y3 – 7 = y–4

e Subtract indices:

x8

x–4 = x8 – –4 = x12

f Subtract indices:

p–5

p2 = p–5 – 2 = p–7

g Subtract indices:

a12 ÷ a

23 = a

36

– 46 = a

– 16

h Multiply indices: (a–2)

4= a–2 × 4 = a–8

i Multiply indices: (y–2)

–7= y–2 × –7 = y14

j Multiply indices: (x5)

3= x5 × 3 = x15

k Multiply indices:

(a–20)35 = a

–20 × 35 = a–12

l Multiply indices:

⎝⎜⎛

x– 1

2⎠⎟⎞–4

= x– 1

2 × –4= x2

m Multiply indices:

(n10)15 = n

10 × 15 = n2

n Multiply the coefficients and add the

indices:

2x12 × 4x3 = (2 × 4)x

12

+ 3= 8x

72

o Multiply the first two indices and add

the third:

(a2)52 × a–4 = a

2 × 52 × a–4

= a5 + –4

= a1 = a

p 1x–4 = x

1 ÷ 14 = x4

q ⎝⎜⎛

2n– 2

5⎠⎟⎞5

÷ (43 n4) = 25 n– 2

5 × 5÷ ((22)

3 n4)

= 25 n–2 ÷ (26 n4) = 25 – 6n–2 – 4

= 2–1n–6 = 12n6

r Multiply the coefficients and add the indices.

x3 × 2x12 × –4x

– 32 = (1 × 2 × –4)x

3 + 12

+ – 32

= –8x2

s (ab3)2× a–2b–4 × 1

a2b–3 = a2b6 × a–2b–4 × a–2b3

= a2 + –2 + –2b6 + –4 + 3

= a–2b5

t (22 p–3 × 43 p5 ÷ ((6p–3))0

= 1 Anything to the power zero is 1.


2

a 2512 = 25 = 5

b 6413 = 3 64 = 4

c ⎝⎜⎛16

9 ⎠⎟⎞

12 =

1612

912

= 169

= 43

d 16– 1

2 =1

1612

= 116

= 14

e ⎝⎜⎛49

36⎠⎟⎞ – 1

2 =1

⎝⎜⎛49

36⎠⎟⎞

12

=14936

= 3649

= 67

f 2713 = 3 27 = 3

g 14412 = 144 = 12

h 6423 = ⎝

⎜⎛

6413⎠⎟⎞2

= 42 = 16

i 932 = ⎝

⎜⎛

912⎠⎟⎞3

= 33 = 27

j ⎝⎜⎛81

16⎠⎟⎞

14 =

8114

1614

= 32

k ⎝⎜⎛23

5 ⎠⎟⎞0

= 1

l 12837 = ⎝

⎜⎛

12817⎠⎟⎞3

= 23 = 8

3 a 4.352 = 18.9225 ≈ 18.92 b 2.45 = 79.62624 ≈ 79.63 c 34.6921 = 5.89 d 0.02–3 = 125 000 e 3 0.729 = 0.9 f 4 2.3045 = 1.23209. . . ≈ 1.23

g (345.64)– 1

3 = 0.14249. . . ≈ 0.14

h (4.558)25 = 1.83607. . . ≈ 1.84

i 1

(0.064)– 1

3= (0.064)

13 = 0.4

4

a a2b3

a–2b–4 = a2 – –2b3 – –4

= a4b7

b 2a2(2b)3

(2a)–2b–4 = 2a2 × 23 b3

2–2a–2b–4

= 24 a2b3

2–2a–2b–4

= 24 – –2a2 – –2b3 – –4

= 26 a4b7 = 64a4b7


c a–2b–3

a–2b–4 = a–2 – –2b–3 – –4

= a0b1 = b

d a2b3

a–2b–4 ×ab

a–1b–1 = a2 + 1b3 + 1

a–2 + –1b–4 + –1

= a3b4

a–3b–5

= a3 – –3b4 – –5 = a6b9

e (2a)2 × 8b3

16a–2b–4 = 4a2 × 8b3

16a–2b–4

= 32a2b3

16a–2b–4

= 3216

a2 – –2b3 – –4

= 2a4b7

f 2a2b3

8a–2b–4 ÷16ab

(2a)–1b–1 = 2a2b3

8a–2b–4 ×(2a)–1b–1

16ab

= 2a2b3

8a–2b–4 ×2–1a–1b–1

16ab

= 21 + –1a2 + –1b3 + –1

8 × 16 × a–2 + 1b–4 + 1

= 20 a1b2

128a–1b–3

= 1128

a1 – –1b2 – –3 = a2b5

128

5 2n × 8n

22n × 16= 2n × (23)

n

22n × 24

= 2n × 23n

22n × 24

= 2n + 3n – 2n

24 = 22n × 2–4

6 2 – x × 3 – x × 62x × 32x × 22x

= (2 × 3) – x × 62x × (2 × 3)2x

= 6 – x × 62x × 62x

= 6 – x + 2x + 2x

= 63x

7 In each case, add the fractional indices.

a 213 × 2

16 × 2

– 23 = 2

26

+ 16

+ – 46

= 2– 1

6 =⎝⎜⎛1

2⎠⎟⎞

16

b a14 × a

25 × a

– 110 = a

520

+ 820

+ – 220

= a1120

c 223 × 2

56 × 2

– 23 = 2

46

+ 56

+ – 46

= 256

d ⎝⎜⎛

213⎠⎟⎞2

× ⎝⎜⎛

212⎠⎟⎞5

= 223 × 2

52

= 246

+ 156 = 2

196

e ⎝⎜⎛2

13⎠⎟⎞2

× 213× 2

– 25 = 2

23× 2

13× 2

– 25

= 223

+ 13

+ – 25 = 2

35

8

a 3 a3b2 ÷ 3 a2b–1 = (a3b2)13 ÷ (a2b–1)

13

= a1b23÷ a

23b

– 13

= a1 – 2

3b23

– – 13 = a

13b

b a3b2 × a2b–1 = (a3b2)12 × (a2b–1)

12

= a32b1

× a1b– 1

2

= a32

+ 1b

1 + – 12 = a

52b

12

c 5 a3b2 × 5 a2b–1 = (a3b2)15 × (a2b–1)

15

= a35b

25× a

25b

– 15

= a35

+ 25b

25

+ – 15 = ab

15


d a–4b2 × a3b–1 = (a–4b2)12 × (a3b–1)

12

= a–2b1 × a32b

– 12

= a–2 + 3

2b1 + – 1

2

= a– 1

2b12

=b

12

a12

=⎝⎜⎛b

a⎠⎟⎞

12

f 5 a3b2 ÷ 5 a2b–1 = (a3b2)15 ÷ (a2b–1)

15

= a35b

25÷ a

25b

– 15

= a35

– 25b

25

– – 15 = a

15b

35

g a3b2

a2b–1c–5 ×a–4b2

a3b–1 × a3b–1

= (a3b2)12

a2b–1c–5 ×(a–4b2)

12

a3b–1 × (a3b–1)12

= a32b1

a2b–1c–5 ×a–2b1

a3b–1 × a32b

– 12

= a32

– 2b1 – –1c0 – –5 × a–2 – 3b1 – –1 × a

32b

– 12

= a– 1

2b2c5 × a–5b2 × a32b

– 12

= a– 1

2+ –5 + 3

2b2 + 2 + – 1

2c5

= a–4b72c5

e a3b2c–3 × a2b–1c–5

= (a3b2c–3)12 × (a2b–1c–5)

12

= a32b1c

– 32 × a1b

– 12c

– 52

= a32

+ 1b

1 + – 12c

– 32

+ – 52

= a52b

12c–4


Exercise 2B Solutions 1 a 47.8 = 4.78 × 101 = 4.78 × 10

b 6728 = 6.728 × 103

c 79.23 = 7.923 × 101 = 7.923 × 10

d 43 580 = 4.358 × 104

e 0.0023 = 2.3 × 10–3

f 0.000 000 56 = 5.6 × 10–7

g 12.000 34 = 1.200 034 × 101

= 1.200 034 × 10 h Fifty million = 50 000 000

= 5.0 × 107

i 23 000 000 000 = 2.3 × 1010

j 0.000 000 0013 = 1.3 × 10–9 k 165 thousand = 165 000

= 1.65 × 105

l 0.000 014 567 = 1.4567 × 10–5 2 a The decimal point moves 8 places to the

right = 1.0 × 10–8 b The decimal point moves 23 places to

the right = 1.66 × 10–23 c The decimal point moves 5 places to the

right = 5.0 × 10–5 d The decimal point moves 3 places to the

left = 1.853 18 × 103

e The decimal point moves 12 places to

the left = 9.463 × 1012

f The decimal point moves 10 places to

the right = 2.998 × 1010

3 a The decimal point move 13 places to the

right = 75 684 000 000 000 b The decimal point move 8 places to the

right = 270 000 000 c The decimal point move 13 places to the

left = 0.000 000 000 000 19 4

a 324 000 × 0.000 000 74000

= 3.24 × 105 × 7 × 10–7

4 × 103

= 3.24 × 74

× 105 + –7 – 3

= 5.67 × 10–5

= 0.000 056 7

b 5 240 000 × 0.842 000 000

= 5.24 × 106 × 8 × 10–1

4.2 × 107

= 41.92 × 105

4.2 × 107

= 4192 × 103

42 000 × 103

= 419242 000

= 2622625

5

a 3 ab4 =

3 2 × 109

3.2154

=3 2 × 3 109

106.8375. . .

= 1.2599. . . × 103

106.8375. . . = 0.011 792. . . × 103 ≈ 11.8

b 4 a4b4 =

4 2 × 1012

4 × 0.054

=4 2 × 4 1012

4 × 0.000 006 25

= 1.189 2. . . × 103

4 × 6.25 × 10–6

= 0.047 568. . . × 109 ≈ 4.76 × 107


Exercise 2C Solutions 1 a 3x + 7 = 15

3x = 15 – 7 = 8x = 8

3

b 8 – x

2= 15

– x2

= 15 – 8

= 7– x

2× –2 = 7 × –2

x = –14

c 42 + 3x = 22

3x = 22 – 42 = –20x = – 20

3

d 2x

3– 15 = 27

2x3

= 27 + 15

= 422x3× 3

2= 42 × 3

2x = 63

e 5(2x + 4) = 13

10x + 20 = 1310x = 13 – 20

= –7x = – 7

10= –0.7

f –3(4 – 5x) = 24

–12 + 15x = 2415x = 24 + 12

= 36x = 36

15 = 12

5= 2.4

g 3x + 5 = 8 – 7x3x + 7x = 8 – 5

10x = 3x = 3

10= 0.3

h 2 + 3(x – 4) = 4(2x + 5)

2 + 3x – 12 = 8x + 203x – 10 = 8x + 203x – 8x = 20 + 10

–5x = 30x = 30

–5= –6

i 2x

5– 3

4= 5x

2x5× 20 – 3

4× 20 = 5x × 20

8x – 15 = 100x8x – 100x = 15

–92x = 15x = – 15

92

j 6x + 4 = x

3– 3

6x × 3 + 4 × 3 = x3× 3 – 3 × 3

18x + 12 = x – 918x – x = –9 – 12

17x = –21x = – 21

17

2 a x

2+ 2x

5= 16

x2× 10 + 2x

5× 10 = 16 × 10

5x + 4x = 1609x = 160

x = 1609


b 3x4

– x3

= 8

3x4× 12 – x

3× 12 = 8 × 12

9x – 4x = 965x = 96

x = 965

= 19.2

c 3x – 2

2+ x

4= –18

3x – 22

× 4 + x4× 4 = –18 × 4

2(3x – 2) + x = –726x – 4 + x = –72

7x = –72 + 4 = –68x = – 68

7

d 5x

4– 4

3= 2x

55x4× 60 – 4

3× 60 = 2x

5× 60

75x – 80 = 24x75x – 24x = 80

51x = 80x = 80

51

e x – 4

2+ 2x + 5

4= 6

x – 42

× 4 + 2x + 54

× 4 = 6 × 4

2(x – 4) + (2x + 5) = 242x – 8 + 2x + 5 = 24

4x = 24 + 8 – 5 = 27x = 27

4= 6.75

f 3 – 3x

10– 2(x + 5)

6= 1

203 – 3x

10× 60 – 2(x + 5)

6× 60 = 1

20× 60

6(3 – 3x) – 20(x + 5) = 318 – 18x – 20x – 100 = 3

–38x = 3 – 18 + 100 = 85x = – 85

38

g 3 – x4

– 2(x + 1)5

= –24

3 – x4

× 20 – 2(x + 1)5

× 20 = –24 × 20

5(3 – x) – 8(x + 1) = –48015 – 5x – 8x – 8 = –480

–13x = –480 – 15 + 8 = –487x = 487

13

h –2(5 – x)

8+ 6

7= 4(x – 2)

3–2(5 – x)

8× 168 + 6

7× 168 = 4(x – 2)

3× 168

–42(5 – x) + 144 = 224(x – 2)–210 + 42x + 144 = 224x – 448

42x – 224x = –448 + 210 – 144–182x = –382

x = 382182

= 19191

3 a 3x + 2y = 2; 2x – 3y = 6 Use elimination. Multiply the first equation

by 3 and the second equation by 2. 9x + 6y = 6

4x – 6y = 12 + :

13x = 18x = 18

13

Substitute into the first equation: 3 × 18

13+ 2y = 2

5413

+ 2y = 2

2y = 2 – 5413

= – 2813

y = – 1413


b 5x + 2y = 4; 3x – y = 6 e 7x – 3y = –6; x + 5y = 10 Use elimination. Multiply the second

equation by 2. Use substitution. Make x the subject of

the second equation. 5x + 2y = 4

6x – 2y = 12 + :

11x = 16x = 16

11

x =10 – 5y Substitute into the first equation: 7(10 – 5y) – 3y = –6

70 – 35y – 3y = –6 –38y = –6 – 70

= –76 y = –76

–38= 2

Substitute into the second, simpler equation:

3 × 1611

– y = 6

4811

– y = 6

– y = 6 – 4811

y = 1811

Substitute into the second equation: x + 5 × 2 = 10

x + 10 = 10x = 0

f 15x + 2y = 27; 3x + 7y = 45 Use elimination. Multiply the second

equation by 5. 15x + 2y = 27

15x + 35y = 225 – :

–33y = –198y = –198

–33= 6

c 2x – y = 7; 3x – 2y = 2 Use substitution. Make y the subject of

the first equation. y = 2x – 7 Substitute into the second equation: 3x – 2(2x – 7) = 2

3x – 4x + 14 = 2– x = 2 – 14

x = 12

Substitute into the second equation: 3x + 7 × 6 = 45

3x + 42 = 453x = 45 – 42

= 3x = 1

Substitute into the equation in which y is the subject:

y = 2 × 12 – 7= 17

d x + 2y = 12; x – 3y = 2 Use substitution. Make x the subject of

the first equation. x = 12 – 2y Substitute into the second equation: 12 – 2y – 3y = 2

–5y = 2 – 12 = –10y = 2

Substitute into the first equation: x + 2 × 2 = 12 x + 4 = 12 x = 8


Exercise 2D Solutions 1 a 4(x – 2) = 60 4x – 8 = 60 4x = 60 + 8 = 68 x = 17 b The length of the square is 2x + 7

4.

⎝⎜⎛2x + 7

4 ⎠⎟⎞2

= 49

2x + 74

= 7

2x + 7 = 7 × 4 = 282x = 28 – 7 = 21

x = 10.5

c The equation is length = twice width. x – 5 = 2(12 – x)

x – 5 = 24 – 2x x + 2x = 24 + 5

3x = 29x = 29

3

d y = 2((2x + 1) + (x – 3)) = 2(2x + 1 + x – 3) = 2(3x – 2) = 6x – 4 e Q = np f If a 10% service charge is added, the total

price will be multiplied by 110%, or 1.1. R = 1.1pS g Using the fact that there are 12 lots of

5 min in an hour (60 ÷ 12 = 5), 60n

5= 2400

h a = circumference × 60

360 = 2π(x + 3) × 60

360 = 2π(x + 3) × 1

6

= π3

(x + 3)

2 Let the value of Bronwyn’s sales in the first week be $s.

s + (s + 500) + (s + 1000) + (s + 1500) + (s + 2000) = 17 5005s + 5000= 17 500

5s = 12 500s = 2500

The value of her first week’s sales is $2500. 3 Let d be the number of dresses bought

and h the number of handbags bought. 65d + 26h = 598

d + h = 11 Multiply the second equation by 26 (the

smaller number). 65d + 26h = 598

26d + 26h = 286 – :

39d = 312d = 312

39= 8

h + 8 = 11h = 3

Eight dresses and three handbags. 4 Let the courtyard’s width be w metres. 3w + w + 3w + w = 67 8w = 67 w = 8.375 The width is 8.375 m. The length = 3 × 8.375 = 25.125 m. 5 Let p be the full price of a case of wine. The merchant will pay 60% (0.6) on the

25 discounted cases. 25p + 25 × 0.6p = 2260 25p + 15p = 2260 40p = 2260 p = 56.5 The full price of a case is $56.50.


6 Let x be the number of houses with an $11 500 commission and y be the number of houses with a $13 000 commission.

9 Let r km/h be the speed Kim can run. Her cycling speed will be (r + 30) km/h. Her time cycling will be 48 + 48 ÷ 3 = 64 min. Converting the times to hours (÷ 60) and

using distance = speed × time gives the following equation:

We only need to find x. x + y = 22

r × 4860

+ (r + 30) × 6460

= 60

48r + 64(r + 30) = 60 × 6048r + 64r + 1920 = 3600

112r + 1920 = 3600112r = 1680

r = 1680112

= 15

11 500x + 13 000y = 272 500 To simplify the second equation, divide

both sides by 500. 23x + 26y = 545 Using the substitution method: 23x + 26y = 545

y = 22 – x23x + 26(22 – x) = 54523x + 572 – 26x = 545

–3x = 545 – 572 = –27x = 9

10 Let c g be the mass of a carbon atom

and x g be the mass of an oxygen atom. (o is too confusing a symbol to use)

He sells nine houses with an $11 500 commission.

2c + 6x = 2.45 × 10–22

x = c3

7 It is easiest to let the third boy have Use substitution. m marbles, in which case the second boy

will have 2m marbles and the first boy will have 2m –14.

2c + 6 × c3

= 2.45 × 10–22

2c + 2c = 2.45 × 10–22

4c = 2.45 × 10–22

c = 2.45 × 10–22

4 = 6.125 × 10–23

x = c3

= 6.125 × 10–23

3 ≈ 2.04 × 10–23

(2m – 14) + 2m + m = 71 5m – 14 = 71 5m = 85 m = 17 The first boy has 20 marbles, the second

boy has 34 and the third boy has 17 marbles, for a total of 71.

8 Let Belinda’s score be b. Annie’s score will be 110% of Belinda’s

or 1.1b. The mass of an oxygen atom is Cassie’s will be 60% of their combined scores: 0.6(1.1b + b) = 0.6 × 2.1b 2.04 × 10–23g.

= 1.26b

1.1b + b + 1.26b = 504 3.36b = 504

b = 5043.36

= 150

Belinda scores 150 Annie scores 1.1 × 150 = 165 Cassie scores 0.6 × (150 + 165) = 189


Exercise 2E Solutions 1 Let k be the number of kilometres

travelled in a day. The unlimited kilometre alternative will become more attractive when 0.32k + 63 > 108.

Solve for 0.32k + 63 = 108: 0.32k = 108 – 63

= 45k = 45

0.32= 140.625

The unlimited kilometre alternative will become more attractive when you travel more than 140.625 km.

2 Let g be the number of guests. Solve for

the equality.

300 + 43g = 450 + 40g43g – 40g = 450 – 300

3g = 150g = 50

Company A is cheaper when there are more than 50 guests.

3 Let a be the number of adults and c the

number of children. 45a + 15c = 525 000 a + c = 15 000 Multiply the second equation by 15. 45a + 15c = 525 000

15a + 15c = 225 000 – :

30a = 300 000a = 10 000

10 000 adults bought tickets. 4 Let $m be the amount the contractor paid a

man and $b the amount he paid a boy. 8m + 3b = 2240 6m + 18b = 4200 Multiply the first equation by 6. 48 m + 18b = 13 440

6m + 18b = 4200 – :

42m = 9240m = 220

Substitute in the first equation:

8 × 220 + 3b = 22401760 + 3b = 2240

3b = 2240 – 1760 = 480

b = 160 He paid the men $220 each and the boys $160. 5 Let the numbers be x and y.

x + y = 212x – y = 42

+ :2x = 254

x = 127127 + y = 212

y = 85 The numbers are 127 and 85. 6 Let x L be the amount of 40% solution

and y L be the amount of 15% solution. Equate the actual substance. 0.4x + 0.15y = 0.24 × 700

= 168x + y = 700

Multiply the second equation by 0.15.

0.4x + 0.15y = 1680.15x + 0.15y = 105

– :0.25x = 63

x = 63 × 4 = 252

252 + y = 700y = 448

Use 252 L of 40% solution and 448 L of 15% solution.

7 Form two simultaneous equations.

x + y = 220x – x

2= y – 40

x2

– y = –40

+ :3x2

= 180

x = 120120 + y = 220

y = 100 They started with 120 and 100 marbles

and ended with 60 each.


8 Let $x be the amount initially invested at 10% and $y the amount initially invested at 7%. This earns $31 000.

9 Let a be the number of adults and s the number of students who attended.

30a + 20s = 37 000a + s = 1600

20a + 20s = 1600 × 20 = 32 000

– :10a = 5000

a = 500500 + s = 1600

s = 1100

0.1x + 0.07y = 31 000 When the amounts are interchanged, she

earns $1000 more, i.e. $32 000. 0.07x + 0.1y = 32 000 Multiply the first equation by 100 and

the second equation by 70.

10x + 7y = 3 100 0004.9x + 7y = 2 240 000

– :5.1x = 860 000

x = 860 0005.1

≈ 168 627.451

500 adults and 1100 students attended the concert.

10 × 168 627.451 + 7y = 3 100 0001 686 274.51 + 7y = 3 100 000

7y = 1 413 725.49y = 201 960.78

The total amount invested is x + y = 168 627.45 + 201 960.78 = $370 588.23


Exercise 2F Solutions 1 a v = u + at

= 15 + 2 × 5= 25

b I = PrT

100

= 600 × 5.5 × 10100

= 330

c V = πr2h

= π × 4.252 × 6≈ 340.47

d S = 2πr(r + h)

= 2π × 10.2 × (10.2 + 15.6)≈ 1653.48

e V = 4

3 πr2h

= 4π × 3.582 × 11.43

≈ 612.01

f s = ut + 1

2 at2

= 25.6 × 3.3 + 12× –1.2 × 3.32

≈ 77.95

g T = 2π lg

= 2π × 1.459.8

= 2π × 0.3846. . .≈ 2.42

h 1

f= 1

v+ 1

u = 1

3+ 1

7= 10

21f = 21

10 = 2.1

i c2 = a2 + b2

= 8.82 + 3.42

= 89c = 89 ≈ 9.43

j v2 = u2 + 2as

= 4.82 + 2 × 2.25 × 13.6 = 91.04v = 91.04 ≈ 9.54

2 a v = u + at

v – u = at∴ a = v – u

t

b S = n

2(a + l)

2S = n(a + l)a + l = 2S

n∴ l = 2S

n– a

c A = 1

2 bh

2A = bh∴ b = 2A

h

d P = I2R

PR

= I2

∴ I = ± PR

e s = ut + 1

2 at2

s – ut = 12 at2

2(s – ut) = at2

∴ a = 2(s – ut)t2


f E = 12

mv2

2E = mv2

v2 = 2Em

∴ v = ± 2Em

g Q = 2gh

Q2 = 2gh

∴ h = Q2

2g

h – xy – z = xy + z

– xy – xy = z + z–2xy = 2z

∴ x = 2z–2y

= – zy

i ax + by

c= x – b

ax + by = c(x – b)ax + by = cx – bcax – cx = – bc – by

x(a – c) = – b(c + y)∴ x = – b(c + y)

a – c = b(c + y)

c – a

j mx + b

x – b= c

mx + b = c(x – b)mx + b = cx – bc

mx – cx = – bc – bx(m – c) = – b(c + 1)

∴ x = – b(c + 1)m – c

3 a F = 9C

5+ 32

= 9 × 285

+ 32

= 82.4°

b F = 9C5

+ 32

F – 32 = 9C5

9C = 5(F – 32)∴ C = 5(F – 32)

9

Substitute F = 135. C = 5(135 – 32)

9 = 515

9 ≈ 57.22°

4 a S = 180(n – 2)

= 180(8 – 2)= 1080°

b S = 180(n – 2)

S180

= n – 2

∴ n = S180

+ 2

= 1260180

+ 2

= 7 + 2 = 9

Polygon has 9 sides (a nonagon). 5 a V = 1

3 πr2h

= 13× π × 3.52 × 9

≈ 115.45 cm3

b V = 13

π2h

3V = πr2h∴ h = 3V

πr2

= 3 × 210π 42

≈ 12.53 cm


c V = 13

πr2h

3V = πr2hr2 = 3V

πh

∴ r = 3Vπh

= 3 × 262π × 10

≈ 5.00 cm

6 a S = n

2 (a + l)

= 72(–3 + 22)

= 66.5

b S = n

2 (a + l)

2S = n(a + l)2Sn

= a + l

∴ a = 2Sn

– l

= 2 × 104013

– 156

= 4

c S = n

2 (a + l)

2S = n(a + l)∴ n = 2S

a + l

= 2 × 11025 + –5

= 11

There are 11 terms.


Exercise 2G Solutions 1 a 2x

3+ 3x

2= 4x + 9x

6= 13x

6

b 3a

2– a

4= 6a – a

4= 5a

4

c 3h

4+ 5h

8– 3h

2= 6h + 5h – 12h

8= – h

8

d 3x

4– y

6– x

3= 9x – 2y – 4x

12= 5x – 2y

12

e 3

x+ 2

y= 3y + 2x

xy

f 5

x – 1+ 2

x= 5x + 2(x – 1)

x(x – 1)= 5x + 2x – 2

x(x – 1)= 7x – 2

x(x – 1)

g 3

x – 2+ 2

x + 1= 3(x + 1) + 2(x – 2)

(x – 2)(x + 1)= 3x + 3 + 2x – 4

(x – 2)(x + 1)= 5x – 1

(x – 2)(x + 1)

h 2xx + 3

– 4xx – 3

– 32

= 4x(x – 3) – 8x(x + 3) – 3(x + 3)(x – 3)2(x + 3)(x – 3)

= 4x2 – 12x – 8x2 – 24x – 3(x2 – 9)2(x + 3)(x – 3)

= 4x2 – 12x – 8x2 – 24x – 3x2 + 272(x + 3)(x – 3)

= –7x2 – 36x + 272(x + 3)(x – 3)

i 4

x + 1+ 3

(x + 1)2 = 4(x + 1) + 3(x + 1)2

= 4x + 4 + 3(x + 1)2

= 4x + 7(x + 1)2

j a – 2a

+ a4

+ 3a8

= 8(a – 2) + 2a2 + 3a2

8a

= 5a2 + 8a – 168a

k 2x – 6x2 – 45x

= 10x2 – (6x2 – 4)5x

= 10x2 – 6x2 + 45x

= 4x2 + 45x

= 4(x2 + 1)5x

l 2

x + 4– 3

x2 + 8x + 16= 2

x + 4– 3

(x + 4)2

= 2(x + 4) – 3(x + 4)2

= 2x + 8 – 3(x + 4)2

= 2x + 5(x + 4)2


m 3x – 1

+ 2(x – 1)(x + 4)

= 3(x + 4) + 2(x – 1)(x + 4)

= 3x + 12 + 2(x – 1)(x + 4)

= 3x + 14(x – 1)(x + 4)

n 3

x – 2– 2

x + 2+ 4

x2 – 4= 3

x – 2– 2

x + 2+ 4

(x – 2)(x + 2) = 3(x + 2) – 2(x – 2) + 4

(x – 2)(x + 2) = 3x + 6 – 2x + 4 + 4

(x – 2)(x + 2) = x + 14

(x – 2)(x + 2)

o 5

x – 2– 3

x2 + 5x + 6+ 2

x + 3

= 5x – 2

– 3(x + 2)(x + 3)

+ 2x + 3

= 5(x + 3)(x + 2) – 3(x – 2) + 2(x – 2)(x + 2)(x – 2)(x + 2)(x + 3)

= 5(x2 + 5x + 6) – 3x + 6 + 2(x2 – 4)(x – 2)(x + 2)(x + 3)

= 5x2 + 25x + 30 – 3x + 6 + 2x2 – 8(x – 2)(x + 2)(x + 3)

= 7x2 + 22x + 28(x – 2)(x + 2)(x + 3)

p x – y – 1

x – y= (x – y)(x – y) – 1

x – y

= (x – y)2 – 1x – y

q 3

x – 1– 4x

1 – x= 3

x – 1+ 4x

x – 1= 4x + 3

x – 1

r 3

x – 2+ 2

2 – x= 3

x – 2– 2x

x – 2= 3 – 2x

x – 2

2

a x2

2y× 4y3

x= 4y3x2

2yx= 2xy2

b 3x2

4y× y2

6x= 3x2y2

24yx= xy

8

c 4x3

3× 12

8x4 = 48x3

24x4

= 2x

d x2

2y÷ 3xy

6= x2

2y× 6

3xy

= 6x2

6xy2

= xy2

e 4 – x3a

× a2

4 – x= a2(4 – x)

3a(4 – x)= a

3

f 2x + 5

4x2 + 10x= 2x + 5

2x(2x + 5)

= 12x

g (x – 1)2

x2 + 3x – 4= (x – 1)2

(x – 1)(x + 4)

= x – 1x + 4

h x2 – x – 6x – 3

= (x – 3)(x + 2)x – 3

= x + 2

i x2 – 5x + 4x2 – 4x

= (x – 1)(x – 4)x(x – 4)

= x – 1x


j 5a2

12b2 ÷10a6b

= 5a2

12b2 ×6b

10a

= 30a2 b120ab2

= a4b

k x – 2x

÷ x2 – 42x2 = x – 2

x× 2x2

x2 – 4

= x – 2x

× 2x2

(x – 2)(x + 2)

= 2x2

x(x + 2)= 2x

x + 2

l x + 2

x(x – 3)÷ 4x + 8

x2 – 4x + 3 = x + 2

x(x – 3)÷ 4(x + 2)

(x – 1)(x – 3) = x + 2

x(x – 3)× (x – 1)(x – 3)

4(x + 2) = 1

x× x – 1

4= x – 1

4x

m 2xx – 1

÷ 4x2

x2 – 1= 2x

x – 1× x2 – 1

4x2

= 2xx – 1

× (x – 1)(x + 1)4x2

= 2x(x + 1)4x2

= x + 12x

n x2 – 9x + 2

× 3x + 6x – 3

÷ 9x

= (x – 3)(x + 3)x + 2

× 3(x + 2)x – 3

× x9

= 3x(x – 3)(x + 3)(x + 2)9(x + 2)(x – 3)

= x(x + 3)3

o 3x9x – 6

÷ 6x2

x – 2× 2

x + 5= 3x

3(3x – 2)× x – 2

6x2 × 2x + 5

= 2x(x – 2)6x2(3x – 2)(x + 5)

= x – 23x(3x – 2)(x + 5)

3 a 1

x – 3+ 2

x – 3= 3

x – 3

b 2

x – 4+ 2

x – 3= 2(x – 3) + 2(x – 4)

(x – 4)(x – 3)= 2x – 6 + 2x – 8

x2 – 7x + 12= 4x – 14

x2 – 7x + 12

c 3

x + 4+ 2

x – 3= 3(x – 3) + 2(x + 4)

(x + 4)(x – 3)= 3x – 9 + 2x + 8

x2 + x – 12= 5x – 1

x2 + x + 12

d 2x

x – 3+ 2

x + 4= 2x(x + 4) + 2(x – 3)

(x – 3)(x + 4)

= 2x2 + 8x + 2x – 6x2 + x – 12

= 2x2 + 10x – 6x2 + x + 12

e 1

(x – 5)2 + 2x – 5

= 1 + 2(x – 5)(x – 5)2

= 1 + 2x – 10x2 – 10x + 25

= 2x – 9x2 – 10x + 25

f 3x

(x – 4)2 + 2x – 4

= 3x + 2(x – 4)(x – 4)2

= 3x + 2x – 8x2 – 8x + 16

= 5x – 8x2 – 8x + 16


g 1x – 3

– 2x – 3

= –1x – 3

= 13 – x

h 2

x – 3– 5

x + 4= 2(x + 4) – 5(x – 3)

(x – 3)(x + 4)= 2x + 8 – 5x + 15

x2 + x – 12= 23 – 3x

x2 + x – 12

i 2x

x – 3+ 3x

x + 3= 2x(x + 3) + 3x(x – 3)

(x – 3)(x + 3)

= 2x2 + 6x + 3x2 – 9xx2 – 9

= 5x2 – 3xx2 – 9

j 1

(x – 5)2 – 2x – 5

= 1 – 2(x – 5)(x – 5)2

= 1 – 2x + 10x2 – 10x + 25

= 11 – 2xx2 – 10x + 25

k 2x

(x – 6)3 – 2(x – 6)2 = 2x – 2(x – 6)

(x – 6)3

= 2x – 2x + 12(x – 6)3

= 12(x – 6)3

l 2x + 3

x – 4– 2x – 4

x – 3= (2x + 3)(x – 3) – (2x – 4)(x – 4)

(x – 4)(x – 3)

= (2x2 – 3x – 9) – (2x2 – 12x + 16)x2 – 7x + 12

= 2x2 – 3x – 9 – 2x2 + 12x – 16x2 – 7x + 12

= 9x – 25x2 – 7x + 12

4

a 1 – x + 21 – x

= 1 – x 1 – x + 21 – x

= 1 – x + 21 – x

= 3 – x1 – x

b 2x – 4

+ 23

= 2 x – 4 + 63 x – 4

c 3

x + 4+ 2

x + 4= 5

x + 4

d 3x + 4

+ x + 4 = 3 + x + 4 x + 4x + 4

= 3 + x + 4x + 4

= x + 7x + 4

e 3x3

x + 4– 3x2 x + 4 = 3x3 – 3x2 x + 4 x + 4

x + 4

= 3x3 – 3x2(x + 4)x + 4

= 3x3 – 3x3 – 12x2

x + 4

= – 12x2

x + 4

f 3x3

2 x + 3+ 3x2 x + 3 = 3x3 + 6x2 x + 3 x + 3

x + 3

= 3x3 + 6x2(x + 3)x + 3

= 3x3 + 6x3 + 18x2

x + 3

= 9x3 + 18x2

x + 3

= 9x2(x + 2)x + 3


5 c (3 – x)

13 – 2x(3 – x)

– 23

= (3 – x)13 –

2x

(3 – x)23

=(3 – x)

13(3 – x)

23 – 2x

(3 – x)23

=3 – x – 2x

(3 – x)23

=3 – 3x

(3 – x)23

a (6x – 3)

13 – (6x – 3)

– 23

= (6x – 3)13 –

1

(6x – 3)23

=(6x – 3)

13(6x – 3)

23 – 1

(6x – 3)23

=6x – 3 – 1

(6x – 3)23

=6x – 4

(6x – 3)23

Since (3 the answer is

equivalent to

– x)2 = (x – 3)2,3 – 3x

(x – 3)23 .

b (2x + 3)13 – 2x(2x + 3)

– 23

= (2x + 3)13 –

2x

(2x + 3)23

=(2x + 3)

13(2x + 3)

23 – 2x

(2x + 3)23

=2x + 3 – 2x

(2x + 3)23

=3

(2x + 3)23


Exercise 2H Solutions 1 a ax + n = m

ax = m – nx = m – n

a

b ax + b = bx

ax – bx = – bx(a – b) = – b

x = – ba – b

This answer is correct, but to avoid a negative sign, multiply numerator and denominator by –1.

x = – ba – b

× –1–1

= bb – a

c ax

b+ c = 0

axb

= – c

ax = – bcx = – bc

a

d px = qx + 5

px – qx = 5x(p – q) = 5

x = 5p – q

e mx + n = nx – m

mx – nx = – m – nx(m – n) = – m – n

x = – m – nm – n

= m + nn – m

f 1x + a

= bx

Take reciprocals of both sides:

x + a = xb

x – xb

= – a

xb

– x = a

x – xbb

= a

x – xbb

× b = ab

x – xb = abx(1 – b) = ab

x = aba – b

g b

x – a= 2b

x + a

Take reciprocals of both sides:

x – ab

= x + a2b

x – ab

× 2b = x + a2b

× 2b

2(x – a) = x + a2x – 2a = x + a

2x – x = a + 2ax = 3a

h x

m+ n = x

n+ m

xm× mn + n × mn = x

n× mn + m × mn

nx + mn2 = mx + m2nnx – mx = m2n – mn2

x(n – m) = mn(m – n)x = mn(m – n)

n – m

Note that n – m = – m + n= –1(m – n)

∴ x = – mn(n – m)n – m

= – mn


i – b(ax + b) = a(bx – a)– abx – b2 = abx – a2

– abx – abx = – a2 + b2

–2abx = – a2 + b2

x = – ( – a2 + b2)2ab

= a2 – b2

2ab

j p2(1 – x) – 2pqx = q2(1 + x)

p2 – p2x – 2pqx = q2 + q2x– p2x – 2pqx – q2x = q2 – p2

– x(p2 + 2pq + q2) = q2 – p2

x = – (q2 – p2)p2 + 2pq + q2

= p2 – q2

(p + q)2

= (p – q)(p + q)(p + q)2

= p – qp + q

k x

a– 1 = x

b+ 2

xa× ab – ab = x

b× ab + 2ab

bx – ab = ax + 2abbx – ax = 2ab + ab

x(b – a) = 3abx = 3ab

b – a

l x

a – b+ 2x

a + b= 1

a2 – b2

x(a – b)(a + b)a – b

+ 2x(a + b)(a – b)a + b

= (a + b)(a – b)a2 – b2

x(a + b) + 2x(a – b) = 1ax + bx + 2ax – 2bx = 1

3ax – bx = 1x(3a – b) = 1

x = 13a – b

m p – qxt

+ p = qx – 1p

pt(p – qx)t

+ p × pt = pt(qx – 1)p

p(p – qx) + p2t = t(qx – 1)p2 – pqx + p2t = qtx – t

– pqx – qtx = – t – p2 – p2t– qx(p + t) = – (t + p2 + p2t)

x = t + p2 + p2tq(p + t)

or p2 + p2t + tq(p + t)

n 1

x + a+ 1

x + 2a= 2

x + 3a Multiply each term by (x + a)(x + 2a)(x + 3a).

(x + 2a)(x + 3a) + (x + a)(x + 3a) = 2(x + a)(x + 2a)x2 + 5ax + 6a2 + x2 + 4ax + 3a2 = 2x2 + 6ax + 4a2

2x2 + 9ax + 9a2 = 2x2 + 6ax + 4a2

2x2 – 9ax – 2x2 – 6ax = 4a2 – 9a2

3ax = –5a2

x = –5a2

3a = – 5a

3 2 ax + by = p ; bx – ay = q Multiply the first equation by a and the

second equation by b. a2x + aby = ap

b2x – aby = bp + :

x(a2 + b2) = ap + bqx = ap + bq

a2 + b2

Substitute into ax + by = p:

a × ap + bqa2 + b2 + by = p

a(ap + bq) + by(a2 + b2) = p(a2 + b2)a2p + abq + by(a2 + b2) = a2p + b2p

by(a2 + b2) = a2p + b2p – a2p – abqby(a2 + b2) = b2p – abq

y = b(bp – aq)b(a2 + b2)

= bp – aqa2 + b2


3 xa

+ yb

= 1; xb

+ ya

= 1

First, multiply both equations by ab, giving the following:

bx + ay = abax + by = ab

Multiply the first equation by b and the second equation by a:

b2x + aby = ab2

a2x + aby = a2b – :

x(b2 – a2) = ab2 – a2b

x = ab2 – a2bb2 – a2

= ab(b – a)(b – a)(b + a)

= aba + b

Substitute into bx + ay = ab:

b × aba + b

+ ay = ab

ab2(a + b)a + b

+ ay(a + b) = ab(a + b)

ab2 + ay(a + b) = a2b + ab2

ay(a + b) = a2b + ab2 – ab2

ay(a + b) = a2b

y = a2ba(a + b)

= aba + b

4 a Multiply the first equation by b.

abx + by = bcx + by = d

– :x(ab – 1) = bc – d

x = bc – dab – 1

= d – bc1 – ab

It is easier to substitute in the first equation for x:

a × bc – dab – 1

+ y = c

a(bc – d)(ab – 1)ab – 1

+ y(ab – 1) = c(ab – 1)

abc – ad + y(ab – 1) = abc – cy(ab – 1) = abc – c – abc + ady(ab – 1) = – c + ad

y = ad – cab – 1

= c – ad1 – ab

b Multiply the first equation by a and the

second equation by b.

a2x – aby = a3

b2x – aby = b3

– :x(a2 – b2) = a3 – b3

x = a3 – b3

a2 – b2

= (a – b)(a2 + ab + b2)(a – b)(a + b)

= a2 + ab + b2

a + b In this case it is easier to start again, but

eliminate x. Multiply the first equation by b and the

second equation by a.

abx – b2y = a2babx – a2y = ab2

– :y( – b2 + a2) = a2b – ab2

y(a2 – b2) = ab(a – b)y = ab(a – b)

a2 – b2

= ab(a – b)(a – b)(a + b)

= aba + b


c Add the starting equations:

ax + by + ax – by = t + s2ax = t + s

x = t + s2a

Subtract the starting equations: ax + by – (ax – by) = t – s

2by = t – sy = t – s

2b

d Multiply the first equation by a and the

second equation by b. a2x + aby = a3 + 2a2b – ab2

b2x + aby = a2b + b3

– :x(a2 – b2) = a3 + a2b – ab2 – b3

x = a3 + a2b – ab2 – b3

a2 – b2

= a2(a + b) – b2(a + b)a2 – b2

= (a2 – b2)(a + b)a2 – b2

= a + b

Substitute into the second, simpler equation.

b(a + b) + ay = a2 + b2

ab + b2 + ay = a2 + b2

ay = a2 + b2 – ab – b2

ay = a2 – ab

y = a2 – aba

= a – b

e Rewrite the second equation, then multiply the first equation by b + c and the second equation by c.

(a + b)(b + c)x + c(c + c)y = bc(b + c)acx + c(b + c)y = – abc

– :x((a + b)(b + c) – ac) = bc(b + c) + abc

x(ab + ac + b2 + bc – ac) = bc(b + c + a)x(ab + b2 + bc) = bc(a + b + c)

xb(a + b + c) = bc(a + b + c)x = bc(a + b + c)

b(a + b + c) = c

Substitute into the first equation. (It has the simpler y term.)

c(a + b) + cy = bcac + bc + cy = bc

cy = bc – ac – bccy = – acy = – ac

c = – a

f First simplify the equations.

3x – 3a – 2y – 2a = 5 – 4a3x – 2y = 5 – 4a + 3a + 2a3x – 2y = a + 5

2x + 2a + 3y – 3a = 4a – 12x + 3y = 4a – 1 – 2a + 3a2x + 3y = 5a – 1

Multiply ① by 3 and ② by 2.

9x – 6y = 3a + 154x + 6y = 10a – 2

+ :13x = 13a + 13

x = a + 1 Substitute into ②:

2(a + 1) + 3y = 5a – 12a + 2 + 3y = 5a – 1

3y = 5a – 1 – 2a – 23y = 3a – 3

y = a – 1


e s = h2 + ah= (3a2)

2+ a(3a2)

= 9a4 + 3a3

= 3a3(3a + 1)

5 a s = ah

= a(2a + 1)

b Make h the subject of the second equation. h = a(2 + h)

= 2a + ahh – ah = 2a

h(1 – a) = 2ah = 2a

1 – a

f as = a + 2h

= a + 2(a – s) = a + 2a – 2s

as + 2s = 3as(a + 2) = 3a

s = 3aa + 2

Substitute into the first equation. s = ah

= a × 2a1 – a

= 2a2

1 – a

g s = 2 + ah + h2

= 2 + a⎝⎜⎛a – 1

a⎠⎟⎞ +

⎝⎜⎛a – 1

a⎠⎟⎞2

= 2 + a2 – 1 + a2 – 2 + 1a2

= 2a2 – 1 + 1a2

c h + ah = 1

h(1 + a) = 1h = 1

(1 + a)= 1

a + 1as = a + h

= a + 1a + 1

= a(a + 1) + 1a + 1

= a2 + a + 1a + 1

s = a2 + a + 1a(a + 1)

h Make h the subject of the second

equation.

as + 2h = 3a2h = 3a – ash = 3a – as

2 Substitute into the first equation. 3s – ah = a2

3s – a(3a – as)2

= a2

6s – a(3a – as) = 2a2

6s – 3a2 + a2s = 2a2

a2s + 6s = 2a2 + 3a2

s(a2 + 6) = 5a2

s = 5a2

a2 + 6

d Make h the subject of the second equation.

ah = a + hah – h = a

h(a – 1) = ah = 1

a – 1 Substitute into the first equation.

as = s + has = s + a

a – 1as – s = a

a – 1s(a – 1) = a

a – 1s(a – 1)(a – 1) = a(a – 1)

a – 1s(a – 1)2 = a

s = a(a – 1)2


Exercise 2I Solutions Use your CAS calculator to find the solutions to these problems. The exact method will vary depending on the calculator used. 1 a x = a – b b x = 7

c x = – a ± a2 + 4ab – 4b2

2

d x = a + c

2

2 a (x – 1)(x + 1)(y – 1)(y + 1) b (x – 1)(x + 1)(x + 2) c (a2 – 12b)(a2 + 4b) d (a – c)(a – 2b + c)

3 a axy + b = (a + c)y

bxy + a = (b + c)yDividing by y yields:

ax + by

= a + c

bx + ay

= b + c

let n = 1y and the equations become:

ax + bn = a + cbx + an = b + c

∴ x = a + b + ca + b

y = a + bc

b x(b – c) + by – c = 0

y(c – a) – ax + c = 0

(b – c)x + by = c– ax + (c – a)y = – c

∴ x = – (a – b – c)a + b – c

y = a – b + ca + b – c


Solutions to Multiple-choice Questions 1 5x + 2y = 0

2y = –5x y

x= – 5

2 A

2 Multiply both sides of the second

equation by 2. 3x + 2y = 36

6x – 2y = 24 + :

9x = 60x = 20

33 × 20

3– y = 12

20 – y = 12

y = 8 A 3 t – 9 = 3t – 17

t – 3t = 9 – 17–2t = –8

t = 4 C 4 m = n – p

n + pm(n + p) = n – pmn + mp = n – p

mp + p = n – mnp(m + 1) = n(1 – m)

p = n(1 – m)1 + m

A

5 3

x – 3– 2

x + 3= 3(x + 3) – 2(x – 3)

(x – 3)(x + 3)= 3x + 9 – 2x + 6

x2 – 9

= x + 15x2 – 9

B

6 9x2y3 ÷ 15(xy)3 = 9x2y3

15(xy)3

= 9x2y3

15x3y3

= 915x

= 35x

E

7 V = 13 h(l + w)

3V = h(l + w)3V = hl + hwhl = 3V – hw

l = 3V – hwh

= 3Vh

– w B

8 (3x2y3)2

2x2y= 9x4y6

2x2y

= 9x2y5

2

= 92

x2y5 B

9 Y = 80% × Z = 4

5 Z

X = 150% × Y = 32 Y

= 32× 4Z

5 = 12Z

10 = 1.2 Z

= 20% greater than Z B 10 Let the other number be n. x + n

2= 5x + 4

x + n = 2(5x + 4) = 10x + 8

n = 10x + 8 – x

= 9x + 8 B


Solutions to Short-answer Questions 1 a (x3)

4= x3 × 4

= x12

b (y–12)34 = y

–12 × 34

= y–9

c 3x32 × –5x4 = (3 × –5)x

32

+ 4

= –15x112

d (x3)43 × x–5 = x

3 × 43 × x–5

= x4 – 5

= x–1

2 32 × 1011 × 12 × 10–5 = (32 × 12) × 1011 – 5

= 384 × 106

= 3.84 × 108

3 a 3x

5+ y

10– 2x

5= 6x + y – 4x

10= 2x + y

10

b 4

x– 7

y= 4y – 7x

xy

c 5

x + 2+ 2

x – 1= 5(x – 1) + 2(x + 2)

(x + 2)(x – 1)= 5x – 5 + 2x + 4

(x + 2)(x – 1)= 7x – 1

(x + 2)(x – 1)

d 3

x + 2+ 4

x + 4= 3(x + 4) + 4(x + 2)

(x + 2)(x + 4)= 3x + 12 + 4x + 8

(x + 2)(x + 4)= 7x + 20

(x + 2)(x + 4)

e 5xx + 4

+ 4xx – 2

– 52

= 10x(x – 2) + 8x(x + 4) – 5(x + 4)(x – 2)2(x + 4)(x – 2)

= 10x2 – 20x + 8x2 + 32x – 5(x2 + 2x – 8)2(x + 4)(x – 2)

= 10x2 – 20x + 8x2 + 32x – 5x2 – 10x + 402(x + 4)(x – 2)

= 37x2 + 2x + 402(x + 4)(x – 2)

f 3x – 2

– 6(x – 2)2 = 3(x – 2) – 6

(x – 2)2

= 3x – 6 – 6x – 2

= 3x – 12x – 2

= 3(x – 4)x – 2

4

a x + 52x – 6

÷ x2 + 5x4x – 12

= x + 52x – 6

× 4x – 12x2 + 5x

= x + 52(x – 3)

× 4(x – 3)x(x + 5)

= 42x

= 2x

b 3xx + 4

÷ 12x2

x2 – 16= 3x

x + 4× x2 – 16

12x2

= 3xx + 4

× (x – 4)(x + 4)12x2

= 3x(x – 4)12x2

= x – 44x

c x2 – 4x – 3

× 3x – 9x + 2

÷ 9x + 2

= x2 – 4x – 3

× 3x – 9x + 2

× x + 29

= (x – 2)(x + 2)x – 3

× 3(x – 3)x + 2

× x + 29

= (x + 2)(x – 2)3

= x2 – 43

d 4x + 209x – 6

× 6x2

x + 5÷ 2

3x – 2= 4(x + 5)

3(3x – 2)× 6x2

x + 5× 3x – 2

2

= 4 × 6x2

3 × 2= 4x2


5 Let t seconds be the required time. The number of red blood cells to be

replaced is 12× 5 × 1012 = 2.5 × 1012

2.5 × 106 × t = 2.5 × 1012

t = 2.5 × 1012

2.5 × 106

= 106

Time = 106 seconds = 106 ÷ 3600 ÷ 24 days ≈ 11.57 or 11 31

54 days

6 1.5 × 108

3 × 106 = 0.5 × 102

= 50 times further

7 Let g be the number of games the team

lost. They won 2g games and drew one third of 54 games, i.e. 18 games.

g + 2g + 18 = 543g = 54 – 18

= 36g = 12

They have lost 12 games. 8 Let b be the number of blues CDs sold.

The store sold 1.1b classical and 1.5(b + 1.1b) heavy metal CDs, totalling

420 CDs.

b + 1.1b + 1.5 × 2.1b = 4205.25b = 420

b = 4205.25

= 80

1.1b = 1.1 × 80 = 88

1.5 × 2.1b = 1.5 × 2.1 × 80 = 252 80 blues, 88 classical and 252 heavy

metal (totalling 420)

9 a V = πr2h

= π × 52 × 12 = 300π ≈ 942 cm3

b h = V

πr2

= 585π × 52

= 1175π

≈ 7.4 cm

c r2 = V

πh

r = Vπh

(use positive root)

= 786π × 6

= 128π

≈ 40.7 cm

10 a xy + ax = b

x(y + a) = bx = b

a + y

b a

x+ b

x= c

axx

+ bxx

= cx

a + b = cxx = a + b

c

c x

a= x

b+ 2

xaba

= xabb

+ 2ab

bx = ax + 2abbx – ax = 2ab

x(b – a) = 2abx = 2ab

b – a


d a – dxd

+ b = ax + db

bd(a – dx)d

+ bd × b = bd(ax + d)b

b(a – dx) + b2d = d(ax + d)ab – bdx + b2d = adx + d2

– bdx – adx = d2 – ab – b2d– x(bd + ad) = – (ab + b2d – d2)

x = – (ab + b2d – d2)– (bd + ad)

= ab + b2d – d2

bd + ad

11 a p

p + q+ q

p – q= p(p – q) + q(p + q)

(p + q)(p – q)

= p2 – qp + qp + q2

p2 – pq + pq – q2

= p2 + q2

p2 – q2

b 1x

– 2yxy – y2 = (xy – y2) – 2xy

x(xy – y2)

= – xy – y2

x2y – xy2

= y( – x – y)xy(x – y)

= – x – yx(x – y)

= x + yx(y – x)

c x2 + x – 6x + 1

× 2x2 + x – 1x + 3

= (x – 2)(x + 3)x + 1

× (x + 1)(2x – 1)x + 3

= (x – 2)(2x – 1)

d 2a2a + b

× 2ab + b2

ba2 = 2a2a + b

× b(2a + b)ba2

= 2abba2

= 2a

12 Let A’s age be a, B’s age be b and C’s age be c.

a = 3bb + 3 = 3(c + 3)

a + 15 = 3(c + 15) Substitute for a and simplify:

b + 3 = 3(c + 3)b + 3 = 3c + 9

b = 3c + 63b + 15 = 3(c + 15)3b + 15 = 3c + 45

3b = 3c + 30b = c + 10 = :

3c + 6 = c + 103c – c = 10 – 6

2c = 4c = 2b = 3 × 2 + 6 = 12

a = 3 × 12 = 36

A, B and C are 36, 12 and 2 years old respectively.

13 a Simplify the first equation: a – 5 = 1

7 (b + 3)

7(a – 5) = b + 37a – 35 = b + 37a – b = 38

Simplify the second equation: b – 12 = 1

5 (4a – 2)

5(b – 12) = 4a – 25b – 60 = 4a – 2

–4a + 5b = 58

Multiply the first equation by 5, and add the second equation.

35a – 5b = 190–4a + 5b = 58

+ :31a = 248

a = 8

Substitute in the first equation: 7 × 8 – b = 38

56 – b = 38b = 56 – 38 = 18


b Multiply the first equation by p.

(p – q)x + (p + q)y = (p + q2)

p(p – q)x + p(p + q)y = p(p + q2) Multiply the second by (p + q). qx – py = q2 – pq

q(p + q)x – p(p + q)y = (p + q)(q2 – pq) ① + ②:

(p(p – q) + q(p + q))x = p(p + q)2 + (p + q)(q2 – pq)(p2 – pq + pq + q2)x = p(p2 + 2pq + q2) + pq2 – p2q + q3 – pq2

(p2 + q2)x = p3 + 2p2q + pq2 – p2q + q3

= p3 + p2q + pq2 + q3

= p2(p + q) + q2(p + q) = (p + q)(p2 + q2)x = p + q

Substitute into the second equation, factorising the right side. q(p + q) – py = q2 – pq

pq + q2 – py = q2 – pq– py = q2 – pq – pq – q2

– py = –2pqy = –2pq

– p = 2q

14 Time = distance

speed

Remainder = 50 – 7 – 7 = 36 km

7x

+ 74x

+ 366x + 3

= 4

7x

+ 74x

+ 122x + 1

= 4

(4x(2x + 1)) ×⎝⎜⎛7

x+ 7

4x+ 12

2x + 1⎠⎟⎞ = 4 × 4x(2x + 1)

28(2x + 1) + 7(2x + 1) + 48x = 16x(2x + 1)56x + 28 + 14x + 7 + 48x = 32x2 + 16x

56x + 28 + 14x + 7 + 48x – 32x2 – 16x = 0–32x2 + 102x + 35 = 0

32x2 – 102x – 35 = 0(2x – 7)(16x + 5) = 0

2x – 7 = 0 or 16x + 5 = 0x > 0, so 2x – 7 = 0x = 3.5


15

a 2n2 × 6nk2 ÷ 3n = 2n2 × 6nk2

3n

= 12n3k2

3n= 4n2k2

b 8c2x3y6a2b3c3 ÷

12

xy

15abc2 = 8c2x3y6a2b3c3 ÷

xy30abc2

= 8c2x3y6a2b3c3 ×

30abc2

xy

= 240abc4x3y6a2b3c3xy

= 40cx2

ab2

16 x + 5

15– x – 5

10= 1 + 2x

1530(x + 5)

15– 30(x – 5)

10= 30 ×

⎝⎜⎛1 + 2x

15⎠⎟⎞

2(x + 5) – 3(x – 5) = 30 + 4x2x + 10 – 3x + 15 = 30 + 4x

2x – 3x – 4x = 30 – 10 – 15–5x = 5

x = –1


Chapter 3 – Number systems and sets Exercise 3A Solutions 1

a A' = {4} b B' = {1, 3, 5}

c A ∪ B = {1, 2, 3, 4, 5}, or ξ

d (A ∪ B)' = ∅

e A' ∩ B' = ∅ 2

a P' = {1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16} b Q' = {1, 3, 5, 7, 9, 11, 13, 15}

c P ∪ Q = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16}

d (P ∪ Q)' = {1, 5, 7, 11, 13}

e P' ∩ Q' = {1, 5, 7, 11, 13} 3

a A' = {1, 2, 3, 5, 6, 7, 9, 10, 11} b B' = {1, 3, 5, 7, 9, 11}

c A ∪ B = {2, 4, 6, 8, 10, 12}

d (A ∪ B)' = {1, 3, 5, 7, 9, 11}

e A' ∩ B' = {1, 3, 5, 7, 9, 11} 4

a P' = {10, 11, 13, 14, 15, 17, 18, 19, 21, 22, 23, 25} b Q' = {11, 12, 13, 14, 16, 17, 18, 19, 21, 22, 23, 24}

c P ∪ Q = {10, 12, 15, 16, 20, 24, 25}

d (P ∪ Q)' = {11, 13, 14, 17, 18, 19, 21, 22, 23}

e P' ∩ Q' = {11, 13, 14, 17, 18, 19, 21, 22, 23} 5

a A' = {R} b B' = {G, R}

c A ∩ B = {L, E, A, N}

d A ∪ B = {A, N, G, E, L}

e (A ∪ B)' = {R}

f A' ∪ B' = {G, R}


6

a X' = {p, q, u, v} b Y' = {p, r, w}

c X' ∩ Y' = {p}

d X' ∪ Y' = {p, q, r, u, v, w}

e X ∪ Y = {q, r, s, t, u, v, w}

f (X ∪ Y)' = {p} c and f are equal. 7

a X' = {5, 7, 8, 9, 10, 11} b Y' = {1, 3, 5, 7, 9, 11}

c X' ∪ Y' = {1, 3, 5, 7, 8, 9, 10, 11}

d X' ∩ Y' = {5, 7, 9, 11}

e X ∪ Y = {1, 2, 3, 4, 6, 8, 10, 12}

f (X ∪ Y)' = {5, 7, 9, 11} d and f are equal.

8

a

b

c

d

e

f

9

a A' = {E, H, M , S} b B' = {C, H, I, M}

c A ∩ B = {A, T}

d (A ∪ B)' = {H, M}

e A' ∪ B' = {C, E, H, I, M, S}

f A' ∩ B' = {H, M}


Exercise 3B Solutions 1 a Yes, because the sum can be expressed

as a terminating or recurring decimal. For instance, 1

10+ 1

20= 3

20.

b Yes, because the product can be

expressed as a terminating or recurring decimal. For instance, 1

10× 1

20= 1

200.

c Yes, because the quotient can be

expressed as a terminating or recurring decimal, so long as the denominator ≠ 0.

For instance, 110

÷ 120

= 110

× 20 = 2.

2 a No, because the sum cannot be

expressed as a terminating or recurring decimal. For instance, 2 + 3 .

a No, because the sum cannot be

expressed as a terminating or recurring decimal. For instance, 2 × 3 = 6 .

c No, because the sum cannot be

expressed as a terminating or recurring

decimal. For instance, 23

.

3 a 0.

.2

.7 = 0.272727. . .

0..2

.7 × 100 = 27.272727. . .

0..2

.7 × 99 = 27

∴ 0..2

.7 = 27

99= 3

11

b 0.12 = 12

100= 3

25

c 0..28571

.4 = 0.285714285714. . .

0..28571

.4 × 106 = 285714.285714. . .

0..28571

.4 × (106 – 1) = 285714

∴ 0..28571

.4 = 285714

999999= 2

7

d 0.

.3

.6 = 0.363636. . .

0..3

.6 × 100 = 36.3636. . .

0..3

.6 × 99 = 36

∴ 0..3

.6 = 36

99= 4

11

e 0.

.2 = 0.22222. . .

0..2 × 10 = 2.2222. . .

0..2 × 9 = 2

∴ 0..2 = 2

9

f 0.45 = 45

100= 9

20

4 a 2

7= 7 2.000000. . .

= 0.2857142857. . .= 0.

.28571

.4

b 5

11= 11 5.000000. . .

= 0.454545. . .= 0.

.4

.5

c 7

20= 20 7.00

= 0.35

d 4

13= 13 4.000000. . .

= 0.30769230. . .= 0.

.30769

.2

e 1

17= 17 1.00000000000000000. . .

= 0.0588235294117647058. . .= 0.

.058823529411764

.7


5 Assume 3 is a rational number, so 3 is a fraction in its simplest form, a

b.

3 = ab

3 = a2

b2

3b2 = a2

∴ a is a multiple of 3.∴ a = 3k, where k is an integer.

3b2 = (3k)2

3b2 = 9k2

b2 = 3b2

∴ b is a multiple of 3.

But this contradicts the assumption that ab

is a fraction in its simplest form,

as a and b are both multiples of 3. Therefore the initial assumption must be

incorrect and 3 is not a rational number.


Exercise 3C Solutions 1 a 8 = 4 × 2

= 2 2

b 12 = 4 × 3

= 2 3

c 27 = 9 × 3

= 3 3

d 50 = 25 × 2

= 5 2

e 45 = 9 × 5

= 3 5

f 1210 = 121 × 10

= 11 10

g 98 = 49 × 2

= 7 2

h 108 = 36 × 3

= 6 3

i 25 = 5 j 75 = 25 × 3

= 5 3

k 512 = 256 × 2

= 16 2

2 a 8 + 18 – 2 2

= 4 × 2 + 9 × 2 – 2 2 = 2 2 + 3 2 – 2 2 = 3 2

b 75 + 2 12 – 27

= 25 × 3 + 2 4 × 3 – 9 × 3= 5 3 + 4 3 – 3 3= 6 3

c 28 + 175 – 63= 4 × 7 + 25 × 7 – 9 × 7

= 2 7 + 5 7 – 3 7 = 4 7

d 1000 – 40 – 90

= 100 × 10 – 4 × 10 – 9 × 10 = 10 10 – 2 10 – 3 10 = 5 10

e 512 + 128 + 32

= 256 × 2 + 64 × 2 + 16 × 2 = 16 2 + 8 2 + 4 2 = 28 2

f 24 – 3 6 – 216 + 294

= 4 × 6 – 3 6 – 36 × 6 + 49 × 6 = 2 6 – 3 6 – 6 6 + 7 6 = 0

3 a 75 + 108 + 14

= 25 × 3 + 36 × 3 + 14 = 5 3 + 6 3 + 14 = 11 3 + 14

b 847 – 567 + 63

= 121 × 7 – 81 × 7 + 9 × 7 = 11 7 – 9 7 + 3 7 = 5 7

c 720 – 245 – 125

= 144 × 5 – 49 × 5 – 25 × 5 = 12 5 – 7 5 – 5 5 = 0

d 338 – 288 + 363 – 300

= 169 × 2 – 144 × 2 + 121 × 3 – 100 × 3 = 13 2 – 12 2 + 11 3 – 10 3 = 2 + 3


e 12 + 8 + 18 + 27 + 300= 4 × 3 + 4 × 2 + 9 × 2

+ 9 × 3 + 100 × 3 = 2 3 + 2 2 + 3 2 + 3 3 + 10 3 = 5 2 + 15 3

f 2 18 + 3 5 – 50 + 20 – 80

= 2 9 × 2 + 3 5 – 25 × 2 + 4 × 5 – 16 × 5

= 6 2 + 3 5 – 5 2 + 2 5 – 4 5 = 2 + 5

4

a 15× 5

5= 5

5

b 17× 7

7= 7

7

c – 12× 2

2= – 2

2

d 23× 3

3= 2 3

3

e 36× 6

6= 3 6

6

f 12 2

× 22

= 24

g 12 + 1

× 2 – 12 – 1

= 2 – 12 – 1

= 2 – 11

= 2 – 1

h 12 – 3

× 2 + 32 + 3

= 2 + 34 – 3

= 2 + 3

i 14 – 10

× 4 + 104 + 10

= 4 + 1016 – 10

= 4 + 106

j 26 + 2

× 6 – 26 – 2

= 2 6 – 46 – 4

= 2 6 – 42

= 6 – 2

k 15 – 3

× 5 + 35 + 3

= 5 + 35 – 3

= 5 + 32

l 16 – 5

× 6 + 56 + 5

= 6 + 56 – 5

= 6 + 5

m 13 – 2 2

× 3 + 2 23 + 2 2

= 3 + 2 29 – 8

= 3 + 2 2

5

a 23 – 2 2

× 3 + 2 23 + 2 2

= 6 + 4 29 – 8

= 6 + 4 2

b ( 5 + 2)2 = ( 5 )2 + 4 5 + 4= 5 + 4 5 + 4= 9 + 4 5

c (1 + 2 )(3 – 2 2 ) = 3 – 2 2 + 3 2 – 4= –1 + 2

d ( 3 – 1)2 = 3 – 2 3 + 1= 4 – 2 3

e 13

– 127

= 13× 27

27– 1

27× 3

3

= 3 3 – 39

= 2 39

f 3 + 22 3 – 1

= 3 + 22 3 – 1

× 2 3 + 12 3 + 1

= 6 + 3 + 4 3 + 212 – 1

= 8 + 5 311


6 g 5 + 1

5 – 1= 5 + 1

5 – 1× 5 + 1

5 + 1

= 5 + 2 5 + 15 – 1

= 6 + 2 54

= 3 + 52

a (2 a – 1)2 = (2 a – 1)(2 a – 1)

= 4a – 2 a – 2 a + 1= 4a – 4 a + 1

b ( x + 1 + x + 2 )2

= ( x + 1 + x + 2 )( x + 1 + x + 2 ) = x + 1 + 2 (x + 1)(x + 2) + x + 2 = 2x + 3 + 2 (x + 1)(x + 2)

h 8 + 318 + 2

= 2 2 + 33 2 + 2

= 2 2 + 33 2 + 2

× 3 2 – 23 2 – 2

= 12 – 4 2 + 9 2 – 618 – 4

= 6 + 5 214

7 a (5 – 3 2 ) – (6 2 – 8) = 5 – 3 2 – 6 2 + 8

= 13 – 9 2= 169 – 162> 0

5 – 3 2 is larger.

b (2 6 – 3) – (7 – 2 6 ) = 2 6 – 3 – 7 + 2 6= 4 6 – 10= 96 – 100< 0

7 – 2 6 is larger.


Exercise 3D Solutions 1 a 2 68 640

2 34 320 2 17 160 2 8580 2 4290 3 2145 5 71511 14313 13 1

Prime decomposition = 25 × 3 × 5 × 11 × 13

b 2 96 096

2 48 048 2 24 024 2 12 012 2 6006 3 3003 7 100111 14313 13 1


c 2 32 032

2 16 016 2 8008 2 4004 2 2002 7 100111 14313 13 1

Prime decomposition = 25 × 7 × 11 × 13

d 2 544 544 2 272 272 2 136 136 2 68 068 2 34 034 7 17 01711 243113 22117 17 1


2 For each part, first find the prime

decomposition of each number. a 4361 = 72 × 89 Neither 7 nor 89 are factors of 9281. HCF = 1 b 999 = 33 × 37

2160 = 24 × 33 × 5HCF = 33 = 27

c 5255 = 5 × 1051 716 845 is divisible by 5 but not 1051. HCF = 5 d 1271 = 31 × 41

3875 = 53 × 31HCF = 31

e 804 = 22 × 3 × 67

2358 = 2 × 32 × 131HCF = 2 × 3 = 6


Exercise 3E Solutions 1 a

9 5H E

x

Since all students do at least one of these subjects, 9 + 5 + x = 28

x = 14 b i 5 + 14 = 19 ii 9 iii 9 + 14 = 23 or 28 – 5 = 23 2

a

3 9

7

5

6 42

14

C

A B

b i n(A' ∩ C') = 9 + 14 = 23 ii n(A ∪ B') = 3 + 6 + 5 + 2 + 7 + 14 = 37 iii n(A' ∩ B ∩ C') = 9

3

yx

B (60%) G

20%

Since 40% don’t speak Greek, y + 20% = 40% y = 20% Since 40% speak Greek, x + 20% = 40% x = 20% 20% speak both languages.

4

yx6

C (25) D (16)

Since 40 – 25 = 15 don’t own a cat, y + 6 = 15 y = 9 Since 16 own a dog, x + 9 = 16 x = 7 Seven students own both. 5

a

a

b cx

E (70) F (50)

J (50)

We must assume every delegate spoke at least one of these languages.

If 70 spoke English, and 25 spoke English and French, 45 spoke English but not French.

∴45 + 50 = 95 spoke either English or French or both.

∴105 – 95 = 10 spoke only Japanese.

If 50 spoke French, and 15 spoke French and Japanese, 35 spoke French but not Japanese.

∴35 + 50 = 85 spoke either French or Japanese or both.

∴105 – 85 = 20 spoke only English.

If 50 spoke Japanese, and 30 spoke Japanese and English, 20 spoke Japanese but not English.

∴20 + 70 = 90 spoke either Japanese or English or both.

∴105 – 90 = 15 spoke only French.


We can now fill in more of the Venn diagram.

a

b cx

E (70) F (50)

J (50)

20 15

10

c is the number who don’t speak English. 105 – 70 = 10 + c + 15

c + 25 = 35c = 10

x + c = 15x = 5

5 delegates speak all five languages. b We have already found that 10 spoke

only Japanese. 6 Enter the information into a Venn

diagram.

50 70

60

30

10 4540

P G

F

Number having no dessert = 350 – 50 – 30 – 70 – 10 – 40 – 45 – 60 = 45

7 Insert the given information on a Venn diagram. Place y as the number taking a bus only, and z as the number taking a car only.

z y

x

x

C B ( 33 )

T ( 20 )

4

82

a Using n(T ) = 20, 2x + 10 = 20 x = 5 b Using n(B) = 33 and x = 5, 12 +5 + y = 33 y = 16 c Assume they all used at least one of

these forms of transport. z + 4 + 8 + 16 + 2 + 5 + 5 = 40 z = 0 8 a

b i (X ∩ Y ∩ Z) = intersection of all sets = 36 (from diagram) ii n(X ∩ Y) = number of elements in both X and Y = 5 (from diagram)


9 The following information can be placed on a Venn diagram.

102

5 33

R G

B

x

x

The additional information gives 5 > x and x > 3. ∴x = 4

Number of students = 10 + 2 + 4 + 5 + 3 + 3 + 4 = 31 20 bought red pens, 12 bought green

pens and 15 bought black pens. 10 Enter the given information as below. B ∩ M is shaded.

x y

12

z

B M

F

5 5

5

5 + 12 + 5 + 5 + x + y + z = 28

27 + x + y + z = 28x + y + z = 1

This means that exactly one of x, y and z must equal 1, and the other two will equal zero.

Since n(F ∩ B) > n(M ∩ F), the Venn diagram shows that this means x > y.

∴x = 1, y = z = 0

5 5

5

1 0

0

B M

F

a + b = 12

a

b

n(M ∩ F ∩ B) = n(F') ∴b = a + 10 Substitute in a + b = 12: a + (a + 10) = 12

2a = 12a = 6b = a + 10 = 16

n(M ∩ F) = b + 0 = 16 11 Enter the given information as below.

p q

r

a

b cx

A (23) S (22)

F (18)46

a + x = n(A ∩ S) = 10 The shaded area is given by n(A ∩ S ′ ) = n(A) – (a + x) = 23 – 10 = 13 n(A U S) = 10 + 22 = 32 ∴ r + 46 = 80 – 32 = 48 r = 2 Use similar reasoning to show c + r = 18 – (b + x) = 18 – 11 = 7 Since r = 2, c = 5 Since x + c = n(S U F) = 6 and c = 5, x = 1 One person plays all three sports.


12 Enter the information into a Venn diagram. 13 Enter the given information into a Venn diagram.

3

5 42

I F

G

x x

x + 1

35

25

7 204

M P

C

x1.5 x

1.5x + 25 + x + 7 + 4 + 20 + 35 = 201

2.5x + 91 = 2012.5x = 110

x = 1102.5

= 44

Since they are all proficient in at least one language,

x + 3 + x + 5 + 2 + 4 + x + 1 = 333x + 15 = 33

3x = 18 x = 6 The number studying Mathematics

The number proficient in Italian = 1.5x + 25 + 7 + 4 = 6 + 3 + 2 + 5 = 66 + 25 + 7 + 4 = 16 = 102


Solutions to Multiple-choice Questions

1 43 + 2 2

= 43 + 2 2

× 3 – 2 23 – 2 2

= 12 – 8 29 – 8

= 12 – 8 2 A 2 2 86 400

2 43 2002 21 6002 10 8002 54002 27002 13503 6753 2253 755 255 5 1

Prime decomposition D = 27 × 33 × 52

3 ( 6 + 3)( 6 – 3)

= ( 6 )2 + 3 6 – 3 6 – 9 = 6 – 9

= –3 D 4 B' ∩ A = numbers in set A that are not also in set B = {1, 2, 4, 5, 7, 8} D 5 ( 3,∞) ∩ ( – ∞, 5]

∞ – ∞ cancels out the middle = (3, 5] C 6 The next time will be both a multiple of 6 and a multiple of 14.

LCM = 6 × 143

= 42

The next time is in 42 minutes. D

7 X ∩ Y ∩ Z = set of numbers that are multiples of 2, 5 and 7 LCM = 2 × 5 × 7 = 35 B 8 Draw a Venn diagram.

x

30 %

F ( 50 % ) T ( 40 % )

Since 50% don’t play football, x + 30% = 50% x = 20% Since 40% play tennis, it can be

seen that 20% play both sports. B

9 7 – 67 + 6

= 7 – 67 + 6

× 7 – 67 – 6

= 7 – 2 42 + 67 – 6

= 13 – 2 42 C 10 Draw a Venn diagram.

15 50

L E

x

15 + 5 + x = 40 x = 20 20 students take only Economics. A


Solutions to Short-answer Questions 1 a 0.0

.7 = 0.07777. . .

0.0.7 × 10 = 0.7777. . .

0.0.7 × 9 = 0.7 = 7

100.0

.7 = 7

90

b 0.

.4

.5 = 0.454545. . .

0..4

.5 × 100 = 45.4545. . .

0..4

.5 × 99 = 45

0..4

.5 = 45

99= 5

11

c 0.005 = 5

1000= 1

200

d 0.405 = 405

1000= 81

200

e 0.2

.6 = 0.26666. . .

0.2.6 × 10 = 2.6666. . .

0.2.6 × 9 = 2.4 = 24

100.2

.6 = 24

90= 4

15

f 0.1

.71428

.5 = 0.1714825714. . .

0.1.71428

.5 × 106 = 171 428.5714285. . .

0.1.71428

.5 × (106 – 1) = 171 428.4

= 1 714 28410

0.1.71428

.5 = 1 714 284

9 999 990 = 6

35 2 2 504

2 2522 1263 633 217 7 1504 = 23 × 32 × 7

3

a 2 3 – 12

= 2 3 – 12

× 22

= 2 6 – 22

b 5 + 25 – 2

= 5 + 25 – 2

× 5 + 25 + 2

= 5 + 4 5 + 45 – 4

= 4 5 + 9

c 3 + 23 – 2

= 3 + 23 – 2

× 3 + 23 + 2

= 3 + 2 6 + 23 – 2

= 2 6 + 5

4 3 + 2 753 – 12

= 3 + 2 25 × 33 – 4 × 3

= 3 + 2 × 5 33 – 2 3

= 3 + 10 33 – 2 3

× 3 + 2 33 + 2 3

= 9 + 6 3 + 30 3 + 609 – 12

= 69 + 36 3–3

= –23 – 12 3

5

a 6 23 2 – 2 3

= 6 23 2 – 2 3

× 3 2 + 2 33 2 + 2 3

= 36 + 12 618 – 12

= 36 + 12 66

= 6 + 2 6


b a + b – a – ba + b + a – b

= a + b – a – ba + b + a – b

× a + b – a – ba + b – a – b

= a + b – 2 (a + b)(a – b) + a – b(a + b) – (a – b)

= 2a – 2 a2 – b2

2b

= a – a2 – b2

b

6 First enter the information on a Venn

diagram.

15

20 105

G (55 ) BE (45 )

Bl (40 )

a It is obvious to make up the 40 blonds that 5 must be blond only, so the number of boys (not girls) who are blond is

5 +10 = 15. b The rest of the Venn diagram can be

filled in the same way:

15 15

5

15

20 105

G (55 ) BE (45 )

Bl (40 )

Boys not blond or blue-eyed = 100 – 15 – 15 – 15 – 20 – 5 – 10 – 5 = 15

7 Fill in a Venn diagram.

2 5

2

14

42

E M

F

a 30 – 2 – 14 – 5 – 4 – 2 – 2 = 1 (since all received at least one prize.)

b 14 + 5 + 2 + 1 = 22 c 2 + 14 + 4 + 2 = 22 8 Enter the given information on a Venn

diagram as below.

3

9 62

X ( 23 ) Y ( 25 )

Z ( 19 )

The numbers liking X only, Y only and Z only are 9, 14 and 2 respectively.

The number who like none of them = 50 – 9 – 3 – 14 – 9 – 2 – 6 – 2 = 5


9 The rectangles can be represented by circles for clarity. Enter the data:

p

q

a

b cx

A ( 20 ) B (10 )

C ( 16 )

Note: a + x = 3, b + x = 6 and c + x = 4 p + b + a + x = 20

p + b + 3 = 20p + b = 17

q + (p + b) + n(B) = 35 q + 17 + 10 = 35

∴ q = 8q + (b + x) + c = n(C) = 16

8 + 6 + c = 16∴ c = 2

c + x = 4∴ x = 2

There is 2 cm2 in common. 10 112 – 63 – 224

28

= 16 × 7 – 9 × 7 – 2244 × 7

= 4 7 – 3 7 – 2242 7

× 77

= 4 7 – 3 7 – 224 714

= 4 7 – 3 7 – 16 7 = –15 7

11 Cross multiply: ( 7 – 3 )( 7 + 3 ) = x2

7 – 3 = x2

4 = x2

x = ± 2

12 1 + 25 + 3

+ 1 – 25 – 3

= 1 + 25 + 3

× 5 – 35 – 3

+ 1 – 25 – 3

× 5 + 35 + 3

= 5 – 5 + 10 – 65 – 3

+ 5 + 5 – 10 – 65 – 3

= 2 5 – 2 62

= 5 – 6

13 27 – 12 + 2 75 – 4825

= 9 × 3 – 4 × 3 + 2 25 × 3 – 16 × 325

= 3 3 – 2 3 + 10 3 – 4 35

= 15 3 – 10 3 + 50 3 – 4 35

= 51 35

14 a A ∪ B = 32 + 7 + 15 + 3 = 57 b C = 3 c B' ∩ A = 32 15 17 + 6 8 = 17 + 2 × 9 × 8

= 17 + 2 72

a + b = 17; ab = 72 a = 8, b = 9 (or a = 9, b = 8, giving the

same answer.) ( 8 + 9 )2 = 17 + 6 8 So the square root of

17 + 6 8 = 8 + 9= 2 2 + 3


Chapter 4 – Variation Exercise 4A Solutions 1 a

y = kx2

8 = k × 22

8 = 4k∴ k = 2

x = 6: y = 2 × 62

= 72y = 128: 2x2 = 128

x2 = 64x = 8 (take the positive root)

x 2 4 6 8 y 8 32 72 128

b y = kx

16

= k × 12

1 = 3k∴ k = 1

3

x = 1: y = 13× 1

= 13

y = 23

: 13 x = 2

3x = 2

x 12 1

32

2

y 16

13

12

23

c y = k x

6 = k × 46 = 2k

∴ k = 3

x = 49: y = 3 × 49 = 21

y = 90: 3 x = 90x = 30x = 900

x 4 9 49 900 y 6 9 21 90

d y = kx15

15

= k ×⎝⎜⎛ 1

32⎠⎟⎞

15

15

= k × 12

∴ k = 25

x = 32: y = 25× 32

15

= 45

y = 85 : 2

5× x

15 = 8

5

x15 = 4

⎝⎜⎛

x15⎠⎟⎞5

= 45

x = 1024

x 132

1 32 1024

y 15

25

45

85

2 V = kr3

125 = k × 2.53

125 = 15.625k∴ k = 125

15.625 = 8

a V = 8 × 3.23

= 262.144 b 200 = 8r3

r3 = 25r = 3 25 ≈ 2.924


3 a = b23

23

= k × 123

∴ k = 23

a a = 23× 2

23

≈ 1.058

b 2 = 23× b

23

b23 = 2 × 3

2= 3

⎝⎜⎛

b23⎠⎟⎞

32

= 332

b ≈ 5.196

4 A = kh

60 = k × 10∴ k = 6

a A = 6 × 12= 72 cm2

b 120 = 6h

h = 20 cm 5 E = kw

3.2 = k × 452∴ k = 3.2

452 = 8

1130

a E = 81130

× 810

= 648113

cm

b 10 = 8

1130× w

w = 10 × 11308

= 28252

= 1412.5 g

6 W = kL2

18 = k × 202

18 = 400k∴ k = 18

400 = 9

200L = 225 = 15

W = 9200

× 152

= 10.125 kg

7 V = kr3

4188.8 = k × 103

∴ k = 4.18881 m3 = 1 000 000 cm3

1 000 000 = 4.1888r3

r3 = 1 000 0004.1888

≈ 238 731.85r ≈ 62.035 cm

8 T = k L

1.55 = k × 60∴ k = 1.55

60T = 1.55

60× 90

= 1.55 × 1.5 ≈ 1.898 seconds

9 a d = k h

4.8 = k × 1.8∴ k = 4.8

1.8

Person’s height above ground = 4 + 1.8 = 5.8 m d = 4.8

1.8× 5.8

≈ 8.616 km

b Height difference between person and

yacht = 5.8 + 10 = 15.8 m d = 4.8

1.8× 15.8

≈ 14.221 km


10 In each case set the initial value of x to 1. Initial value of y = k (when x = 1). a i y = k × 22

= 4k (300% increase)

ii y = k × 2 ≈ 1.41k (41% increase)

iii y = k × 23

= 8k (700% increase) b i y = k × 0.52

= 0.25k (75% decrease)

ii y = k × 0.5 ≈ 0.71k (29% decrease)

iii y = k × 0.53

= 0.125k (87.5% decrease) c i y = k × 0.82

= 0.64k (36% decrease)

ii y = k × 0.8 ≈ 0.89k (11% decrease)

iii y = k × 0.83

= 0.512k (48.8% decrease) d i y = k × 1.42

= 1.96k (96% increase)

ii y = k × 1.4≈ 1.18k (18% increase)

iii y = k × 1.43

= 2.744k (174.4% increase)


Exercise 4B Solutions 1 a y = 2

x2 = k

1∴ k = 2

x = 6: y = 26

= 13

y = 116

: 2x

= 116

x = 2 × 16 = 32

x 2 4 6 8 16

y 1 12

13

116

132

b y = k

x12

= k1

∴ k = 12

y = 12 x

y = 14 : 1

2 x= 1

42 x = 4

x = 4x = 9: y = 1

2 9 = 1

6

x 14

1 4 9

y 1 12

14

16

c y = kx2

3 = k12

∴ k = 3

x = 3: y = 332 = 1

3

y = 112

: 3x2 = 1

12

x2 = 36x = 6 (take the positive root)

x 1 2 3 6

y 334

13

112

d y =k

x13

13

=k

113

∴ k = 13

y =1

3x13

y = 19

: 1

3x13

= 19

x13 = 3x = 33 = 27

x =125: y =1

3 × 12513

= 115

x 18

1 27 125

y23

13

19

115


2 a = kb3

4 = k( 2 )3

∴ k = 4 × ( 2 )2 × 2 = 8 2

a a = 8 2b3

= 8 2(2 2 )3

= 8 28 × 8

= 14

= 12

b a = 8 2b3

116

= 8 2b3

b3 = 8 2 × 16 = 128 2 ≈ 181.01

b ≈ 5.657

3 a = k

b4

5 = k24

∴ k = 5 × 24 = 80

a a = 8044

= 80256

= 0.3125

b 20 = 80

b4

b4 = 8020

= 4 = 22

b = (22)14

= 212 = 2

4 V = kP

22.5 = k1.9

∴ k = 1.9 × 22.5 = 42.75

15 = 42.75P

P = 42.7515

= 2.85 kg/ cm2

5 I = k

R3 = k

80∴ k = 3 × 80

= 240

a I = 240100

= 2.4 amperes

b 80% of 3 = 2.4 2.4 = 240

RR = 240

2.4 = 100 ohms

6 I = k

d2

100 = k202

∴ k = 100 × 400 = 40 000I = 40 000

252

= 64 candela

7 r = k

h5.64 = k

10∴ k = 5.64 10

r = 5.64 1012

= 5.15 cm


8 In each case set the initial value of x to 1. Initial value of y = k (when x = 1). a i y = k

4 = 0.25k (75% decrease)

ii y = k2

≈ 0.71k (29% decrease)

iii y = k23

= 0.125k (87.5% decrease)

b i y = k

0.52


ii y = k0.5

≈ 1.41k (41% increase)

iii y = k0.53


c i y = k

0.82

= 1.5625k (56.25% increase)

ii y = k0.8


iii y = k0.83


d i y = k

1.42


ii y = k1.4


iii y = k1.43



Exercise 4C Solutions 1 a From the table, y = 2

3 x,

which is a direct relationship.

b From the table, y = 4x2, which is a direct square relationship. c From the table, y = 1

x/5 ,

which is an inverse relationship. d From the table, y = 2 x , which is a direct square root relationship. e From the table, y = 4

x2 ,

which is an inverse square relationship.

2 If direct variation exists, then the graph of y vs xn will be a straight line through the origin.

Graphs b and e fit this criteria. Graph f is a straight line but does not pass through the origin.

3 If inverse variation exists, then the graph

of y vs 1xn will be a straight line that is

undefined at the origin. Graphs a, b and e fit this criteria. Graph a

is a curve when showing y vs x, but will straighten out when showing y vs 1

xn .

4 a Gradient = 3

1= 3

y = 3x

b Gradient = 6

2= 3

y = 3x

c Gradient = 10

3y = 10

3 x

d Gradient = 2

1= 2

y = 2x

e Gradient = 9

3= 3

y = 3x

f Gradient = 6

1= 6

y = 6x

5

The graph is a straight line through the origin.

Gradient = 21.69

= 2.4

y = 2.4x2

Note: Any point can be used to calculate the gradient.

6

The graph is a straight line through the origin.

Gradient = 7.55

= 1.5

y = 1.5 x



7 9 a Use your calculator's power regression

function to determine a and b.

a = 100 b = 0.2 b C = atb

= 100 × 100.2

≈ 158.49 The graph is a straight line through the

origin. 10

Gradient = 21

= 2

y = 2x2

a Use your calculator's power regression function to determine a and b.

a = 1500 b = –0.5


b I = atb

= 1500 × 10–0.5

≈ 474.34

8 Methods will vary depending on the

calculator used.

a y = 14 x

b y = 2x54

c y = 3.5x0.4

d y = 10x23

e y = 2x– 5

2 f y = 3.2x–0.4


Exercise 4D Solutions 1 y = kx

z1 = 2k

10∴ k = 5

When y = 12

and z = 60:

12

= 5x60

x = 6When x = 10 and y = 4:

4 = 5 × 10z

z = 504

= 12.5

x 2 4 6 10 z 10 2 60 12.5 y 1 10 0.5 4

2 y = kxz

10 = k × 2 × 10∴ k = 1

2

When y = 25 and z = 50:25 = 50x

2x = 1

When x = 10 and y = 15:15 = 10z

2z = 3

x 2 4 1 10 z 10 8 50 3 y 10 16 25 15

3 y = kz

x2

152

= 10k22

∴ k = 15 × 42 × 10

= 3

When y = 6 and z = 50:

6 = 3 × 50x2

x2 = 1506

= 25

x = 5When x = 10 and y = 4:

4 = 3z102

3z = 400z = 400

3

x 2 3 5 10

z 10 4 50400

3

y152

43

6 4

4 a = kb2

c

0.54 = k × 1.22

2

k = 0.54 × 21.22

= 0.75

a = 0.75 × 2.62

3.5 ≈ 1.449

5 z = k xy3

146 = k × 51.53

k = 1.46 × 1.53

5 = 4.9275

5

z = 4.9275 xy3 5

= 4.9275 4.82.33 5

≈ 0.397


b W = k × (2a)2

3L

= 4ka2

3L = 4C

3

6 I = krt

130 = k × 6.5 × 2k = 130

13= 10

I = 10 × 5.8 × 3 = $174

The weight has increased by a factor of 7 E = kmv2

281.25 = k × 25 × 152

k = 2812.5 × 225

= 0.5E = 0.5 × 1.8 × 202

= 360 joules

43≈ 1.33or approximately 33%.

10 In both cases, set the initial values of p

and q to 1. Initial value of y = k (when p = 1, q = 1).

a y = k × 22

2= 2.83k (183% increase)

8 In both cases, set the initial length and

diameter to 1. Initial value of y = k (when l = 1, d = 1).

b y = k × 22

0.5= 5.66k (466% increase)

a y = k × 1.5

0.52


11 b y = k × 0.5

1.52


a T = kxl

For the first spring, T = k × 13

= k3

For the second spring, T = k × 0.9

2.7= k

3 9

a W = kd2

L The tensions will both be the same.

Let the diameter be a and the length b for a supported weight of C. b T = kx2

l

C = ka2

L

For the first spring, T = k × 12

3= k

3

Let the new diameter be x. For the second spring,

T = k × 0.92

2.7= 3k

10

If the length doubles and the weight remains the same, then

C = kd2

2Lkx2

2L= ka2

L

∴ x2 = ka2

L× 2l

k = 2a2

x = 2a

The ratio of tension = T (second spring)T (first spring)

=0.3k

k3

= 3k10

× 3k

= 910

= 0.9

The diameter has increased by a factor of 2 ≈ 1.41or approximately 41%. This is a 10% decrease; the tension in

the second spring is 90% that in the first.


Exercise 4E Solutions 1 C = b + kd

42.4 = b + 22k47.8 = b + 25k

– :5.4 = 3k

∴ k = 1.842.4 = b + 22 × 1.842.4 = b + 39.6

b = 2.8C = 2.8 + 1.8 × 17 = $33.40

2 a C = b + kd

2625 = b + 50k3575 = b + 70k

– :950 = 20k

∴ k = 47.52625 = b + 50 × 47.52625 = b + 2375

b = 250The fixed charge is $250 and the cost per guest is $47.50

b C = b + kd

= 250 + 4.75 × 100 = $5000

3 p = k1 x + k2 y

2

14 = 3k1 + 16k2

14.5 = 5k1 + 9k2

Multiply ① by 5 and ② by 3: 70 = 15k1 + 80k2

43.5 = 15k1 + 27k2

– :26.5 = 53k2

k2 = 0.514 = 3k1 + 16 × 0.514 = 3k1 + 83k1 = 6

k1 = 2p = 2 × 4 + 0.5 × 25 = 20.5

4 C = k1 n + k2

n32 = 200k1 + k2

20061 = 400k1 + k2

400

Multiply ① by 0.5:

16 = 100k1 + k2

40061 = 400k1 + k2

400 – :

45 = 300k1

k1 = 45300

= 0.15

32 = 200 × 0.15 + k2

20032 = 30 + k2

200k2

200= 2

k2 = 400C = 0.15 × 360 + 400

360 ≈ $55.11

5 a s = k1 t + k2 t

2

142.5 = 3k1 + 9k2

262.5 = 5k1 + 25k2

Multiply ① by 5 and ② by 3: 712.5 = 15k1 + 45k2

787.5 = 15k1 + 75k2

– :75 = 30k2

k2 = 7530

= 2.5

142.5 = 3k1 + 9 × 2.53k1 = 120

k1 = 40s = 40 × 6 + 2.5 × 36 = 330 m

b The sixth second is the time from t = 5 to t = 6. Distance travelled = 330 – 262.5 = 67.5 m


6 t = k1 b + k2

m45 = 10k1 + k2

145 = 10k1 + k2

30 = 8k1 + k2

2

Multiply ② by 2:

45 = 10k1 + k2

60 = 16k1 + k2

– :15 = 6k1

k1 = 156

= 2.5

45 = 10 × 2.5 + k2

k2 = 20t = 2.5 × 16 + 20

4 = 45 minutes


Solutions to Multiple-choice Questions 1 y = kx2

3 = k × 9

k = 13

C

2 y = k

x14

= k2

k = 24

= 12

A

3 a = kb3

32 = k × 8k = 4a = 4 × 64

= 256 B 4 p = k

q2

13

= k9

k = 93

= 3

1 = 3q2

q2 = 3

q = 3 C 5 Gradient = 6

2= 3

y = 3 B x2

6 Gradient = 41

= 4

y = 4 x D 7 y = kx

z2

13

= k × 222 = k

2

k = 23

E

8 a = kp2

q

8 = k × 45

k = 404

= 10

a = 10 × 96

= 15 D 9 Set the initial value of q to 1. Initial value of p = k p = k × 1.12

= 1.21k

21% increase D 10 Set the initial value of q to 1. Initial value of p = k p = k

0.8= 1.25k

25% increase B


Solutions to Short-answer Questions 1 a a = kb2

32

= k × 22

∴ k = 38

b = 4: a = 38× 42

= 6a = 8: 3

8× b2 = 8

b2 = 643

b = ± 83

b y = kx13

10 = k × 213

∴ k =10

213

x = 27: y =10

213× 27

13

=30

213

y = 18 :

10

213× x

13 = 1

8

x13 = 2

13

80x = 2

803

= 1256 000

c y = k

x2

13

= k22

∴ k = 43

x = 12

: y = 43x2

= 43 × 0.52

= 163

y = 427

: 43x2 = 4

27

3x2 = 27x2 = 9x = ± 3

d a = kb

c14

= k × 14

∴ k = 24

= 12

a = b2 c

=

49

2 169

= 49× 1

2× 3

4 = 1

6

2 a d = kt2

78.56 = k × 42

k = 78.5616

= 4.91d = 4.91t2

b d = 4.91 × 102

= 491 m c 78.56 = 19.64 × t2

t2 = 78.5619.64

= 4

t = 2 seconds


3 a v = k s

7 = k 2.5∴k = 7

2.5

v = 7 s2.5

= 7 s2.5

= 7 102.5

= 14 m/s

b 28 = 7 s2.5

s2.5

= 4

s2.5

= 16

s = 40

4 t = k

v4 = k

30∴ k = 4 × 30 = 120

t = 12050

= 2.4 hours

5 y ∝ 1

x

a y ∝ 12x

∴ y is halved.

b 2y ∝ 1

xx ∝ 1

2y∴ x is halved.

c y ∝1x2

2y ∝ 1x

∴ y is doubled.

d y2∝ 1

xx ∝ 2

y∴ x is doubled.

6 C = ktRI2

9 = k × 2.5 × 60 × 16∴ k = 0.00375

C = 0.00375 × 1.5 × 80 × 9 = 4.05 cents

7 C = a + kn

20 = a + 100k30 = a + 500k

– :10 = 400k

∴ k = 10400

= 140

20 = a + 10040

20 = a + 2.5a = 17.5C = 17.5 + 700

40 = $35

8 v = kI

24 = k × 6∴ k = 4

72 = 4II = 18 amps

9 I = kd2

Let the initial distance be d1. The final distance will be d2.

I1 = k(d1)

2

I2 = k(2d1)

2

= k4(d1)

2

= 14

I1

10 Set the initial values of x and z to 1. Initial value of y = k

y = k × 1.12

0.9 ≈ 1.34 (34% increase)


Chapter 5 – Sequences and series Exercise 5A Solutions 1 a t1 = 3

t2 = 3 + 4 = 7t3 = 7 + 4 = 11t4 = 11 + 4 = 15t5 = 15 + 4 = 19

b t1 = 5

t2 = 3 × 5 + 4 = 19t3 = 3 × 19 + 4 = 61t4 = 3 × 61 + 4 = 187t5 = 3 × 187 + 4 = 565

c t1 = 1

t2 = 5 × 1 = 5t3 = 5 × 5 = 25t4 = 5 × 25 = 125t5 = 5 × 125 = 625

d t1 = –1

t2 = –1 + 2 = 1t3 = 1 + 2 = 3t4 = 3 + 2 = 5t5 = 5 + 2 = 7

e t1 = 1

t2 = 3t3 = 2 × 3 + 1 = 7t4 = 2 × 7 + 3 = 17t5 = 2 × 17 + 7 = 41

2 a tn = 1

nt1 = 1

1= 1

t2 = 12

t3 = 13

t4 = 14

b tn = n2 + 1t1 = 12 + 1 = 2t2 = 22 + 1 = 5t3 = 32 + 1 = 10t4 = 42 + 1 = 17

c tn = 2n

t1 = 2 × 1 = 2t2 = 2 × 2 = 4t3 = 2 × 3 = 6t4 = 2 × 4 = 8

d tn = 2n

t1 = 21 = 2t2 = 22 = 4t3 = 23 = 8t4 = 24 = 16

e tn = 3n + 2

t1 = 3 × 1 + 2 = 5t2 = 3 × 2 + 2 = 8t3 = 3 × 3 + 2 = 11t4 = 3 × 4 + 2 = 14

f tn = (–1)n n3

t1 = (–1)1 × 13 = –1t2 = (–1)2 × 23 = 8t3 = (–1)3 × 33 = –27t4 = (–1)4 × 43 = 64

g tn = 2n + 1

t1 = 2 × 1 + 1 = 3t2 = 2 × 2 + 1 = 5t3 = 2 × 3 + 1 = 7t4 = 2 × 4 + 1 = 9

h tn = 2 × 3n – 1

t1 = 2 × 30 = 2t2 = 2 × 31 = 6t3 = 2 × 32 = 18t4 = 2 × 33 = 54


3 a i t1 = 3 = 3 × 1

t2 = 6 = 3 × 2t3 = 9 = 3 × 3t4 = 12 = 3 × 4

∴ tn = 3n ii t 2 = t1 + 3

t3 = t2 + 3∴ tn = tn – 1 + 3, t1 = 3

b i t1 = 1 = 20

t2 = 2 = 21

t3 = 4 = 22

t4 = 8 = 23

∴ tn = 2n – 1

ii t 2 = 2t1

t3 = 2t2

∴ tn = 2tn – 1, t1 = 1 c i t1 = 1 = 1

12

t2 = 14

= 122

t3 = 19

= 132

t4 = 116

= 142

∴ tn = 1n2

ii t2 = 14× t1

t3 = 49× t2

t4 = 916

× t3

∴ tn = (n – 1)2

n2 × tn – 1, t1 = 1

d i t1 = 3 = 3 × (–2)0

t2 = –6 = 3 × (–2)1

t3 = 12 = 3 × (–2)2

t4 = –24 = 3 × (–2)3

∴ tn = 3 × (–2)n – 1

ii t 2 = –2 × t1

t3 = –2 × t2

∴ tn = –2tn – 1, t1 = 3

e i t1 = 4 = 3 × 1 + 1t2 = 7 = 3 × 2 + 1t3 = 10 = 3 × 3 + 1t4 = 13 = 3 × 4 + 1

∴ tn = 3n + 1 ii t 2 = t1 + 3

t3 = t2 + 3∴ tn = tn – 1 + 3, t1 = 4

f i t1 = 4 = 5 × 1 – 1

t2 = 9 = 5 × 2 – 1t3 = 14 = 5 × 3 – 1t4 = 19 = 5 × 4 – 1

∴ tn = 5n – 1 ii t 2 = t1 + 5

t3 = t2 + 5∴ tn = tn – 1 + 5, t1 = 4

4 tn = 3n + 1

tn + 1 = 3(n + 1) + 1 = 3n + 4

t2n = 3(2n) + 1 = 6n + 1

5 a tn = tn – 1 + 3, t1 = 15 b t1 = 15

t2 = 15 + 3t3 = (15 + 3) + 3 = 15 + 2 × 3

∴ tn = 15 + (n – 1) × 3 = 3n + 12

c t13 = 3 × 13 + 12

= 51

6 a 4% reduction is equivalent to 96% of the

original. tn = 0.96tn – 1 b t1 = 94.3

t2 = 0.96 × 94.3t3 = 0.96 × (0.96 × 94.3) = 0.962 × 94.3

∴ tn = 94.3 × 0.96n – 1

c t 9 = 94.3 × 0.968

≈ 68.03 seconds


b un = 12

(n2 – n) + 1

u1 = 12

(12 – 1) + 1 = 1

u2 = 12

(22 – 2) + 1 = 2

u3 = 12

(32 – 3) + 1 = 4

7 a tn = 1.8tn – 1 + 20 b t1 = 1.8 × 100 + 20 = 200

t2 = 1.8 × 200 + 20 = 380t3 = 1.8 × 380 + 20 = 704t4 = 1.8 × 704 + 20 = 1287t5 = 1.8 × 1287 + 20 = 2336

c The sequences are the same for the first three terms. 8

a t1 = 2000 × 1.06 = $2120

t2 = (2120 + 400) × 1.06 = $2671.20

t3 = (2671.2 + 400) × 1.06 = $3255.47

t1 = u1

t2 = u2

t3 = u3

d t4 = 23 = 8

u4 = 12 (42 – 4) + 1 = 7

b tn = (tn – 1 + 400) × 1.06

= 1.06(tn – 1 + 400), t1 = 2120 The sequences diverge after the first

three terms.

11 S1 = a × 12 + b × 1 = a + bS2 = a × 22 + b × 2 = 4a + 2bS3 = a × 32 + b × 3 = 9a + 3b

Sn + 1 – Sn = a(n + 1)2 + b(n + 1) – an2 – bn = a(n2 + 2n + 1) + bn + b – an2 – bn = an2 + 2an + a + b – an2

= 2an + a + b

c Method will depend on the calculator or spreadsheet used.

t10 = $8454.02 9 Method will depend on the calculator used. a 1, 4, 7, 10, 13, 16 b 3, 1, –1, –3, –5, –7 c 1

2, 1, 2, 4, 8, 16 12 t2 = 1

2 ⎝⎜⎛1 + 2

1⎠⎟⎞ = 3

2

t3 = 12⎝⎜⎛3

2+ 2

3/ 2⎠⎟⎞ = 17

12

t4 = 12⎝⎜⎛17

12+ 2

17/ 12⎠⎟⎞ = 577

408

d 32, 16, 8, 4, 2, 1 e 1.1, 1.21, 1.4641, 2.144, 4.595, 21.114 Comparing the terms to real numbers

between 1 and 1.5, it can be seen that the sequence gives an approximation of 2 .

f 27, 18, 12, 8, 163

, 329

g –1, 3, 11, 27, 59, 123 13 t3 = t2 + t1

= 1 + 1 = 2t4 = t3 + t2

= 2 + 1 = 3t5 = t4 + t3

= 3 + 2 = 5tn + 2 = tn + 1 + tn

∴ tn + 1 = tn + tn – 1

∴ tn + 2 = (tn + tn – 1) + tn

= 2tn + tn – 1

h –3, 7, –3, 7, –3, 7 10 a tn = 2n – 1

t1 = 20 = 1t2 = 21 = 2t3 = 22 = 4


Exercise 5B Solutions 1 tn = a + (n – 1)da t1 = 0 + (1 – 1) × 2 = 0

t2 = 0 + (2 – 1) × 2 = 2t3 = 0 + (3 – 1) × 2 = 4t4 = 0 + (4 – 1) × 2 = 6

b t1 = –3 + (1 – 1) × 5 = –3

t2 = –3 + (2 – 1) × 5 = 2t3 = –3 + (3 – 1) × 5 = 7t4 = –3 + (4 – 1) × 5 = 12

c t1 = – 5 + (1 – 1) × – 5 = – 5

t2 = – 5 + (2 – 1) × – 5 = –2 5t3 = – 5 + (3 – 1) × – 5 = –3 5t4 = – 5 + (4 – 1) × – 5 = –4 5

d t1 = 11 + (1 – 1) × –2 = 11

t2 = 11 + (2 – 1) × –2 = 9t3 = 11 + (3 – 1) × –2 = 7t4 = 11 + (4 – 1) × –2 = 5

2 a a + (1 – 1)d = 3

a = 33 + (2 – 1)d = 7

d = 7 – 3 = 4 ∴ tn = 3 + 4(n – 1)

= 4n – 1 b a + (1 – 1)d = 3

a = 33 + (2 – 1)d = –1

d = –1 – 3 = –4 ∴ tn = 3 + –4(n – 1)

= 7 – 4n

c a + (1 – 1)d = – 12

a = – 12

– 12

+ (2 – 1)d = 32

d = 32

– – 12

= 2

tn = – 12

+ 2(n – 1)

= 2n – 52

d a + (1 – 1)d = 5 – 5a = 5 – 5

(5 – 5 ) + (2 – 1)d = 5d = 5 – (5 – 5 ) = 5

tn = (5 – 5 ) + 5 (n – 1) = 5 n + 5 – 2 5

3 a t13 = a + 12d

= 5 + 12 × –3 = –31 b t10 = a + 9d

= –12 + 9 × 4 = 24 c t9 = a + 8d

= 25 + 8 × –2.5 = 5 d t5 = a + 4d

= 2 3 + 4 × 3 = 6 3

4 a = 4, d = 2 a t6 = a + 5d

= 4 + 5 × 2 = 14 b t10 = a + 9d

= 4 + 9 × 4 = 22 c tn = 4 + 2(n – 1) = 10

2(n – 1) = 6n – 1 = 3

n = 4 The fourth day. 5 a P is the 16th row. a = 25, d = 3 t16 = a + 15d

= 25 + 15 × 3 = 70 seats

b X is the 24th row. a = 25, d = 3 t24 = a + 23d

= 25 + 23 × 3 = 94 seats


c tn = 25 + 3(n – 1) = 403(n – 1) = 15

n – 1 = 5n = 6

Row F 6 a a = 6, d = 4 t5 = 6 + 4 × 4 = 22 b a = 5, d = –3 t12 = 5 + 11 × –3 = –28 c a = 16, d = –3 tn = 16 – 3(n – 1) = –41

–3(n – 1) = –57n – 1 = 19

n = 20

d a = 7, d = 4 tn = 7 + 4(n – 1) = 227

4(n – 1) = 220n – 1 = 55

n = 56 7 t30 = 7 + 29d = 108 1

229d = 101 1

2d = 3 1

2

8 t6 = 3 + 5d = 98

5d = 95d = 19t7 = t6 + 19 = 117

9 a + 2d = 18

a + 5d = 486 – :

3d = 468d = 156

a + 2 × 156 = 18a + 312 = 18

a = –294∴ tn = –294 + 156(n – 1)

= 156n – 450

10 a + 4d = 24a + 9d = 39 – :

5d = 15d = 3

a + 4 × 3 = 24a + 12 = 24

a = 12∴ t15 = 12 + 14 × 3

= 54 11 a + 6d = 0.6

a + 11d = –0.4 – :

5d = –1.0d = –0.2

a + 6 × –0.2 = 0.6a – 1.2 = 0.6

a = 1.8∴ t20 = 1.8 + 19 × –0.2

= –2 12 4 + 9d = 30

9d = 26d = 29

9

t2 = 4 + 1 × 269

= 629

t3 = 4 + 2 × 269

= 889

t4 = 4 + 3 × 269

= 383

t5 = 4 + 4 × 269

= 1409

t6 = 4 + 5 × 269

= 1669

t7 = 4 + 6 × 269

= 643

t8 = 4 + 7 × 269

= 2189

t9 = 4 + 8 × 269

= 2449

13 5 + 5d = 15

5d = 10d = 2

t2 = 5 + 1 × 2 = 7t3 = 5 + 2 × 2 = 9t4 = 5 + 3 × 2 = 11t5 = 5 + 4 × 2 = 13


17 3x – 2 = 5x + 1 + 112

6x – 4 = 5x + 12x = 16

14 a + (m – 1)d = 0(m – 1)d = – a

d = – am – 1

tn = a – a(n – 1)m – 1

18 Use the fact that the difference is

constant. This could be simplified as follows: tn = a(m – 1) – a(n – 1)

m – 1 = a(m – 1 + n + 1)

m – 1 = a(m – n)

m – 1

(8a – 13) – (4a – 4) = (4a – 4) – a8a – 13 – 4a + 4 = 4a – 4 – a

4a – 9 = 3a – 4a = 5

19 tx = a + (x – a)d = y

ty = a + (y – a)d = x

15 a + 14d = 3 + 9 3

a + 19d = 38 – 3 – :

5d = 35 – 10 3d = 7 – 2 3

Subtract: (x – y)d = y – x

= –1(x – y)d = –1(x – y)

x – y = –1

a + 14 × (7 – 2 3 ) = 3 + 9 3a + 98 – 28 3 = 3 + 9 3

a = 37 3 – 95

Substitute:

a + (x – a) × –1 = ya = y + x – 1

tx + y = a + (x + y – 1)d = y + x – 1 + (x + y – 1) × –1 = y + x – 1 – x – y + 1 = 0

t6 = 37 3 – 95 + 5 × (7 – 2 3 ) = 37 3 – 95 + 35 – 10 3 = 27 3 – 60

20 Use the fact that the difference is

constant. 16 a c = a + b

2 = 8 + 15

2= 11.5

a2 – 2a = 2a – a

a2 – 3a = 0a(a – 3) = 0

a = 3 (since a ≠ 0) b c = a + b

2

= 12

⎝⎜⎛ 1

2 2 – 1+ 1

2 2 + 1⎠⎟⎞

= 2 2 + 1 + 2 2 – 12(2 2 – 1)(2 2 + 1)

= 4 22 × (8 – 1)

= 2 27


Exercise 5C Solutions 1 a a = 8, d = 5, n = 12 t12 = 8 + 11 × 5 = 63

S12 = 122

(8 + 63)

= 6 × 71 = 426

b a = –3.5, d = 2, n = 10 t10 = –3.5 + 9 × 2 = 14.5

S10 = 102

(–3.5 + 14.5)

= 5 × 11 = 55

c a = 12

, d = 12

, n = 15

t15 = 12

+ 14 × 12

= 152

S15 = 152

⎝⎜⎛ 1

2+ 15

2 ⎠⎟⎞

= 60 2

d a = –4, d = 5, n = 8 t8 = –4 + 7 × 5 = 31

S8 = 82 (–4 + 31)

= 108

2 a = 7, d = 3, n = 7 S7 = 7

2 (14 + 6 × 3)

= 112

3 Sn = n2

(30 + (n – 1) × –1)

= 110

n(30 – n + 1) = 220– n2 + 31n – 220 = 0

n2 – 31n + 220 = 0(n – 11)(n – 20) = 0

n = 11 or n = 20Reject any value of n > 15, as this would involve a negative number of logs in a row. There will be 11 layers.

4 a = 5, d = 5, n = 16 S16 = 16

2 (10 + 15 × 5)

= 680

5 There will be half of 98 = 49 numbers: a = 2, d = 2, n = 49 S49 = 49

2 (4 + 48 × 2)

= 2450

6 a a = 6

t15 = 6 + 14d = 2714d = 21

d = 1.5t8 = 6 + 7 × 1.5 = 16.5 km

b S5 = 5

2 (12 + 4 × 1.5)

= 45 km

c Total distance: S15 = 15

2 (12 + 14 × 1.5)

= 247.5

Distance missed = 18 + 19.5 + 21 = 58.5 km (8th day = 16.5 km) Distance Dora walks = 247.5 – 58.5 = 189 km 7 a a = 30, d = 5

Sn = n2

(60 + (n – 1) × 5)

= 500n(60 + 5n – 10) = 1000

5n2 + 50n – 1000 = 0n2 + 10n – 200 = 0

(n – 10)(n + 20) = 0n = 10, as n > 0

10 days


b a = 50, n = 5 S5 = 5

2 (100 + 4d)

= 500100 + 4d = 200

d = 200 – 1004

= 25 pages per day

8 Sn = n

2 (8 + (n – 1) × 4)

= 180n(8 + 4n – 4) = 360

4n2 + 4n – 360 = 0n2 + n – 90 = 0

(n – 9)(n + 10) = 0n = 9 as n ≠ – 10

9 a = –5, d = 4 Sm = m

2 (–10 + (m – 1) × 4)

= 660m(–10 + 4m – 4) = 1320

4m2 – 14m – 1320 = 0(m – 20)(4m + 66) = 0

m = 20 as m > 0

10 a Row J = t10

= 50 + 9 × 4 = 86 b S26 = 26

2 (100 + 25 × 4)

= 2600

c 50 + 54 + 58 + 62 = 224 d 2600 – 224 = 2376 e Sn = n

2 (100 + (n – 1) × 4)

= 3410n(100 + 4n – 4) = 6820

4n2 + 96n – 6820 = 0n2 + 24n – 1705 = 0(n – 31)(n + 55) = 0

n = 31 as n ≠ – 55

There are 5 extra rows (from 26 to 31).

11 Total members S12 = 12

2 (80 + 11 × 15)

= 1470

Total fees = 1470 × $120 = $176 400 12 a + d = –12

6(2a + 11d) = 182a + 11d = 3

Substitute a = –12 – d:

–24 – 2d + 11d = 39d – 24 = 3

d = 3a + 3 = –12

a = –15t6 = –15 + 5 × 3 = 0

S6 = 62

(–30 + 5 × 3)

= –45

13 5(2a + 9d) = 120

2a + 9d = 2410(2a + 19d) = 840

2a + 19d = 84 – :

10d = 60d = 6

2a + 9 × 6 = 24a = –15

S30 = 302

(–30 + 29 × 6)

= 2160

14 a = 54, d = –6, n = 19 (9 + 9 + 1) S19 = 19

2 (108 + 18 × –6)

= 0

You may also observe that every positive term will be balanced by a negative term, so the sum will be zero.


15 a + 5d = 16a + 11d = 28

– :6d = 12

d = 2a + 10 = 16

a = 6S14 = 14

2 (12 + 13 × 2)

= 266

16 a a + 2d = 6.5

4(2a + 7d) = 67a + 3.5d = 67

8= 8.375

– :1.5d = 1.875

d = 1.25a + 1.25 × 2 = 6.5

a = 4tn = 4 + 1.25(n – 1) = 2.75 + 1.25n = 5

4 n + 11

4

b a + 3d = 6

5

= 65× 5

5

= 6 55

52

(2a + 4d) = 16 5

5(a + 2d) = 16 5

a + 2d = 16 55

– :

d = 6 55

– 16 55

= – 10 55

a + 2 × –10 55

= 16 55

a = 16 55

+ 20 55

= 36 55

tn = 36 55

– 10 55

(n – 1)

= 46 55

– 10 55

n

= 46 55

– 2 5 n

17 a tn + 1 – tn = b(n + 1) – bn

= b b Sn = n

2 (2b + (n – 1)b)

= n2 (2b + nb – b)

= n2 (nb + b)

This can be factorised to nb(n + 1)2

.

18 a = 10, d = –5 t5 = 10 + 4 × –5

= –10S25 = 25

2 (20 + 24 × –5)

= –1250

19 S20 = 10(2a + 19d)

= 25a20a + 190d = 25a

190d = 5aa = 38d

S30 = 15 (76d + 29d) = 1575d


21 Let the terms be a, a + d, a + 2d. 20 Sum = 3a + 3d = 36 a Sn – 1 = 17(n – 1) – 3(n – 1)2

= 17n – 17 – 3(n2 – 2n + 1) = 17n – 17 – 3n2 + 6n – 3 = 23n – 3n2 – 20

a + d = 12 Product = a(a + d)(a + 2d) = 1428 Substitute d = 12 – a. a(a + 12 – a)(a + 24 – 2a) = 1428

12a(24 – a) = 1428a(24 – a) = 11924a – a2 = 119

a2 – 24a + 119 = 0(a – 7)(a – 17) = 0

a = 7 or a = 17

b tn = Sn – Sn – 1

= 17n – 3n2 – 23n + 3n2 + 20 = 20 – 6n

c tn + 1 – tn = 20 – 6(n + 1) – (20 – 6n)

= 20 – 6n – 6 – 20 + 6n = –6 ∴ d = 12 – 7 = 5

or d = 12 – 17 = –5

The sequence has a constant difference of –6 and so is arithmetic. The three terms are either 7, 12, 17

a = t1

= 20 – 6 × 1 = 14d = –6

or 17, 12, 7. Note: in cases like this, it is sometimes

easier to call the terms a – d, a, a + d.

22 The middle terms will be t n and tn + 1.

tn = a + (n – 1)dtn + 1 = a + nd

tn + tn + 1 = 2a + (2n – 1)dn(tn + tn + 1) = n(2a + (2n – 1)d)

S2n = 2n2

(2a + (2n – 1)d)

= n(2a + (2n – 1)d) = n(tn + tn + 1)


Exercise 5D Solutions 1 tn = arn – 1 a t1 = 3 × 21 – 1 = 3

t2 = 3 × 22 – 1 = 6t3 = 3 × 23 – 1 = 12t4 = 3 × 24 – 1 = 24

b t1 = 3 × –21 – 1 = 3

t2 = 3 × –22 – 1 = –6t3 = 3 × –23 – 1 = 12t4 = 3 × –24 – 1 = –24

c t1 = 10 000 × 0.11 – 1 = 10 000

t2 = 10 000 × 0.12 – 1 = 1000t3 = 10 000 × 0.13 – 1 = 100t4 = 10 000 × 0.14 – 1 = 10

d t1 = 3 × 31 – 1 = 3

t2 = 3 × 32 – 1 = 9t3 = 3 × 33 – 1 = 27t4 = 3 × 34 – 1 = 81

2 a a = 15

7r = 1

3

t6 = 157×⎝⎜⎛1

3⎠⎟⎞5

= 5567

b a = 1

r = – 14

t5 = 1 ×⎝⎜⎛ – 1

4⎠⎟⎞4

= 1256

c a = 2

r = 2t10 = 2 × ( 2 )9 = 32

d a = ax

r = at 6 = ax × a5 = ax + 5

3 a a = 3

r = 23

tn = 3 ×⎝⎜⎛2

3⎠⎟⎞n – 1

b a = 2

r = –42

= –2

tn = 2 × –2n – 1

c a = 2

r = 2 52

= 5

tn = 2 × ( 5 )n – 1

4 a = 25 t5 = 25r4 = 16

25r4 = 16

625r = ± 2

5

5 1

4 2n – 1 = 64

2n – 1 = 64 × 4 = 28

n = 9

6 a a = 2, r = 3 2 × 3n – 1 = 486

3n – 1 = 243 = 35

n = 6

b a = 5, r = 2 5 × 2n – 1 = 1280

2n – 1 = 256 = 28

n = 9


c a = 768, r = 12

768 ×⎝⎜⎛1

2⎠⎟⎞n – 1

= 3

12n – 1 = 3

768

= 1256

= 128

n = 9

d a = 8

9 , r = 3

2

89× 3n – 1

2n – 1 = 274

3n – 1

2n – 1 = 274× 9

8

= 35

25

n = 6

e a = – 4

3 , r = – 1

2

– 43×⎝⎜⎛ – 1

2⎠⎟⎞n – 1

= 196

⎝⎜⎛ – 1

2⎠⎟⎞n – 1

= 196

× – 34

= – 132 × 4

= – 127 =

⎝⎜⎛ – 1

2⎠⎟⎞7

n = 8

7 a At the end of 10 years, it will have

increased 10 times. 2500 × 1.089 = $5397.31 b 2500 × 1.08n ≥ 100 000

1.08n ≥ 100 0002500

= 40

nlog10 1.08 ≥ log10 40n ≥ 47.93. . .

It will take 48 years until the value exceeds $100 000.

8 a At the end of 7 days, it will have

increased 7 times. 10 × 37 = 2187 m2

b 10 × 3n ≥ 200 000

3n ≥ 20 000nlog10 3 ≥ log10 20 000

n ≥ 9.014. . . It will cover the lake early in the tenth day. 9 2 × 0.755 ≈ 0.4746 m

≈ 47.46 cm 10 a 120 × 0.97 ≈ 57.4 km b 120 × 0.9n – 1 = 30.5

0.9n – 1 = 30.5120

= 0.251. . .(n – 1)log10 0.9 = log10 0.251. . .

n – 1 = 13.0007. . .n = 14

The 14th day. 11 0.01 × 229 = $5 369 000 12 a ar5 = 768

ar2 = 96r3 = 768

96= 8

r = 2a × 22 = 96

a = 24 fish

b 24 × 29 = 12 888 fish


13 ar14 = 54ar11 = 2

r3 = 542

= 27

r = 3a × 311 = 2

a = 2311

t7 = 2311 × 36

= 235

14 ar1 = 1

2 2ar3 = 2

r2 = 2 ÷ 12 2

= 4r = 2

a × 2 = 12 2

a = 14 2

t8 = 14 2

× 27

= 322

Rationalise the denominator:

t8 = 322× 2

2

= 32 22

= 16 2

15 a = 4, r = 2 4 × 2n – 1 > 2000

2n – 1 > 50029 = 512

The tenth term.

16 a = 3, r = 3

3 × 3n – 1 > 5003n > 50035 = 243 and 36 = 729

The sixth term. 17 a At the end of 10 years: value = 5000 × 1.066

= $7092.60 b 1.06n ≥ 2

nlog10 1.06 ≥ log10 2n ≥ 11.89. . .

In the 12th year. 18 A × 1.08512 = 8000

A × 2.6616. . . = 8000A = $3005.61

19 Let the rate be r. r10 = 3

r = 30.1 = 0.11612. . . Approximately 11.6%. 20 Let the number of weeks be n.

40 960 ×⎝⎜⎛1

2⎠⎟⎞n – 1

= 40 × 2n – 1

40 96040

= 2n – 1 × 2n – 1

1024 = 22n – 2 = 210

2n – 2 = 10n = 6

21 a 5 × 720 = 3600 = 60 b 1 × 6.25 = 6.25 = 2.5

c 13× 3 = 1 = 1

d x2y3 × x6y11 = x8y14

= x4y7


22 r = t7

t4= t16

t7

a + 6da + 3d

= a + 15da + 6d

(a + 6d)2 = (a + 15d)(a + 3d)a2 + 12ad + 36d2 = a2 + 18ad + 45d2

9d2 + 6ad = 03d(3d + 2a) = 0

3d + 2a = 0d = – 2

3 a

r = a + 6da + 3d

= a – 4aa – 2a

= –3a– a

= 3

Note: d = 0 gives the trivial case r = a

a= 1.

(All the terms are the same.)


Exercise 5E Solutions

1 Sn = a(rn – 1)r – 1

a a = 5r = 10

5= 2

S10 = 5(210 – 1)2 – 1

= 5115

b a = 1

r = –31

= –3

S6 = 1(–36 – 1)–3 – 1

= –182

c a = – 4

3r = 2

3÷ – 4

3= – 1

2

S9 =– 4

3⎝⎜⎛⎝⎜⎛ – 1

2⎠⎟⎞9

– 1⎠⎟⎞

– 12

– 1

= – 5764

2 a a = 2

r = –62

= –3

tn = 1458 = 2 × –3n – 1

–3n – 1 = 729n = 7

S7 = 2 × (–37 – 1)–3 – 1

= 1094

b a = –4

r = 8–4

= –2

tn = –1024 = –4 × –2n – 1

–2n – 1 = 256n = 9

S9 = –4 × (–29 – 1)–2 – 1

= –684

c a = 6250r = 1250

6250= 0.2

tn = 2 = 6250 × (0.2)n – 1

(0.2)n – 1 = 26250

= 13125

n = 6

S6 = 6250 × ((0.2)6 – 1)0.2 – 1

= 7812

3 a = 600, r = 1.1 a t 7 = 600 × 1.16

= 1062.9366 About 1062.9 mL

b S7 = 600 × (1.17 – 1)1.1 – 1

= 5692.3026

About 5692.3 mL 4 a = $15 000, r = 1.05 a t 5 = 15 000 × 1.054

= 18 232.593. . . $18 232.59

b S5 = 15 000 × (1.055 – 1)1.05 – 1

= 82 844.4686

$82 884.47 5 a = 20, r = 25

20 = 2.5

a t5 = 20 × 1.254

= 48.828125 49 minutes (to the nearest minute)

b S5 = 20 × (1.255 – 1)1.25 – 1

= 164.140625

164 minutes, or 2 hours and 44 minutes


c Sn > 15 × 60 = 90020 × (1.25n – 1)

0.25> 900

1.25n – 1 > 900 × 0.2520

= 11.25

1.25n > 12.25nlog10 1.25 > log10 12.25

n > 11.228

12 – 7 = 5, so Friday. 6 a = 15, r = 2

3

S10 =15 ×

⎝⎜⎛ 1 –

⎝⎜⎛2

3⎠⎟⎞10

⎠⎟⎞

1 – 23

= 3 × 15 × 310 – 210

310

= 5 × 310 – 210

38

= 5 × 58 0256561

= 290 1256561

The bounces will all be doubled (up and down) except for the first (down only).

Distance = 2 × 290 125

6561– 15

= 481 8356561

= 73 28826561

m

7 Andrew: Interest = 1000 × 0.20 × 10 = $2000 His investment is worth $1000 + $2000 = $3000. Bianca's investment is worth

1000 × 1.12510 = $3247.32 Bianca’s investment is worth more.

8 a ar2 = 20

ar5 = 160r3 = 160

20= 8

r = 2a × 22 = 20

a = 5

S5 = 5 × (25 – 1)2 – 1

= 155

b ar2 = 2

ar7 = 8r5 = 8

2

= 642

= 32 = ( 2 )5

r = 2a × ( 2 )2 = 2

a = 12

S8 =

12× (( 2 )8 – 1)

2 – 1

=

12× 15

2 – 1× 2 + 1

2 + 1

=

152× ( 2 + 1)

2 – 1

= 15 + 15 22


9 a = 1, r = 2 10 a = 1, r = – x2

a Sn = 2551 × (2n – 1)

2 – 1= 255

2n – 1 = 2552n = 256n = 8

Note that there are (m + 1) terms.

Sm + 1 = 1 × ( – x2)m + 1

– 1– x2 – 1

= – x2(m + 1) – 1– x2 – 1

= x2m + 2 + 1x2 + 1

b Sn > 1 000 000

1 × (2n – 1)2 – 1

> 1 000 000

2n – 1 > 1 000 0002n > 1 000 001

nlog10 2 > log10 1 000 001n > 19.931. . .

{n: n > 19} or {n: n ≥ 20}, since n is a positive integer.


Exercise 5F Solutions 1 S∞ = a

1 – r

a a = 1r = 1

5÷ 1 = 1

5

S∞ =1

1 – 15

= 54

b a = 1

r = – 23÷ 1 = – 2

3

S∞ =1

1 – – 23

= 35

2 Each side, and hence each perimeter,

will be half the larger side. r = 1

2, a = p

Perimeter of nth triangle = p ×⎝⎜⎛1

2⎠⎟⎞n – 1

= p2n – 1

S∞ =p

1 – 12

= 2p

Find original area using A = s(s – a)(s – b)(s – c) For first triangle, s = 3p

2

A = 3p2 ⎝⎜⎛3p

2– p

⎠⎟⎞⎝⎜⎛3p

2– p

⎠⎟⎞⎝⎜⎛3p

2– p

⎠⎟⎞

= 3p2× p

2× p

2× p

2

= p2 34

Area of nth triangle (r = 14

):

p2 34

×⎝⎜⎛ 1

4⎠⎟⎞n – 1

= p2 34n

S∞ =

p2 34n

1 – 12

= 2 p2 34

= p2 32

3 a = 200, r = 0.94

S∞ = 2001 – 0.94

= 3333 13 m

4 a = 450, r = 0.65 S∞ = 450

1 – 0.65 ≈ 1285.7

Yes, it will kill him. 5 a = 3, r = 0.5 S∞ = 3

1 – 0.5= 6

He can only make the journey if he walks for an infinite time (which isn't very likely).

6 a = 2, r = 3

4

S∞ = 21 – 0.75

= 8

The frog will approach a limit of 8 m. 7 a = 1

3 , r = 1

3

S∞ =

13

1 – 13

= 12

or 50%


8 r = 70% = 0.7 11 S4 = a(1 – r4)

1 – r= 30

s∞ = a1 – r

= 32

a = 32(1 – r)

S∞ = a

1 – 0.7= 40

a = 0.3 × 40 = 12 m

Substitute for a: 9 Note: all distances will be double (up

and down) except the first (down only). 32(1 – r)(1 – r4)1 – r

= 30

32(1 – r4) = 301 – r4 = 30

32r4 = 1 – 30

32 = 2

32= 1

16r = 1

2

a = 32⎝⎜⎛1 – 1

2⎠⎟⎞

= 16

Use a = 30, r = 2

3 and subtract 15 m

from the answer.

S∞ =

30

1 – 23

= 90

Distance = 90 – 15 = 75 m 10 a a = 0.4, r = 0.1 S∞ = 0.4

1 – 0.1= 4

9

The first two terms are 16 and 8. b a = 0.03, r = 0.1 S∞ = 0.03

1 – 0.1 = 3

90= 1

30

12 S∞ =

a

1 + 14

= 4a5

= 8

a = 10

t3 = 10 ×⎝⎜⎛ – 1

4⎠⎟⎞2

= 58

c a = 0.3, r = 0.1 S∞ = 0.3

1 – 0.1 = 3

9= 1

3

Decimal = 10 13

13 5

1 – r= 15

5 = 15(1 – r)1 – r = 1

3r = 1 – 1

3 = 2

3

d a = 0.035, r = 0.01 S∞ = 0.035

1 – 0.01 = 35

990= 7

198

e a = 0.9, r = 0.1 S∞ = 0.9

1 – 0.1 = 9

9= 1

f a = 0.1, r = 0.1 S∞ = 0.1

1 – 0.1= 1

9

Decimal = 4 19


Exercise 5G Solutions 1 a i x1 = 3

x2 = 34

+ 2 = 2.75

x3 = 2.754

+ 2 = 2.688

x4 = 2.6884

+ 2 = 2.672

x5 = 2.6724

+ 2 = 2.668

x6 = 2.6684

+ 2 = 2.667

ii x1 = 1x2 = 12 – 3 = –2x3 = –22 – 3 = 1x4 = 12 – 3 = –2x5 = –22 – 3 = 1x6 = 12 – 3 = –2

iii x1 = 2x2 = 3 × 22 + 1 = 13x3 = 3 × 132 + 1 = 508x4 = 3 × 5082 + 1 = 774 193x5 = 3 × 774 1932 + 1 = 1.8 × 1012

x6 = 3 × (1.8 × 1012)2

+ 1 = 9.7 × 1024

iv x1 = 3x2 = 3 + 2 + 1 = 3.236x3 = 3.236 + 2 + 1 = 3.288x4 = 3.288 + 2 + 1 = 3.300x5 = 3.300 + 2 + 1 = 3.302x6 = 3.302 + 2 + 1 = 3.303

b i and iv converge.

2 a f(x) = x3 + 4x – 3 = 0

4x = 3 – x3

∴ x = 3 – x3

4

Successive iterations: 1 0.5 0.71875 0.657173157 0.67904558 0.671722529 0.674227825 0.673376846 0.673666611 0.673568026 0.673601576 0.67359016 0.673594045 0.673592723

Solution is 0.6736 (to 4 decimal places). b f(x) = x3 + x – 1 = 0

x3 = 1 – x2x3 = 1 – x + x3

x3 = 1 – x + x3

2

∴ x = 3 1 – x + x3

2

Successive iterations: 0.7937010.7068380.6862320.6828930.6824080.6823390.6823290.6823280.682328

Solution is 0.6823 (to 4 decimal places).


c f(x) = x2

3– x – 1 = 0

∴ x = x2

3– 1

Successive iterations: –1 –0.666666667 –0.851851852 –0.758116141 –0.808419972 –0.782152383 –0.796079217 –0.788752627 –0.792623098 –0.790582875 –0.791659573 –0.791091707 –0.791391304 –0.791233268 –0.791316638 –0.791272659 –0.79129586

Solution is –0.7913 (to 4 decimal places). d f(x) = x4 – x – 2 = 0

x4 = x + 2∴ x = 4 x + 2

Successive iterations: 1 1.316074013 1.349447689 1.352830232 1.353171652 1.353206099 1.353209574 1.353209925 1.35320996


e f(x) = 2x – 4x = 04x = 2x

∴ x = 2x

4

Successive iterations: 0.5 0.353553391 0.319425942 0.311958482 0.310347942 0.310001681 0.309927286 0.309911305 0.309907872 0.309907134 0.309906976

Solution is 0.3099 (to 4 decimal places). f f(x) = – x + log10 x + 2

∴ x = log10 x + 2

Successive iterations: 5 2.698970004 2.431198058 2.38582034 2.377637737 2.376145685 2.375873064 2.375823234 2.375814125 2.37581246 2.375812156 2.3758121



g f(x) = 4x × 2x – 3 = 04x = 3

2x

∴ x = 34 × 2x

h f(x) = x3 – 3x + 1 = 03x = x3 + 1

∴ x = x3 + 13

Successive iterations: Successive iterations: 0.5

0.375 0.3509110.3477370.34735 0.3473030.3472970.3472960.3472960.347296

1 0.375 0.57832906 0.502304269 0.52948372 0.519601961 0.523173198 0.521879741 0.522347845 0.522178389 0.522239726 0.522217523 0.52222556 0.522222651




Solutions to Multiple-choice Questions 1 t1 = 3 × 1 + 2 = 5

t2 = 3 × 2 + 2 = 8 D t3 = 3 × 3 + 2 = 11 2 t2 = 3 + 3 = 6

t3 = 6 + 3 = 9 B t4 = 9 + 3 = 12 3 a = 10

d = 8 – 10 = –2t10 = 10 + (9 × –2)

= –8 A 4 a = 10, d = –2

S10 = 102

(10 + –8)

= 10 A 5 a = 8

d = 13 – 8 = 5tn = 8 + 5(n – 1) = 58

5(n – 1) = 50n – 1 = 10

n = 11 B 6 a = 12

r = 812

= 23

t6 = 12 ×⎝⎜⎛2

3⎠⎟⎞5

= 12881

D

7 a = 8r = 4

8= 1

2

S6 =8 ×

⎝⎜⎛ 1 –

⎝⎜⎛1

2⎠⎟⎞6

⎠⎟⎞

1 – 12

= 15 34

E

8 a = 8

r = 48

= 12

S∞ =8

1 – 12

= 16 C 9 Value = 2000 × 1.0556

= $2757.69 E

10 a

1 – 13

= 37.5

a = 37.5 × 23

= 25 D


Solutions to Short-answer Questions 1 a t1 = 3

t2 = 3 – 4 = –1t3 = –1 – 4 = –5t4 = –5 – 4 = –9t5 = –9 – 4 = –13t6 = –13 – 4 = –17

b t1 = 5

t2 = 2 × 5 + 2 = 12t3 = 2 × 12 + 2 = 26t4 = 2 × 26 + 2 = 54t5 = 2 × 54 + 2 = 110t6 = 2 × 110 + 2 = 222

2 a t1 = 2 × 1 = 2

t2 = 2 × 2 = 4t3 = 2 × 3 = 6t4 = 2 × 4 = 8t5 = 2 × 5 = 10t6 = 2 × 6 = 12

b t1 = –3 × 1 + 2 = –1

t2 = –3 × 2 + 2 = –4t3 = –3 × 3 + 2 = –7t4 = –3 × 4 + 2 = –10t5 = –3 × 5 + 2 = –13t6 = –3 × 6 + 2 = –16

3 a End of first year: $5000 × 1.05 = $5250 Start of second year: $5250 + $500 = $5750 End of second year: $5750 × 1.05 = $6037.50 b t n = 1.05(tn – 1 + 500)

4 a + 3d = 19a + 6d = 43 – :

3d = 24d = 8

a + 3 × 8 = 19a = –5

t20 = –5 + 19 × 8 = 147

5 a + 4d = 0.35

a + 8d = 0.15 – :

4d = –0.2d = –0.05

a + 4 × –0.05 = 0.35a = 0.35 + 0.2 = 0.55

t14 = 0.55 + 13 × –0.55 = –0.1

6 a + 5d = –24

a + 13d = 6 – :

8d = 30d = 3.75

a + 5 × 3.75 = –24a = –24 – 18.75 = –42.75

S10 = 5 × (–85.5 + 9 × 3.75) = –258.75

7 a = –5, d = 7

Sn = n

2 (–10 + 7(n – 1))

= 402n(–10 + 7(n – 1)) = 804

7n2 – 10n – 7n = 8047n2 – 17n – 804 = 0

(7n + 67)(n – 12) = 0n = 12 (since n > 0)


8 ar5 = 9ar9 = 729

r4 = 81r = 3

a × 35 = 9a = 9

243= 1

27t4 = 1

27× 33 = 1

9 a = 1000

r = 1.035tn = arn

= 1000 × 1.035n

10 9r2 = 4

r2 = 49

r = ± 23

t2 = ar = ± 6t4 = ar3 = ± 8

3

Terms = 6, 83 or – 6, – 8

3

11 a + ar + ar2 = 24

ar3 + ar4 + ar5 = 24r3(a + ar + ar2) = 24

r3 = 1r = 1

All terms will be the same: tn = 243

= 8

S12 = 12 × 8 = 96

12 Sn = a(rn – 1)r – 1

S8 = 6 × (–38 – 1)–3 – 1

= –9840

13 a = 1, r = – 1

3

S∞ =1

1 – – 13

= 34

14 x + 4

x= 2x + 2

x + 4(x + 4)2 = x(2x + 2)

x2 + 8x + 16 = 2x2 + 2x2x2 + 2x – x2 – 8x – 16 = 0

x2 – 6x – 16 = 0(x – 8)(x + 2) = 0

x = 8 or x = –2


Chapter 6 – Algebra II Exercise 6A Solutions 1 ax2 + bx + c = 10x2 – 7

= 10x2 + 0x – 7 a = 10, b = 0, c = –7 2 2a – b = 4

a + 2b = –34a – 2b = 8 + :

5a = 5a = 1

a × 1 – b = 4b = –2

3 2a – 3b = 7

3a + b = 5 + 3 × :

11a = 22a = 2

3 × 2 + b = 5b = –1c = 7

4 a(x + b)2 + c = ax2 + 2abx + ab2 + c

a = 22ab = 4

b = 1ab2 + c = 5

2 + c = 5c = 3

5 c(x + 2)2 + a(x + 2) + 2

= cx2 + 4cx + 4c + ax + 2a + dc = 1

4c + a = 0a = –4

4c + 2a + d = 04 – 8 + d = 0

d = 4∴ x2 = (x + 2)2 – 4(x + 2) + 4

6 (x + 1)3 + a(x + 1)2 + b(x + 1) + c = x3 + 3x2 + 3x + 1 + ax + a + bx + b + c

3 + a = 0a = –3

3 + 2a + b = 03 – 6 + b = 0

b = 31 + a + b + c = 0

c = –1∴ x3 = (x + 1)3 – 3(x + 1)2 + 3(x + 1) – 1

7 ax2 + 2ax + a + bx + c = x2

a = 12a + b = 0

b = –2a + c = 0

c = –1 8 a a(x + b)3 + c = ax3 + 3abx2 + 3ab2x + ab3 + c

= 3x3 – 9x2 + 8x + 12a = 3

3ab = –93 × 3 × b = –9

b = –1 Equating x terms:

3ab2 = 83ab2 = 3 × 3 × (–1)2 = 9

The equality is impossible. b Clearly this expression can be expressed

in this form, if a = 3, b = –1 and ab3 + c = 2

–3 + c = 2c = 5


9 Expanding gives the following:

n3 = an3 + 6an2 + 11an + 6a + bn2

+ 3bn + 2b + cn + c + da = 1

6a + b = 0b = –6

11a + 3b + c = 011 – 18 + c = 0

c = 76a + 2b + c + d = 0

6 – 12 + 7 + d = 0d = –1

10 a Expanding gives the following: n2 = an2 + 3an + 2a + bn2 + 5bn + 6b

a + b = 13a + 5b = 02a + 6b = 0a + 3b = 0 – :

2b = –1b = – 1

2a + – 1

2= 1

a = 1 12

These do not satisfy the second equation, as 3 × 1 1

2 + 5 × –1

2 = 2.

b n2 = an2 + 3an + 2a + bn + b + c

a = 13a + b = 0

b = –32a + b + c = 0

2 – 3 + c = 0c = 1

∴ n2 = (n + 1)(n + 2) – 3(n + 1) + 1

11 a a(x2 + 2bx + b2) + c = ax2 + 2abx + ab2 + c b ax2 + bx + c = A(x + B)2 + C

= Ax2 + 2ABx + AB2 + CA = a

2AB = bB = b

2aAB2 + C = c

a × b2

4a2 + C = c

C = c – b2

4a

∴ax2 + bx + x = a⎝⎜⎛ x + b

2a⎠⎟⎞2

+ 4ac – b2

4a

12 (x – 1)2(px + q) = (x2 – 2x + 1)(px + q)

= px3 + (q – 2p)x2 + (p – 2q)x + q Equating 3x and 2x terms:

p = aq – 2p = bq – 2a = b

q = 2a + b Equating x and constant terms: q = d

p – 2q = cp = c + 2d

Equating the two different expressions for p and q gives:

d = 2a + b (q)∴ b = d – 2a

a = c + 2d (p)∴ c = a + 2d


15 (x – a)2(x – b) = (x2 – 2ax + a2)(x – b) = x3 – 2ax2 – bx2 + a2x + 2abx – a2b

–2a – b = 3a2 + 2ab = –9

13 c(x – a)(x – b) = cx2 – acx – bcx + abc = 3c = 3

– ac – bc = 10–3a – 3b = 10

abc = 33ab = 3ab = 1

b = 1a

Substitute b = –2a – 3:

a2 + 2a(–2a – 3) = –9a2 – 4a2 – 6a = –9

–3a2 – 6a + 9 = 0a2 + 2a – 3 = 0

(a + 3)(a – 1) = 0a = –3 or a = 1b = –2a – 3b = 3 or b = –5

–3a – 3a

= 10

3a2 + 3 = –10a3a2 + 10a + 3 = 0

(3a + 1)(a + 3) = 0a = – 1

3 , b = –3, c = 3

or a = –3, b = – 13 , c = 3

Comparing the constant terms: c = – a2b

c = (–3)2 × 3 = –27or c = (–1)2 × –5 = 5

14 n2 = a(n – 1)2 + b(n – 2)2 + c(n – 3)2

= an2 – 2an + a + bn2 – 4bn + 4b + cn2 + 9c

So a = 1, b = –5, c = 5 or a = –3, b = 3, c = –27

a + b + c = 1–2a – 4b – 6c = 0

a + 2b + 3c = 0a + 4b + 9c = 0 – :

b + 2c = –1 – :

2b + 6c = 0b + 3c = 0

– :c = 1

b + 3 × 1 = 0b = –3

a + b + c = 1a – 3 + 1 = 1

a = 3

∴ n2 = 3(n – 1)2 – 3(n – 2)2 + (n – 3)2


Exercise 6B Solutions 1 a x2 – 2x = –1

x2 – 2x + 1 = 0(x – 1)2 = 0

x = 1

b x2 – 6x + 9 = 0

(x – 3)2 = 0x = 3

c Divide both sides by 5: x2 – 2x = 1

5x2 – 2x + 1 = 6

5(x – 1)2 = 6

5= 30

25

x – 1 = ± 305

x = 1 ± 305

d Divide both sides by –2: x2 – 2x = – 1

2x2 – 2x + 1 = 1

2(x – 1)2 = 1

2= 2

4

x – 1 = ± 22

x = 1 ± 22

e Divide both sides by 2:

x2 + 2x = 72

x2 + 2x + 1 = 92

(x + 1)2 = 92

= 9 × 24

x + 1 = ± 3 22

x = –1 ± 3 22

f 6x2 + 13x + 1 = 0

x = –13 ± 169 – 4 × 6 × 112

= –13 ± 14512

2 a 2x2 – x – 4t = 0

x = 1 ± 1 – 4 × 2 × –4t4

= 1 ± 32t + 14

32t + 1 ≥ 032t ≥ – 1

t ≥ – 132

b 4x2 + 4x – t – 2 = 0

x = –4 ± 16 – 4 × 4 × – (t + 2)8

= –4 ± 16 + 32 + 16t8

= –4 ± 16t + 488

= –4 ± 4 t + 38

= –1 ± t + 32

t + 3 ≥ 0t ≥ – 3


c 5x2 + 4x – t + 10 = 0

x = –4 ± 16 – 4 × 5 × ( – t + 10)10

= –4 ± 16 + 20t – 20010

= –4 ± 20t – 18410

= –4 ± 4(5t – 46)10

= –4 ± 2 5t – 4610

= –2± 5t – 465

5t – 46 ≥ 05t ≥ 46t ≥ 46

5

d tx2 + 4tx – t + 10 = 0

x = –4t ± 16t2 – 4 × t × ( – t + 10)2t

= –4t ± 16t2 + 4t2 – 40t2t

= –4t ± 20t2 – 40t2t

= –4t ± 2 5t2 – 10t2t

= –2t ± 2t(t – 5)t

2t(t – 5) ≥ 0

This is a quadratic with a minimum and solutions t = 0, t = 5.

∴ t < 0, t ≥ 5Note: t = 0 gives denominator zero, so it must be checked by substituting t = 0 in the original equation. In this case it gives 10 = 0, and so is not a solution, but it should be checked. (e.g. tx gives a solution with t on the denominator, but substituting t = 0 gives 5x + 4 = 0, which has a solution.)

2 + 5x + 4 = t

3

a x = –3 ± 9 – 4 × 1 × –92

= –3 ± 452

= –3 ± 3 52

= –3(1 ± 5 )2

b i x = – p ± p2 – 4 × 1(–16)2

= – p ± p2 + 642

ii p = 0 gives x = 0 + 642

= 4

p = 6 gives x = –6 + 1002

= 2

4 Use Pythagoras’ Theorem: (8 – x)2 + (6 + x)2 = 100

64 – 16x + x2 + 36 + 12x + x2 = 1002x2 – 4x = 0

2x(x – 4) = 0x = 2 since x ≠ 0

5 a 6(x + 3) – 6x

x(x + 3)= 18

x(x + 3)

b 18

x(x + 3)= 1

18 – x(x + 3)x(x + 3)

= 0

18 – x(x + 3) = 018 – x2 – 3x = 0

Re-arrange and divide by –1: x2 + 3x – 18 = 0

(x – 3)(x + 6) = 0x = 3 or x = –6


6 Let the numbers be n and n + 2. 1

n+ 1

n + 2= 36

3231n

+ 1n + 2

– 36323

= 0

323(n + 2) + 323n – 36n(n + 2)323n(n + 2)

= 0

323n + 646 + 323n – 36n2 – 72n = 0

Re-arrange and divide by –1: 36n2 – 574n – 646 = 0

18n2 – 287 – 323 = 0(n – 17)(18n + 19) = 0

n = 17 The numbers are 17 and 19. 7 a Car = 600

x km/h

Plane = 600x

+ 220 km/h

b 600

x+ 220 = 600

x – 5.5600(x – 5.5) + 220x(x – 5.5) = 600x

600x – 3300 + 220x2 – 1210x = 600x220x2 – 1210x – 3300 = 0

2x2 – 11x – 30 = 0(2x – 15)(x + 2) = 0

x = 7.5

Average speed of car = 6007.5

= 80 km/h

Average speed of plane = 80 + 220 = 300 km/h 8 Time taken by car = 200

x h

Time taken by train = 200x + 5

h = 200x

– 2 h

200x + 5

= 200x

– 2

200x + 5

× x(x + 5) = 200x

× x(x + 5) – 2 × x(x + 5)

200x = 200(x + 5) – 2x(x + 5) = 200x + 1000 – 2x2 – 10x

2x2 + 10x – 1000 = 0x2 + 5x – 500 = 0

(x – 20)(x + 25) = 0x = 20 since x > 0

9 Let his average speed be x km/h. His time for the journey is 108

x h.

108x

– 4 12

= 108x + 2

108 × 2(x + 2) – 4 12× 2x(x + 2) = 108 × 2x

216x + 432 – 9x2 – 18x = 216x–9x2 – 18x + 432 = 0

x2 + 2x – 48 = 0(x – 6)(x + 8) = 0

x = 6 since x > 0

His average speed is 6 km/h. 10 a Usual time = 75

x h.

75x

– 1860

= 75x + 12.5

75x

– 310

= 75x + 12.5

75(x + 12.5) – 0.3x(x + 12.5) = 75x75x + 937.5 – 0.3x2 – 3.75x = 75x

–0.3x2 – 3.75x + 937.5 = 0

Divide by 0.15: 2x2 + 25x – 6250 = 0

(x – 50)(2x + 125) = 0x = 50

b Average speed = x + 12.5 = 62.5 Time = 75

62.5 = 1.2 h,

or 1 hour 12 minutes, or 72 minutes. 11 Let the speed of the slow train be x km/h. The slow train takes 3 1

2– 10

60= 7

2– 1

6 = 20

6= 10

3 hours longer.

Compare the times: 250

x + 20+ 10

3= 250

x750x + 10x(x + 20) = 750(x + 20)750x + 10x2 + 200x = 750x + 15 000

10x2 + 200x – 15 000 = 0x2 + 20x – 1500 = 0(x – 30)(x + 50) = 0

x = 30

Slow train: 30 km/h Fast train: 50 km/h


12 Let the original speed of the car be x km/h. Compare the times: 105

x + 10= 105

x– 1

4420x = 420(x + 10) – x(x + 10)420x = 420x + 4200 – x2 – 10x

x2 + 10x – 4200 = 0(x – 60)(x + 70) = 0

x = 60 km/h

13 Let x min be the time the larger pipe

takes, and C the capacity of the tank. Form an equation using the rates:

Cx

+ Cx + 5

=C

11 19

Cx

+ Cx + 5

= 9C100

1x

+ 1x + 5

= 9100

100(x + 5) + 100x = 9x(x + 5)100x + 500 + 100x = 9x2 + 45x

200x + 500 = 9x2 + 45x9x2 – 155x – 500 = 0

(x – 20)(9x + 25) = 0x = 20 since x > 0

The larger pipe takes 20 min and the smaller pipe takes 25 min.

14 Let x min be the original time the first

pipe takes, and y min be the original time the second pipe takes.

Let C be the capacity of the tank. The original rates are C

x and C

y.

The combined rate is Cx

+ Cy

.

Total time taken = capacity ÷ rate

C ÷⎝⎜⎛C

x+ C

y ⎠⎟⎞ = C ÷ Cy + Cx

xy = C × xy

Cx + Cy = xy

x + y = 20

3

New rates are Cx – 1

and Cy + 2

.

The combined rate is Cx – 1

+ Cy + 2

.

C ÷⎝⎜⎛ C

x – 1+ C

y + 2⎠⎟⎞ = C ÷ C(y + 2) + C(x – 1)

(x – 1)(y + 2) = C × (x – 1)(y + 2)

Cx + Cy + C = (x – 1)(y + 2)

x + y + 1= 7

Solve the simultaneous equations: xy

x + y= 20

3(x – 1)(y + 2)

x + y + 1= 7

Multiply both sides of the first equation by 3(x + y):

3xy = 20x + 20y3xy – 20y = 20x

y(3x – 20) = 20xy = 20x

3x – 20

Substitute into the second equation, after multiplying both sides by x + y + 1:

(x – 1)(y + 2) = 7x + 7y + 7

(x – 1)⎝⎜⎛ 20x3x – 20

+ 2⎠⎟⎞ = 7x + 140x

3x – 20+ 7

(x – 1)20x + 2(3x – 20)3x – 20

= 7x + 140x3x – 20

+ 7

(x – 1)26x – 403x – 20

= 7x + 140x3x – 20

+ 7

(x – 1)(26x – 40) = 7x(3x – 20) + 140x + 7(3x – 20)26x2 – 66x + 40 = 21x2 – 140x + 140x + 21x – 1405x2 – 87x + 180 = 0

(5x – 12)(x – 15) = 0x = 2.4 or x = 15

y = 20x3x – 20

< 0 if x = 2.4

∴ x = 15

y = 20 × 153 × 15 – 20

= 12

The first pipe now takes one minute less, i.e. 15 – 1 = 14 minutes. The second pipe now takes two minutes more, i.e. 12 + 2 = 14 minutes.


15 Let the average speed for rail and sea be x + 25 km/h and x km/h respectively. Time for first route = 233

x + 25+ 126

x hours.

Time for second route = 405x + 25

+ 39x

hours.

233x + 25

+ 126x

= 405x + 25

+ 39x

+ 56

233 × 6x + 126 × 6(x + 25) = 405 × 6x + 39 × 6(x + 25) + 5x(x + 25)1398x + 756x + 18 900 = 2430x + 234x + 5850 + 5x2 + 125x

–5x2 – 635x + 13 050 = 0x2 + 127x – 2625 = 0

x = –127 + 1272 – 4 × 1 × 26252

≈ 18.09

(Ignore negative square root as x > 0.) Speed by rail is 18 + 25 = 43 km/h and by sea is 18 km/h. 16 After 15 min, the freighter has travelled 3 km, bringing it to 12 km from where the

cruiser was. Let x km be the distance the cruiser has travelled in 15 minutes and y km the original distance apart of the ships. The distance the cruiser has travelled can be calculated using Pythagoras’ theorem.

x2 + (y – 3)2 = 102 = 100

After a further 15 minutes, the distances will be 2x km and (y – 6) km. (2x)2 + (y – 6)2 = 132

4x2 + (y – 6)2 = 169

Multiply the first equation by 4 and subtract:

4(y – 3)2 – (y – 6)2 = 400 – 1694y2 – 24y + 36 – y2 + 12y – 36 = 231

3y2 – 12y – 231 = 0y2 – 4y – 77 = 0

(y – 11)(y + 7) = 0y = 11

x2 + 82 = 102

x = 6

The speed of the cruiser is 6 ÷ 0.25 = 24 km/h. The cruiser will be due east of the freighter when the freighter has travelled 11 km. This will take 11

12 hours. During that

time the cruiser will have travelled 24 × 11

12= 22 km.

They will be 22 km apart.


17 Let x be the amount of wine first taken out of cask A. After water is added, the concentration of wine in cask B is x

20.

If cask A is filled, it will receive x litres at concentration x

20.

The amount of wine in cask A will be (20 – x) + x × x20

= 20 – x + x2

20.

The concentration of wine in cask A will be 20 – x + x2

2020

= 1 – x20

+ x2

400.

The amount of wine in cask B will be (20 – x) × x20

= x – x2

20.

Mixture is transferred again.

The amount of wine transferred is ⎝⎜⎛ 1 – x

20+ x2

400⎠⎟⎞ × 20

3= 20

3– x

3+ x2

60.

Amount of wine in A =⎝⎜⎛ 20 – x + x2

20⎠⎟⎞ –

⎝⎜⎛ 20

3– x

3+ x2

60⎠⎟⎞ .

Amount of wine in B =⎝⎜⎛ x – x2

20⎠⎟⎞ +

⎝⎜⎛ 20

3– x

3+ x2

60⎠⎟⎞

⎝⎜⎛ 20 – x + x2

20⎠⎟⎞ –

⎝⎜⎛ 20

3– x

3+ x2

60⎠⎟⎞ =

⎝⎜⎛ x – x2

20⎠⎟⎞ +

⎝⎜⎛ 20

3– x

3+ x2

60⎠⎟⎞

20 – x + x2

20– 20

3+ x

3– x2

60= x – x2

20+ 20

3– x

3+ x2

60

– 4x2

60– 4x

3+ 20

3= 0

x2

15+ 4x

3– 20

3= 0

x2 + 20x – 100 = 0(x – 10)2 = 0

10 litres was first taken out of cask A.


Exercise 6C Solutions 1 a 5x + 1

(x – 1)(x + 2)= A

x – 1+ B

x + 2 = A(x + 2) + B(x – 1)

(x – 1)(x + 2) = Ax + Bx + 2A – B

(x – 1)(x + 2)A + B = 5

2A – B = 1 + :

3A = 6A = 2

2 + B = 5B = 3

∴ 5x + 1(x – 1)(x + 2)

= 2x – 1

+ 3x + 2

b –1

(x + 1)(2x + 1)= A

x + 1+ B

2x + 1 = A(2x + 1) + B(x + 1)

(x + 1)(2x + 1) = 2Ax + Bx + A + B

(x + 1)(2x + 1)2A + B = 0

A + B = –1 – :

A = 11 + B = –1

B = –2∴ –1

(x + 1)(2x + 1)= 1

x + 1– 2

2x + 1

c 3x – 2

(x + 2)(x – 2)= A

x + 2+ B

x – 2 = A(x – 2) + B(x + 2)

(x + 2)(x – 2) = Ax + Bx – 2A + 2B

(x + 2)(x – 2)A + B = 3

2A + 2B = 6–2A + 2B = –2

+ :4B = 4

B = 1A + 1 = 3

A = 2∴ 3x – 2

(x + 2)(x – 2)= 2

x + 2+ 1

x – 2

d 4x + 7(x + 3)(x – 2)

= Ax + 3

+ Bx – 2

= A(x – 2) + B(x + 3)(x + 3)(x – 2)

= Ax + Bx – 2A + 3B(x + 3)(x – 2)

A + B = 42A + 2B = 8

–2A + 3B = 7 + :

5B = 15B = 3

A + 3 = 4A = 1

∴ 4x + 7(x + 3)(x – 2)

= 1x + 3

+ 3x – 2

e 7 – x

(x – 4)(x + 1)= A

x – 4+ B

x + 1 = A(x + 1) + B(x – 4)

(x – 4)(x + 1) = Ax + Bx + A – 4B

(x – 4)(x + 1)A + B = –1

A – 4B = 7 – :

5B = –8B = – 8

5A – 8

5= –1

A = 35

∴ 7 – x(x – 4)(x + 1)

= 35(x – 4)

– 85(x + 1)

2 a 2x + 3

(x – 3)2 = Ax – 3

+ B(x – 3)2

= A(x – 3) + B(x – 3)2

= Ax – 3A + B(x – 3)2

A = 2–3A + B = 3

–6 + B = 3B = 9

∴ 2x + 3(x – 3)2 = 2

x – 3+ 9

(x – 3)2


b 9(1 + 2x)(1 – x)2 = A

1 + 2x+ B

1 – x+ C

(1 – x)2

= A(1 – x)2 + B(1 + 2x)(1 – x) + C(1 + 2x)(1 + 2x)(1 – x)2

= A – 2Ax + Ax2 + B + Bx – 2Bx2 + C + 2Cx(1 + 2x)(1 – x)2

A – 2B = 0–2A + B + 2C = 0

A + B + C = 92A + 2B + 2C = 18 – :

4A + B = 18 × 4: 4A – 8B = 0

9B = 18B = 2

4A + 2 = 18A = 4

4 + 2 + C = 9C = 3

∴ 9(1 + 2x)(1 – x)2 = 4

1 + 2x+ 2

1 – x+ 3

(1 – x)2

c 2x – 2

(x + 1)(x – 2)2 = Ax + 1

+ Bx – 2

+ C(x – 2)2

= A(x – 2)2 + B(x + 1)(x – 2) + C(x + 1)(x + 1)(x – 2)2

= Ax2 – 4Ax + 4A + Bx2 – Bx – 2B + Cx + C(x + 1)(x – 2)2

A + B = 0–4A – B + C = 24A – 2B + C = –2

– : 8A – B = –4 + : 9A = –4

A = – 49

A + B = 0B = 4

94A – 2B + C = –2– 16

9– 8

9+ C = –2

C = –2 + 249

= 23

∴ 2x – 2(x + 1)(x – 2)2 = – 4

9(x + 1)+ 4

9(x – 2)+ 2

3(x – 2)2


3 a 3x + 1

(x + 1)(x2 + x + 1)= A

x + 1+ Bx + C

x2 + x + 1

= A(x2 + x + 1) + (Bx + C)(x + 1)(x + 1)(x2 + x + 1)

= Ax2 + Ax + A + Bx2 + Bx + Cx + C(x + 1)(x2 + x + 1)

A + B = 0A + B + C = 3

A + C = 1 – : C = 3

A + 3 = 1A = –2

A + B + C = 3–2 + B + 3 = 3

B = 2∴ 3x + 1

(x + 1)(x2 + x + 1)= – 2

x + 1+ 2x + 3

x2 + x + 1

b 3x2 + 2x + 5(x2 + 2)(x + 1)

= Ax + Bx2 + 2

+ Cx + 1

= (Ax + B)(x + 1) + C(x2 + 2)(x2 + 2)(x + 1)

= Ax2 + Ax + Bx + B + Cx2 + 2C(x2 + 2)(x + 1)

A + C = 3A + B = 2

B + 2C = 5 – :

C – B = 1 + :

3C = 6C = 2

A + 2 = 3A = 1

1 + B = 2B = 1

∴ 3x2 + 2x + 5(x2 + 2)(x + 1)

= x + 1x2 + 2

+ 2x + 1


c Factorise the denominator: 2x3 + 6x2 + 2x + 6 = 2x2(x + 3) + 2(x + 3)

= 2(x2 + 1)(x + 3) The 2 factor can be put with either fraction.

x2 + 2x – 132(x2 + 1)(x + 3)

= Ax + Bx2 + 1

+ C2(x + 3)

= 2(Ax + B)(x + 3) + C(x2 + 1)2(x2 + 1)(x + 3)

= 2Ax2 + 6Ax + 2Bx + 6B + Cx2 + C2(x2 + 1)(x + 3)

2A + C = 16A + 2B = 29A + 3B = 36B + C = –13

– :2A – 6B = 14

A – 3B = 7 + :

10A = 10A = 1

2 + C = 1C = –1

3A + B = 1A + B = 1

B = –2

∴ x2 + 2x – 132(x2 + 1)(x + 3)

= x – 2x2 + 1

– 12(x + 3)

4 (x – 1)(x – 2) = x2 – 3x + 2 First divide:

3x2 – 4x – 2 = 3(x2 – 3x + 2) + 5x – 83x2 – 4x – 2

(x – 1)(x – 2)= 5x – 8

(x – 1)(x – 2)5x – 8

(x – 1)(x – 2)= A

x – 1+ B

x – 2 = A(x – 2) + B(x – 1)

(x – 1)(x – 2) = Ax + Bx – 2A – B

(x – 1)(x – 2)A + B = 5

–2A – B = –8 + :

– A = –3A = 3

3 + B = 5B = 2

∴ 5x – 8(x – 1)(x – 2)

= 3x – 1

+ 2x – 2

Use the previous working:

3x2 – 4x – 2(x – 1)(x – 2)

= 3 + 3x – 1

+ 2x – 2

5 2x + 10(x + 1)(x – 1)2 = A

x + 1+ C

(x – 1)2

= A(x – 1)2 + C(x + 1)(x + 1)(x – 1)2

= Ax2 – 2Ax + A + Cx + C(x + 1)(x – 1)2

A = 0–2A + C = 2

C = 2A + C = 10 0 + 2 ≠ 10

It is impossible to find A and C to satisfy this equation.


6 a 1

(x – 1)(x + 1)= A

x – 1+ B

x + 1 = A(x + 1) + B(x – 1)

(x – 1)(x + 1) = Ax + Bx + A – B

(x – 1)(x + 1)A + B = 0A – B = 1

+ :2A = 1

A = 12

12

+ B = 0

B = – 12

∴ 1(x – 1)(x + 1)

= 12(x – 1)

– 12(x + 1)

b x

(x – 2)(x + 3)= A

x – 2+ B

x + 3 = A(x + 3) + B(x – 2)

(x – 2)(x + 3) = Ax + Bx + 3A – 2B

(x – 2)(x + 3)A + B = 1

2A + 2B = 23A – 2B = 0

+ :5A = 2

A = 25

25

+ B = 1

B = 35

∴ x(x – 2)(x + 3)

= 25(x – 2)

+ 35(x + 3)

c 3x + 1(x – 2)(x + 5)

= Ax – 2

+ Bx + 5

= A(x + 5) + B(x – 2)(x – 2)(x + 5)

= Ax + Bx + 5A – 2B(x – 2)(x + 5)

A + B = 32A + 2B = 65A – 2B = 1

+ :7A = 7

A = 11 + B = 3

B = 2∴ 3x + 1

(x – 2)(x + 5)= 1

x – 2+ 2

x + 5

d 1

(2x – 1)(x + 2)= A

2x – 1+ B

x + 2 = A(x + 2) + B(2x – 1)

(2x – 1)(x + 2) = Ax + 2Bx + 2A – B

(2x – 1)(x + 2)A + 2B = 0

2A + 4B = 02A – B = 1

+ :5B = –1

B = – 15

A + 2B = 0A = 2

5∴ 1

(2x – 1)(x + 2)= 2

5(2x – 1)– 1

5(x + 2)

e 3x + 5

(3x – 2)(2x + 1)= A

3x – 2+ B

2x + 1 = A(2x + 1) + B(3x – 2)

(3x – 2)(2x + 1) = 2Ax + 3Bx + A – 2B

(3x – 2)(2x + 1)2A + 3B = 3

A – 2B = 52A – 4B = 10

– :7B = –7B = –1

A – 2 × –1 = 5A = 3

∴ 3x + 5(3x – 2)(2x + 1)

= 33x – 2

– 12x + 1


f 2x(x – 1)

= Ax

+ Bx – 1

= A(x – 1) + Bxx(x – 1)

= Ax + Bx – Ax(x – 1)

A + B = 0– A = 2

A = –2–2 + B = 0

B = 2∴ 2

x(x – 1)= 2

x – 1– 2

x

g 3x + 1

x(x2 + 1)= A

x+ Bx + C

x2 + 1

= A(x2 + 1) + x(Bx + C)x(x2 + 1)

= Ax2 + A + Bx2 + Cxx(x2 + 1)

A + B = 0C = 3A = 1

1 + B = 0B = –1

∴ 3x + 1x(x2 + 1)

= 1x

+ 3 – xx2 + 1

h 3x2 + 8x(x2 + 4)

= Ax

+ Bx + Cx2 + 4

= A(x2 + 4) + x(Bx + C)x(x2 + 4)

= Ax2 + 4A + Bx2 + Cxx(x2 + 4)

A + B = 3C = 0

4A = 8A = 2

2 + B = 3B = 1

∴ 3x2 + 8x(x2 + 4)

= 2x

+ xx2 + 4

i 1x(x – 4)

= Ax

+ Bx – 4

= A(x – 4) + Bxx(x – 4)

= Ax + Bx – 4Ax(x – 4)

A + B = 0–4A = 1

A = – 14

– 14

+ B = 0

B = 14

∴ 1x(x – 4)

= 14(x – 1)

– 14x

j x + 3x(x – 4)

= Ax

+ Bx – 4

= A(x – 4) + Bxx(x – 4)

= Ax + Bx – 4Ax(x – 4)

A + B = 1–4A = 3

A = – 34

– 34

+ B = 1

B = 74

∴ x + 3x(x – 4)

= 74(x – 4)

– 34x


k First divide x 2 – x2 – 1 by x2 – x. l First divide ( x2 – x2 – 6) by ( – x2 + 2x). – x – 1

– x2 + 2x x3 – x2 – 6 x3 – 2x2

x2 – 6 x2 – 2x 2x – 6

You might observe a pattern in the question.

x3 – x2 – 1x2 – x

= x(x2 – x) – 1x2 – x

= x – 1x2 – x

Express – 1x2 – x

in partial fractions.

– 1x(x – 1)

= Ax

+ Bx – 1

= A(x – 1) + Bxx(x – 1)

= Ax + Bx – Ax(x – 1)

A + B = 0– A = –1

A = 11 + B = 0

B = –1∴ – 1

x(x – 1)= 1

x– 1

x – 1x3 – x2 – 1

x2 – x= x + 1

x– 1

x – 1

∴ (x3 – x2 – 6) ÷ ( – x2 + 2x) = – x – 1 + 2x – 6x(2 – x)

Separate 2x – 6x(2 – x)

into partial fractions.

2x – 6x(2 – x)

= Ax

+ B2 – x

= A(2 – x) + Bxx(2 – x)

= – Ax + Bx + 2Ax(2 – x)

– A + B = 22A = –6

A = –33 + B = 2

B = –1∴ 2x – 6

x(2 – x)= – 3

x– 1

2 – xx3 – x2 – 6

2x – x2 = – x – 1 – 3x

– 12 – x


m x2 – x(x + 1)(x2 + 2)

= Ax + 1

+ Bx + Cx2 + 2

= A(x2 + 2) + (Bx + C)(x + 1)(x + 1)(x2 + 2)

= Ax2 + 2A + Bx2 + Bx + Cx + C(x + 1)(x2 + 2)

A + B = 1B + C = –1

2A + C = 0 – : A – C = 2

+ : 3A = 2A = 2

323

+ B = 1

B = 13

13

+ C = –1

C = – 43

∴ x2 – x(x + 1)(x2 + 2)

= 23(x + 1)

+ x – 43(x2 + 2)

n can be factorised into ( x3 – 3x – 2 x – 2)(x + 1)2.

x2 + 2(x – 2)(x + 1)2 = A

x – 2+ B

x + 1+ C

(x + 1)2

= A(x + 1)2 + B(x + 1)(x – 2) + C(x – 2)(x – 2)(x + 1)2

= Ax2 + 2Ax + A + Bx2 – Bx – 2B + Cx – 2C(x – 2)(x + 1)2

A + B = 12A – B + C = 0

4A – 2B + 2C = 0A – 2B – 2C = 2 + :

5A – 4B = 2 – 4 × :

9A = 6A = 2

3A + B = 1

B = 13

43

– 13

+ C = 0

C = –1

∴ x2 + 2(x – 2)(x + 1)2 = 2

3(x – 2)+ 1

3(x + 1)– 1

(x + 1)2


o 2x2 + x + 8x(x2 + 4)

= Ax

+ Bx + Cx2 + 4

= A(x2 + 4) + x(Bx + C)x(x2 + 4)

= Ax2 + 4A + Bx2 + Cxx(x2 + 4)

A + B = 2 C = 1

4A = 8A = 2

2 + B = 2B = 0

∴ 2x2 + x + 8x(x2 + 4)

= 2x

+ 1x2 + 4

p 2x2 + 7x + 6 = (2x + 3)(x + 2) 1 – 2x

(2x + 3)(x + 2)= A

2x + 3+ B

x + 2 = A(x + 2) + B(2x + 3)

(2x + 3)(x + 2) = Ax + 2Bx + 2A + 3B

(2x + 3)(x + 2)A + 2B = –2

2A + 4B = –42A + 3B = 1

– :B = –5

A + 2 × –5 = –2A = 8

∴ 1 – 2x(2x + 3)(x + 2)

= 82x + 3

– 5x + 2

q 3x2 – 6x + 2(x – 1)2(x + 2)

= Ax + 2

+ Bx – 1

+ C(x – 1)2

= A(x – 1)2 + B(x + 2)(x – 1) + C(x + 2)(x – 1)2(x + 2)

= Ax2 – 2Ax + A + Bx2 + Bx – 2B + Cx + 2C(x – 1)2(x + 2)

A + B = 34A + 4B = 12

–2A + B + C = –6A – 2B + 2C = 2 – :

5A – 4B = 14 + :

9A = 26A = 26

9269

+ B = 3

B = 19

– 529

+ 19

+ C = –6

C = – 13

∴ 3x2 – 6x + 2(x – 1)2(x + 2)

= 269(x + 2)

+ 19(x – 1)

– 13(x – 1)2


r 4(x – 1)2(2x + 1)

= A2x + 1

+ Bx – 1

+ C(x – 1)2

= A(x – 1)2 + B(2x + 1)(x – 1) + C(2x + 1)(x – 1)2(2x + 1)

= Ax2 – 2Ax + A + 2Bx2 – Bx – B + 2Cx + C(x – 1)2(2x + 1)

A + 2B = 0–2A – B + 2C = 0

A – B + C = 42A – 2B + 2C = 8 – :

4A – B = 88A – 2B = 16

+ :9A = 16

A = 169

169

+ 2B = 0

B = – 89

169

+ 89

+ C = 4

C = 43

∴ 4(x – 1)2(2x + 1)

= 169(2x + 1)

– 89(x – 1)

+ 43(x – 1)2

s Divide: x – 2

x2 – 4 x3 – 2x2 – 3x + 9 x3 – 0x2 – 4x –2x2 + x – 2x2 + 8 x + 1

x3 – 2x2 – 3x + 9x2 – 4

= x – 2 + x + 1x2 – 4

x + 1

(x + 2)(x – 2)= A

x + 2+ B

x – 2 = A(x – 2) + B(x + 2)

(x + 2)(x – 2) = Ax + Bx – 2A + 2B

(x + 2)(x – 2)

A + B = 12A + 2B = 2

–2A + 2B = 1 + :

4B = 3B = 3

4A + 3

4= 1

A = 14

∴ x + 1(x + 2)(x – 2)

= 14(x + 2)

+ 34(x – 2)

x3 – 2x2 – 3x + 9x2 – 4

= x – 2 + 14(x + 2)

+ 34(x – 2)


u 2x – 1(x + 1)(3x + 2)

= Ax + 1

+ B3x + 2

= A(3x + 2) + B(x + 1)(x + 1)(3x + 2)

= 3Ax + Bx + 2A + B(x + 1)(3x + 2)

3A + B = 22A + B = –1

– : A = 39 + B = 2

B = –7∴ 2x – 1

(x + 1)(3x + 2)= 3

x + 1– 7

3x + 2

t Divide: x

x2 – 1 x3 + 3 x3 – x x + 3

x3 + 3(x + 1)(x – 1)

= x + x + 3(x + 1)(x – 1)

x + 3(x + 1)(x – 1)

= Ax + 1

+ Bx – 1

= A(x – 1) + B(x + 1)(x + 1)(x – 1)

= Ax + Bx – A + B(x + 1)(x – 1)

A + B = 1– A + B = 3

+ :2B = 4

B = 2A + 2 = 1

A = –1∴ x + 3

(x + 1)(x – 1)= – 1

x + 1+ 2

x – 1x3 + 3

(x + 1)(x – 1)= x – 1

x + 1+ 2

x – 1


Exercise 6D Solutions 1 a A simple start is often to subtract the

equations. x2 – x = 0

x(x – 1) = 0x = 0 or x = 1

If x = 0, y = 0 If x = 1, y = 1 The points of intersection are (0, 0) and (1, 1). b Subtract the equations: 2x2 – x = 0

x(2x – 1) = 0x = 0 or x = 1

2

If x = 0, y = 0 If x = 1

2, y = 1

2

The points of intersection are (0, 0) and (1

2, 1

2).

c Subtract the equations: x2 – 3x – 1 = 0

x = 3 ± 9 – 4 × 1 × –12

= 3 ± 132

= 3 + 132

or 3 – 132

If x = 3 + 132

, y = 2 × 3 + 132

+ 1

= 4 + 13

If x = 3 – 132

, y = 2 × 3 – 132

+ 1

= 4 – 13

The points of intersection are

⎝⎜⎛ 3 + 13

2 , 4 + 13

⎠⎟⎞ and

⎝⎜⎛ 3 – 13

2 , 4 – 13

⎠⎟⎞ .

2 a Substitute y = 16 – x into x2 + y2 = 178. x2 + (16 – x)2 = 178

x2 + 256 – 32x + x2 = 1782x2 – 32x + 78 = 0

x2 – 16x + 39 = 0(x – 3)(x – 13) = 0

x = 3 or x = 13

If x = 3, y = 16 – x = 13 If x = 13, y = 16 – x = 3 The points of intersection are (3, 13) and (13, 3). b Substitute y = 15 – x into x2 + y2 = 125.

x2 + (15 – x)2 = 125x2 + 225 – 30x + x2 = 125

2x2 – 30x + 100 = 0x2 – 15x + 50 = 0

(x – 5)(x – 10) = 0x = 5 or x = 10

If x = 5, y = 15 – x = 10 If x = 10, y = 15 – x = 5 The points of intersection are (5, 10) and (10, 5). c Substitute y = x – 3 into x 2 + y2 = 185.

x2 + (x – 3)2 = 185

x2 + x2 – 6x + 9 = 1852x2 – 6x – 176 = 0

x2 – 3x – 88 = 0(x – 11)(x + 8) = 0

x = 11 or x = –8

If x = 11, y = x – 3 = 8 If x = –8, y = x – 3 = –11 The points of intersection are (11, 8) and (–8, –11). d Substitute y = 13 – x into x2 + y2 = 97.

x2 + (13 – x)2 = 97x2 + 169 – 26x + x2 = 97

2x2 – 26x + 72 = 0x2 – 13x + 36 = 0(x – 4)(x – 9) = 0

x = 4 or x = 9

If x = 4, y = 13 – x = 9 If x = 9, y = 13 – x = 4 The points of intersection are (4, 9) and (9, 4).


e Substitute y = x – 4 into x 2 + y2 = 106. x2 + (x – 4)2 = 106

x2 + x2 – 8x + 16 = 1062x2 – 8x – 90 = 0

x2 – 4x – 45 = 0(x – 9)(x + 5) = 0

x = 9 or x = –5

If x = 9, y = x – 4 = 5 If x = –5, y = x – 4 = –9 The points of intersection are (9, 5) and (–5, –9). 3 a Substitute y = 28 – x into xy = 187.

x(28 – x) = 18728x – x2 = 187

x2 – 28x + 187 = 0(x – 11)(x – 17) = 0

x = 11 or x = 17 If x = 11, y = 28 – x = 17 If x = 17, y = 28 – x = 11

The points of intersection are (11, 17) and (17, 11).

a Substitute y = 51 – x into xy = 518. x(51 – x) = 518

51x – x2 = 518x2 – 51x + 518 = 0

(x – 14)(x – 37) = 0x = 14 or x = 37

If x = 14, y = 51 – x = 37 If x = 37, y = 51 – x = 14

The points of intersection are (14, 37) and (37, 14).

c Substitute y = x – 5 into xy = 126. x(x – 5) = 126

x2 – 5x = 126x2 – 5x – 126 = 0

(x – 14)(x + 9) = 0x = 14 or x = –9

If x = 14, y = x – 5 = 9 If x = –9, y = x – 5 = –14

The points of intersection are (14, 9) and (–9, –14).

4 Substitute y = 2x into the circle. (x – 5)2 + (2x)2 = 25

x2 – 10x + 25 + 4x2 = 255x2 – 10x = 05x(x – 2) = 0

x = 0 or x = 2

If x = 0, y = 2x = 0 If x = 2, y = 2x = 4

The points of intersection are (0, 0) and (2, 4). 5 Substitute y = x into the circle. x = 1

x – 2+ 3

x(x – 2) = 1 + 3(x – 2)x2 – 2x = 1 + 3x – 6

x2 – 5x + 5 = 0

x = 5 ± 25 – 4 × 1 × 52

= 5 ± 52

= 5 + 52

or 5 – 52

Since y = x, the points of intersection are

⎝⎜⎛ 5 + 5

2 , 5 + 5

2 ⎠⎟⎞ and

⎝⎜⎛ 5 – 5

2 , 5 – 5

2 ⎠⎟⎞ .

6 Substitute x = 3y into the circle. 9y2 + y2 – 30y – 5y + 25 = 0

10y2 – 35y + 25 = 02y2 – 7y + 5 = 0

(2y – 5)(y – 1) = 0y = 5

2 or y = 1

If y = 52

, x = 3y = 152

If y = 1, x = 3y = 3 The points of intersection are ( 15

2, 5

2)

and (3, 1).


7 Make y the subject in y4

– x5

= 1.

y4

= x5

+ 1

y = 4x5

+ 4

Substitute into x 2 + 4x + y2 = 12.

x2 + 4x +⎝⎜⎛4x

5+ 4

⎠⎟⎞2

= 12

x2 + 4x + 16x2

25+ 32x

5+ 16 = 12

25x2 + 100x + 16x2 + 160x + 400 = 30041x2 + 260x + 100 = 0

x = –260 ± 67 600 – 4 × 41 × 10082

= –260 ± 51 20082

= –260 ± 25 600 × 282

= –260 ± 160 282

= –130 ± 80 241

If x = –130 + 80 241

,

y = 4 × (–130 + 80 2 )5 × 41

+ 4

= 4 × (–26 + 16 2 )41

+ 4 × 4141

= –104 + 64 2 + 16441

= 60 + 64 241

Likewise, if x = –130 – 80 241

,

y = 60 – 64 241


⎝⎜⎛ –130 + 80 2

41 , 60 + 64 2

41 ⎠⎟⎞ and

⎝⎜⎛ –130 – 80 2

41 , 60 – 64 2

41 ⎠⎟⎞ .

8 Subtract the second equation from the first.

1x + 2

– 3 + x = 0

1 – 3(x + 2) + x(x + 2) = 01 – 3x – 6 + x2 + 2x = 0

x2 – x – 5 = 0

x = 1 ± 1 – 4 × 1 × –52

= 1± 212

If x = 1 + 212

, y = – x = –1 – 212

If x = 1 – 212

, y = – x = –1 + 212


⎝⎜⎛ 1 + 21

2 , –1 – 21

2 ⎠⎟⎞ and

⎝⎜⎛ 1 – 21

2 , –1 + 21

2 ⎠⎟⎞

9 Substitute y = 9x + 4

4 into the parabola.

⎝⎜⎛9x + 4

4 ⎠⎟⎞2

= 9x

(9x + 4)2

16= 9x

(9x + 4)2 = 9x × 1681x2 + 72x + 16 = 144x81x2 – 72x + 16 = 0

(9x – 4)2 = 0x = 4

9y = 9x + 4

4 = 4 + 4

4= 2

⎝⎜⎛4

9 , 2

⎠⎟⎞

Note: Substitute into the linear equation, as substituting into the parabola introduces a second answer that is not actually a solution.


12 Substitute y = x – 1 into y = 2x – 2

. 10 Substitute y = 2x + 3 5 into the circle.

x2 + (2x + 3 5 )2 = 9x2 + 4x2 + 12 5x + 45 = 9

5x2 + 12 5x + 36 = 0

x2 + 12 55

x + 365

= 0

x2 + 2 × 6 55

x + (6 5 )2

25= 0

⎝⎜⎛ x + 6 5

5 ⎠⎟⎞2

= 0

x = – 6 55

y = 2x + 3 5

= – 12 55

+ 15 55

= 3 55

⎝⎜⎛ – 6 5

5 , 3 5

5 ⎠⎟⎞

x – 1 = 2x – 2

(x – 1)(x – 2) = 2x2 – 3x + 2 = 2

x2 – 3x = 0x(x – 3) = 0

x = 0 or x = 3

If x = 0, y = x – 1 = –1 If x = 3, y = x – 1 = 2 The points of intersection are (0, –1) and (3, 2).

11 Substitute y = 1

4 x + 1 into y = – 1

x.

14 x + 1 = – 1

xx + 4

4= – 1

xx(x + 4) = –4

x2 + 4x + 4 = 0(x + 2)2 = 0

x = –2y = – 1

x = 1

2

⎝⎜⎛–2, 1

2⎠⎟⎞


Solutions to Multiple-choice Questions 1 x2 = (x + 1)2 + b(x + 1) + c

= x2 + 2x + 1 + bx + b + cb + 2 = 0

b = –2b + c + 1 = 0

c = 1 C 2 x3 = a(x + 2)3 + b(x + 2)2

+ c(x + 2) + d = ax3 + 6ax2 + 12ax + 8a

+ bx2 + 4bx + 4b + cx + 2c + d

a = 1b + 6a = 0

b = –612a + 4b + c = 0

c = 128a + 4b + 2c + d = 0

d = –8 D 3 a = 3, b = –6, c = 3

x = 6 ± 36 – 4 × 3 × 32 × 3

= 6 ± 06

= 1 D 4 (x – 4)(x + 6) = 0

x2 + 2x – 24 = 0x2 + 2x = 24

2 C x2 + 4x = 48 5 3

x + 4– 5

x – 2= 3(x – 2) – 5(x + 4)

(x + 4)(x – 2) = 3x – 6 – 5x – 20

(x + 4)(x – 2) = –2x – 26

(x + 4)(x – 2)

= –2(x + 13)(x + 4)(x – 2)

E

6 4(x + 3)2 + 2x

x + 1= 4(x + 1) + 2x(x + 3)2

(x + 3)2(x + 1)

= 4x + 4 + 2x3 + 12x2 + 18x(x + 3)2(x + 1)

= 2x3 + 12x + 22x + 4(x + 3)2(x + 1)

= 2(x3 + 6x2 + 11x + 2)(x + 3)2(x + 1)

E

7 7x2 + 13(x – 1)(x2 + x + 2)

= ax – 1

+ bx + cx2 + x + 2

= a(x2 + x + 2) + (bx + c)(x – 1)(x – 1)(x2 + x + 2)

= ax2 + ax + 2a + bx2 – bx + cx – c(x – 1)(x2 + x + 2)

a + b = 7a – b + c = 0

2a – c = 13 + :

3a – b = 13 + :

4a = 20a = 5

5 + b = 7b = 2

a – b + c = 0

c = –3 C 8 4x – 3

(x – 3)2 = Ax – 3

+ B(x – 3)2

= A(x – 3) + B(x – 3)2

= Ax – 3A + B(x – 3)2

A = 4–3 × 4 + B = –3

B = 9

∴ 4x – 3(x – 3)2 = 4

x – 3+ 9

(x – 3)2 D


10 –3x2 + 2x – 1(x2 + 1)(x + 1)

= Ax + Bx2 + 1

+ Cx + 1

= (Ax + B)(x + 1) + C(x2 + 1)(x2 + 1)(x + 1)

= Ax2 + Ax + Bx + B + Cx2 + C(x2 + 1)(x + 1)

A + C = –3A + B = 2B + C = –1

– :C – B = –5

2C = –6C = –3

A + – 3 = –3A = 0

0 + B = 2B = 2

9 2x2 + 5x + 2 = (2x + 1)(x + 2) 8x + 7

(2x + 1)(x + 2)= A

2x + 1+ B

x + 2 = A(x + 2) + B(2x + 1)

(2x + 1)(x + 2) = Ax + 2Bx + 2A + B

(2x + 1)(x + 2)A + 2B = 8

2A + 4B = 162A + B = 7

– :3B = 9

B = 3A + 2B = 8

A = 2

∴ 8x + 7(2x + 1)(x + 2)

= 22x + 1

+ 3x + 2

B

∴ –3x + 2x + 5(x2 + 1)(x + 1)

= 2x2 + 1

– 3x + 1

B


Solutions to Short-answer Questions 1 3a + b = 11

6a + 2b = 22a – 2b = –1

+ :7a = 21

a = 33 × 3 + b = 11

b = 22 + 2c = 4

c = 1 2 x3 = (x – 1)3 + a(x – 1)2 + b(x – 1) + c

= x3 – 3x2 + 3x – 1 + ax2

– 2ax + a + bx – b + c

–3 + a = 0a = 3

3 – 2 × 3 + b = 0b = 3

–1 + 3 – 3 + c = 0c = 1

∴ x3 = (x – 1)3 + 3(x – 1)1

+ 3(x – 1) + 1 3 (x + 1)2(px + q)

= (x2 + 2x + 1)(px + q) = px3 + (q + 2p)x2 + (p + 2q)x + q

a = pb = q + 2pc = p + 2qd = q

2a + d = 2p + q = ba + 2d = p + 2q = c

4 (x – 2)2(px + q) = (x2 – 4x + 4)(px + q)

= px3 + (q – 4p)x2

+ (4p – 4q)x + 4qa = pb = q – 4pc = 4p – 4qd = 4q

–4a + 14

d = –4p + q = b

4a – d = 4p – 4q = c

5 a x2 + x – 12 = 0

(x + 4)(x – 3) = 0x = –4 or x = 3

b x2 – x – 2 = 0

(x + 1)(x – 2) = 0x = –1 or 2

c x2 – 3x – 11 = –1

x2 – 3x – 10 = 0(x + 5)(x – 2) = 0

x = –5 or x = 2 d 2x2 – 4x + 1 = 0

x = 4 ± 16 – 4 × 2 × 14

= 4 ± 84

= 2 ± 22

e 3x2 – 2x + 5 – t = 0

x = 2 ± 4 – 4 × 3 × (5 – t)6

= 2 ± 4 – 60 + 12t6

= 2 ± 12t – 566

= 2 ± 4(3t – 14)6

= 2 ± 2 3t – 146

= 1 ± 3t – 143

f tx2 – tx + 4 = 0

x = t ± t2 – 4 × t × 42t

= t ± t2 – 16t2t


c 7 – x(x – 3)(x + 5)

= Ax – 3

+ Bx + 5

= A(x + 5) + B(x – 3)(x – 3)(x + 5)

= Ax + Bx + 5A – 3B(x – 3)(x + 5)

A + B = –13A + 3B = –35A – 3B = 7

+ :8A = 4

A = 12

12

+ B = –1

B = – 32

∴ 7 – x(x – 3)(x + 5)

= 12(x – 3)

– 32(x + 5)

6 2(x + 2) – 3(x – 1)(x – 1)(x + 2)

= 12

2(2x + 4 – 3x + 3) = (x – 1)(x + 2)2( – x + 7) = x2 + x – 2

–2x + 14 = x2 + x – 2x2 + 3x – 16 = 0

a = 1, b = 3, c = –16

x = –3 ± 9 – 4 × 1 × –162

= –3 ± 732

7 a –3x + 4

(x – 3)(x + 2)= A

x – 3+ B

x + 2 = A(x + 2) + B(x – 3)

(x – 3)(x + 2) = Ax + Bx + 2A – 3B

(x – 3)(x + 2)A + B = –3

3A + 3B = –92A – 3B = 4

+ :5A = –5

A = –1–1 + B = –3

B = –2∴ –3x + 4

(x – 3)(x + 2)= – 1

x – 3– 2

x + 2

d 3x – 9

(x – 5)(x + 1)= A

x – 5+ B

x + 1 = A(x + 1) + B(x – 5)

(x – 5)(x + 1) = Ax + Bx + A – 5B

(x – 5)(x + 1)A + B = 3

5A + 5B = 15A – 5B = –9

+ :6A = 6

A = 11 + B = 3

B = 2∴ 3x – 9

(x – 5)(x + 1)= 1

x – 5+ 2

x + 1

b 7x + 2

(x + 2)(x – 2)= A

x + 2+ B

x – 2 = A(x – 2) + B(x + 2)

(x + 2)(x – 2) = Ax + Bx – 2A + 2B

(x + 2)(x – 2)A + B = 7

2A + 2B = 14–2A + 2B = 2

+ :4B = 16

B = 4A + 4 = 7

A = 3∴ 7x + 2

(x + 2)(x – 2)= 3

x + 2+ 4

x – 2


e 3x – 4(x + 3)(x + 2)2 = A

x + 3+ B

x + 2+ C

(x + 2)2

= A(x + 2)2 + B(x + 3)(x + 2) + C(x + 3)(x + 3)(x + 2)2

= Ax2 + 4Ax + 4A + Bx2 + 5Bx + 6B + Cx + 3C(x + 3)(x + 2)2

A + B = 08A + 8B = 0

4A + 5B + C = 312A + 15B + 3C = 9

4A + 6B + 3C = –4 – :

8A + 9B = 13 – :

B = 13A + 13 = 0

A = –134 × –13 + 5 × 13 + C = 3

C = –10∴ 3x – 4

(x + 3)(x + 2)2 = – 13x + 3

+ 13x + 2

– 10(x + 2)2

f 6x2 – 5x – 16(x – 1)2(x + 4)

= Ax + 4

+ Bx – 1

+ C(x – 1)2

= A(x – 1)2 + B(x + 4)(x – 1) + C(x + 4)(x – 1)2(x + 4)

= Ax2 – 2Ax + A + Bx2 + 3Bx – 4B + Cx + 4C(x – 1)2(x + 4)

A + B = 616A + 16B = 96

–2A + 3B + C = –5–8A + 12B + 4C = –20

A – 4B + 4C = –16 – :

9A – 16B = 4 + :

25A = 100A = 4

4 + B = 6B = 2

–2 × 4 + 3 × 2 + C = –5C = –3

∴ 6x2 – 5x – 16(x – 1)2(x + 4)

= 4x + 4

+ 2x – 1

– 3(x – 1)2


g x2 – 6x – 4(x2 + 2)(x + 1)

= Ax + Bx2 + 2

+ Cx + 1

= (Ax + B)(x + 1) + C(x2 + 2)(x2 + 2)(x + 1)

= Ax2 + Ax + Bx + B + Cx2 + 2C(x2 + 2)(x + 1)

A + C = 1A + B = –6

B + 2C = –4 – : C – B = 7

+ : 3C = 3C = 1

A + 1 = 1A = 0

0 + B = –6B = –6

∴ x2 – 6x – 4(x2 + 2)(x + 1)

= 1x + 1

– 6x2 + 2

h – x + 4

(x – 1)(x2 + x + 1)= A

x – 1+ Bx + C

x2 + x + 1

= A(x2 + x + 1) + (Bx + C)(x – 1)(x – 1)(x2 + x + 1)

= Ax2 + Ax + A + Bx2 – Bx + Cx – C(x – 1)(x2 + x + 1)

A + B = 0A – B + C = –1

A – C = 4 + : 2A – B = 3 + : 3A = 3

A = 1B = –1

1 – C = 4C = –3

∴ – x + 4(x – 1)(x2 + x + 1)

= 1x – 1

– x + 3x2 + x + 1


i –4x + 5(x + 4)(x – 3)

= Ax + 4

+ Bx – 3

= A(x – 3) + B(x + 4)(x + 4)(x – 3)

= Ax + Bx – 3A + 4B(x + 4)(x – 3)

A + B = –43A + 3B = –12

–3A + 4B = 5 + : 7B = 7

B = –1A – 1 = –4

A = –3∴ –4x + 5

(x + 4)(x – 3)= – 3

x + 4– 1

x – 3 = 1

3 – x– 3

x + 4

j –2x + 8(x + 4)(x – 3)

= Ax + 4

+ Bx – 3

= A(x – 3) + B(x + 4)(x + 4)(x – 3)

= Ax + Bx – 3A + 4B(x + 4)(x – 3)

A + B = –23A + 3B = –6

–3A + 4B = 8 + : 7B = 2

B = 27

A + 27

= –2

A = – 167

∴ –2x + 8(x + 4)(x – 3)

= 27(x – 3)

– 167(x + 4)

8

a 14x – 28(x – 3)(x2 + x + 2)

= Ax – 3

+ Bx + Cx2 + x + 2

= A(x2 + x + 2) + (Bx + C)(x – 3)(x – 3)(x2 + x + 2)

= Ax2 + Ax + 2A + Bx2 – 3Bx + Cx – 3C(x – 3)(x2 + x + 2)

A + B = 09A + 9B = 0

A – 3B + C = 143A – 9B + 3C = 42

2A – 3C = –28 + : 5A – 9B = 14 + : 14A = 14

A = 11 + B = 0

B = –11 – 3 × –3 + C = 14

C = 10∴ 14x – 28

(x – 3)(x2 + x + 2)= 1

x – 3+ – x + 10

x2 + x + 2 = 1

x – 3– x – 10

x2 + x + 2


b 1(x + 1)(x2 – x + 2)

= Ax + 1

+ Bx + Cx2 – x + 2

= A(x2 – x + 2) + (Bx + C)(x + 1)(x + 1)(x2 – x + 2)

= Ax2 – Ax + 2A + Bx2 + Bx + Cx + C(x + 1)(x2 – x + 2)

A + B = 0– A + B + C = 0

2A + C = 1 – : 3A – B = 1 + : 4A = 1

A = 14

14

+ B = 0

B = – 14

– 14

– 14

+ C = 0

C = 12

∴ 1(x + 1)(x2 – x + 2)

= 14(x + 1)

+ – x + 24(x2 – x + 2)

= 14(x + 1)

– x – 24(x2 – x + 2)

c First divide 3 x3 by x2 – 5x + 4. 3x + 15

x2 – 5x + 4 3x3 3x3 – 15x2 + 12x

15x2 – 12x 15x2 – 75x + 60 63x – 60

3x3

x2 – 5x + 4= 3x + 15 + 63x – 60

(x – 4)(x – 1) (factorising the denominator)

63x – 60(x – 4)(x – 1)

= Ax – 4

+ Bx – 1

= A(x – 1) + B(x – 4)(x – 4)(x – 1)

= Ax + Bx – A – 4B(x – 4)(x – 1)

A + B = 63– A – 4B = –60

+ : – 3B = 3B = –1

A – 1 = 63A = 64

∴ 63x – 60(x – 4)(x – 1)

= 64x – 4

– 1x – 1

3x3

x2 – 5x + 4= 3x + 15 + 64

x – 4– 1

x – 1


10 Substitute x = 3y – 1 into the circle. 9 a x2 = – x

x2 + x = 0x(x + 1) = 0

x = 0 or x = –1

(3y – 1)2 + 2(3y – 1) + y2 = 99y2 – 6y + 1 + 6y – 2 + y2 = 9

10y2 – 10 = 0y2 – 1 = 0

(y + 1)(y – 1) = 0y = 1 or y = –1

If x = 0, y = 0 If x = –1, y = 1 The points of intersections are (0, 0) If y = –1, x = –4 and (–1, 1). If y = 1, x = 2 The points of intersections are (2, 1) and (–4, –1). b Substitute y = 4 – x into x 2 + y2 = 16. x2 + (4 – x)2 = 16

x2 + 16 – 8x + x2 = 162x2 – 8x = 0

x2 – 4x = 0x(x – 4) = 0

x = 0 or x = 4

If x = 0, y = 4 If x = 4, y = 0 The points of intersections are (0, 4) and (4, 0). c Substitute y = 5 – x into xy = 4. x(5 – x) = 4

5x – x2 – 4 = 0x2 – 5x + 4 = 0

(x – 4)(x – 1) = 0x = 4 or x = 1

If x = 4, y = 1 If x = 1, y = 4 The points of intersections are (4, 1) and (1, 4).


Chapter 7 – Revision of chapters 1–6 Solutions to Multiple-choice Questions 1 PP = 4I 1

4PP = I

( 14

P)P = I

∴ the inverse of P is 14

P A

2 RS = [5 × 0 + 3 × –1 + 2 × 1]

= [0 – 3 + 2] = [–1] B 3 det(A) = 9 × 5 – 8 × –11 = 133 E 4 A is a m × n matrix = 3 × 1 matrix. B is a n × r matrix = 1 × 3 matrix. AB is therefore a m × r matrix = 1 × 1 matrix A 5 AX = C – B =

⎣⎢⎡ 3

252⎦⎥⎤

det(A) = 5 × 1 – 2 × 2 = 1 A–1 = 1

1 ⎣⎢⎡ 1

–2–25⎦⎥⎤ =

⎣⎢⎡ 1

–2–25⎦⎥⎤

X = A–1AX =

⎣⎢⎡ 1

–2–25⎦⎥⎤ ⎣⎢⎡ 3

252⎦⎥⎤

B =⎣⎢⎡ –1

410⎦⎥⎤

6 PQR =

⎣⎢⎡ 2

3–12⎦⎥⎤⎣⎢⎡ 4

625⎦⎥⎤⎣⎢⎡ 2

–312⎦⎥⎤

=⎣⎢⎡ 2

24–116⎦⎥⎤⎣⎢⎡ 2

–312⎦⎥⎤

=⎣⎢⎡ 8

00

56⎦⎥⎤

There are two zero elements. C Note that since the associative law applies for matrix multiplication, it does not matter which two matrices are multiplied first, as long as the order does not change.

7 det(S) = 3 × –2 – 5 × –1 = –1 S–1 = 1

–1 ⎣⎢⎡ –2

1–53⎦⎥⎤

=⎣⎢⎡ 2

–15

–3⎦⎥⎤ A

8 det = ad – bc = 4 × 4 – 6 × 2 = 4 B 9 det(X) = 5 × 2 – 7 × 2 = –4 X–1 = 1

–4 ⎣⎢⎡ 2

–2–75⎦⎥⎤

= 14 ⎣⎢⎡ –2

27

–5⎦⎥⎤ D

10 5 is seven less than 3 times (x + 1) 5 = 3 × (x + 1) – 7 5 = 3x + 3 – 7 5 = 3x – 4 E 11 3

x – 3– 2

x + 3= 3(x + 3) – 2(x – 3)

(x – 3)(x + 3) = 3x + 9 – 2x + 6

x2 – 9

= x + 15x2 – 9

B

12 p = kx

y2

Set both x and y = 1 so that p = k. When x and y are decreased,

p = k × 0.70.82 = 1.09375k

This has increased by approximately 9.4%. C


13 This is an arithmetic progression with a = 1 and d = 2.

Sn = n2 (2 + 2(n – 1)) = 100

n(2 + 2n – 2) = 200n2 = 100n = 10 termstn = 1 + 2 × (10 – 1)

= 19 D 14 a = S1 = 22 – 2 = 2

S2 = 23 – 2 = 6t2 = S2 – S1 = 4r = t2

t1= 2

tn = arn – 1

= 2 × 2n – 1

B = 2n

15 m = kn 9 = 4k k = 9

4 A

16 A ∩ (B U C) = A ∩ {1, 2, 3, 4, 5, 6, 7} = {2, 3, 4} C 17 In the case of the tank, P = krh. When r = 5 and h = 4, P = 60. 60 = 5 × 4 × k

k = 6020

= 3

When r = 4 and h = 6, P = 3 × 4 × 6 = $72 E 18 x = k y 8 = 2k k = 4 x = 4 × 7 = 28 D 19 0.

.7

.2 = 0.727272. . .

0..7

.2 × 100 = 72.7272. . .

0..7

.2 × 99 = 72

0..7

.2 = 72

99 C

20 x = ky2

x

Set both x and y = 1 so that p = k. When y and z are changed,

p = k × 1.252

0.8= 1.953125k

This has increased by approximately 95%. D 21 –4

x – 1– 3

1 – x+ x

x – 1= –4

x – 1+ 3

x – 1+ x

x – 1 = x – 1

x – 1

= 1 A 22 x + 2

3– 5

6= 2x + 4

6– 5

6

= 2x – 16

C

23 a – 1 = 1

1 + b1

a – 1= 1 + b

1a – 1

– 1 = b

b = 1a – 1

– 1 E

24 0.

.3

.6 = 0.363636. . .

0..3

.6 × 100 = 36.3636. . .

0..3

.6 × 99 = 36

0..3

.6 = 36

99= 4

11

Numerator + denominator = 4 + 11 = 15 A 25 Multiply both sides by 4(2x + y). 4(2x – y) = 3(2x + y)

8x – 4y = 6x + 3y8x – 6x = 3y + 4y

2x = 7y2x2y

= 7y2y

xy

= 72

B


26 a = 12 , r = – 1

2

S∞ = a1 – r

=

12

1 – – 12

=1232

= 13

D

27 x = ky

z2

When y = 4 and z = 14, x = 10. 10 = 4k

142

k = 10 × 142

4= 490

When y = 16 and z = 7,

x = 490 × 1672 = 160 B

28 Multiply both sides by (3 + y). 3 = 4(3 + y)

3 = 12 + 4y–9 = 4y

y = – 94

B

29 Multiply the first equation by 5, then subtract. 15x + 5y = –35

2x + 5y = 4 – :

13x = –39x = –3

3 × –3 + y = –7y = 2

(–3, 2) B 30 Multiply both sides by 4.

(m + 2) – (2 – m) = 2m + 2 – 2 + m = 2

2m = 2 m = 1 A

31 2 46 200 2 23 100 2 11 550 3 5575 5 1925 5 385 7 7711 11 1

D = 23 × 3 × 52 × 7 × 11 32 The difference between terms is constant. (y – 1) – y = (2y – 1) – (y – 1)

y – 1 – y = 2y – 1 – y + 1–1 = y

y = –1 A 33 Order is n – 6, n – 5, n – 1, n + 1, n + 4. Middle number is n – 1. B 34 x ∝ 1

yy = 5yx ∝ 1

5y

x = x5

D

35 t4 = a + 3d

= 3 + 3d = 93d = 6

d = 2t11 = a + (n – 1)d

= 3 + 10 × 2 = 23 A 36 4

n + 1+ 3

n – 1= 4(n – 1) + 3(n + 1)

(n + 1)(n – 1) = 4n – 4 + 3n + 3

n2 – 1 = 7n – 1

n2 – 1 = 7n – 1

n2 – 1× –1

–1

= 1 – 7n1 – n2 B


37 Let the first number be x, so the numbers are x, 2x and x

2.

2x + x + x2

= 28

4x + 2x + x = 567x = 56x = 8

(8, 16, 4) A 38 ( 7 + 3)( 7 – 3) = 7 – 9 = –2 A 39 2x2 – 9x + 4 = (x – 4)(2x – 1) 13x – 10

(x – 4)(2x – 1)= P

x – 4+ Q

2x – 1 = P(2x – 1) + Q(x – 4)

(x – 4)(2x – 1)

= 2Px + Qx – P – 4Q(x – 4)(2x – 1)

2P + Q = 13– P – 4Q = –10

–2P – 8Q = –20 + :

–7Q = –7Q = 1

2P + 1 = 132P = 12

P = 6 C 40 a

1 – r= 4a

Multiply both sides by 1 – ra

.

1 = 4(1 – r)1 = 4 – 4r

4r = 4 – 1 r = 3

4 C

41 5x(x + 2)(x – 3)

= Px + 2

+ Qx – 3

= P(x – 3) + Q(x + 2)(x + 2)(x – 3)

= Px + Qx – 3P + 2Q(x + 2)(x – 3)

P + Q = 53P + 3Q = 15

–3P + 2Q = 0 + :

5Q = 15Q = 3

P + 3 = 5

P = 2 A 42 Assuming n is an integer, and , n = m2

then the next largest perfect square is ( m + 1)2. (m + 1)2 = m2 + 2m + 1 Since n = m2, m = n

(m + 1)2 = n + 2 n + 1 The next largest perfect square is n + 2 n + 1. E 43 A = kb and A = 14 when b = 2.4

14 = 2.4kk = 14

2.4 = 140

24= 35

6A = 35b

6

When A = 18,

18 = 35b6

b = 18 × 635

≈ 3.086 A 44 0.4 and 4.125 are terminating decimals. 3

8= 0.125

16 = 4

Only 5 cannot be expressed as a rational number. C


45 x = ba and y = 1

a – b

x + y = ba

+ 1a – b

= b(a – b) + aa(a – b)

= ba – b2 + aa(a – b)

C

46 The perfect square could be (3x – 2)2 or (3x + 2)2.

The middle term of the expansion would be –12x or 12x respectively.

This means m would be 3 or –3, i.e. ± 3. E 47 x = (n + 1)(n + 2)(n + 3), n > 0 When n = 1, x = (1 + 1) × (1 + 2) × (1 + 3)

= 2 × 3 × 4 = 12When n = 2, x = (2 + 1) × (2 + 2) × (2 + 3)

= 3 × 4 × 5 = 601, 2, 3 and 6 are factors in both equations, but not 5. D

48 There are 8 terms, a = –4 and t 8 = 10. a + 7d = 10

–4 + 7d = 107d = 14

d = 2

The required sum is S 7 – a.

S7 – a = 72

(–8 + 6 × 2) – –4

= 14 + 4 = 18 C 49 An odd number plus an odd number is

always an even number, so n + p. A (The other options all produce odd

numbers for all n and p.)


Chapter 8 – Transformations Exercise 8A Solutions 1 a Image = ((3 + 4), (1 + 2)) = (7, 3) b Image = ((4 + 2), (5 + 4)) = (6, 9) c Image = ((–2 + 4), (4 + 3)) = (2, 7) d Image = ((3 + –2), (2 + 3)) = (1, 5) e Image = ((4 + –3), (5 + 2)) = (1, 7) 2 a (1 + a, 2 + b) = (5, 3)

1 + a = 5 2 + b = 3a = 4 b = 1

Vector =⎣⎢⎡ 4

1⎦⎥⎤

b (3 + a, 8 + b) = (2, 9)

3 + a = 2 8 + b = 9a = –1 b = 1

Vector =⎣⎢⎡ –1

1⎦⎥⎤

c (1 + a, 2 + b) = (5, 4)

1 + a = 5 2 + b = 4a = 4 b = 2

Vector =⎣⎢⎡ 4

2⎦⎥⎤

d (–3 + a, 0 + b) = (4, 6)

–3 + a = 4 0 + b = 6a = 7 b = 6

Vector =⎣⎢⎡ 7

6⎦⎥⎤

e (–4 + a, – 3 + b) = (0, 0)

–4 + a = 0 – 3 + b = 0a = 4 b = 3

Vector =⎣⎢⎡ 4

3⎦⎥⎤

3 a (a, b) → (a + 2, b + 3) = (7, 9)

a + 2 = 7 b + 3 = 9a = 5 b = 6A = (5, 6)

b (a, b) → (a + 1, b + 4) = (3, 6)

a + 1 = 3 b + 4 = 6a = 2 b = 2A = (2, 2)

c (a, b) → (a + 2, b + 3) = (0, 6)

a + 2 = 0 b + 3 = 6a = –2 b = 3A = (–2, 3)

4 Compare a single point in the first

triangle to the corresponding point in the second triangle to find the vector.

a A = (2, 3); P = (2, –2) (2 + a, 3 + b) = (2, – 2)

2 + a = 2 3 + b = –2a = 0 b = –5

Vector =⎣⎢⎡ 0

–5⎦⎥⎤

b A = (2, 3); L = (–4, 2)

(2 + a, 3 + b) = (–4, 2)2 + a = –4 3 + b = 2

a = –6 b = –1Vector =

⎣⎢⎡ –6

–1⎦⎥⎤

c X = (–3, –3); A = (2, 3)

(–3 + a, – 3 + b) = (2, 3)–3 + a = 2 – 3 + b = 3

a = 5 b = 6Vector =

⎣⎢⎡ 5

6⎦⎥⎤

d A = (2, 3); A = (2, 3)

(2 + a, 3 + b) = (2, 3)2 + a = 2 3 + b = 3

a = 0 b = 0Vector =

⎣⎢⎡ 0

0⎦⎥⎤


5

a A = (2, 2); P = (1, –2) (2 + a, 2 + b) = (1, – 2)

2 + a = 1 2 + b = –2a = –1 b = –4


–4⎦⎥⎤

b P = (1, –2); A = (2, 2) (1 + a, – 2 + b) = (2, 2)

1 + a = 2 – 2 + b = 2a = 1 b = 4

Vector =⎣⎢⎡ 1

4⎦⎥⎤

c P = (1, –2); X = (–3, 1) (1 + a, – 2 + b) = (–3, 1)

1 + a = –3 – 2 + b = 1a = –4 b = 3


3⎦⎥⎤

d A = (2, 2); A = (2, 2) (2 + a, 2 + b) = (2, 2)

2 + a = 2 2 + b = 2a = 0 b = 0

Vector =⎣⎢⎡ 0

0⎦⎥⎤

6 a (2 + 3 + –1, 3 + 4 + 5) = (4, 12) b (–5 + 3 + –1, 6 + 4 + 5) = (–3, 15) c Add the two vectors:

⎣⎢⎡ 3

4⎦⎥⎤ +

⎣⎢⎡ –1

5⎦⎥⎤ =

⎣⎢⎡ 2

9⎦⎥⎤

7 a Image = ((1 + –5), (3 + 3)) = (–4, 6) b (a, b) → (a – 5, b + 3) = (6, 7)

a – 5 = 6 b + 3 = 7a = 11 b = 4

(11, 4) 8 (x, y) → (x + 1, y)a (0, 0) → (1, 0)

(1, 1) → (2, 1)(2, 4) → (3, 4)(3, 9) → (4, 9)

b

c Original graph: y = x2

Image: y = ( x – 1)2


Exercise 8B Solutions 1 A(1, 2) → A'(–1, 2)

B(3, 2) → B'(–3, 2)C(3, 5) → C'(–3, 5)

2 P(1, – 1) → P'(1, 1)

Q(5, – 1) → Q'(5, 1)R(4, 0) → R'(4, 0)

3 W(–3, – 1) → W'(–1, – 3)

X(–3, – 2) → X'(–2, – 3)Y(–5, – 2) → Y'(–2, – 5)Z(–5, – 1) → Z'(–1, – 5)

4 A(2, 1) → A'(1, 2)B(5, 1) → B'(1, 5)C(7, 3) → C'(3, 7)D(4, 3) → D'(3, 4)

5 a A(–6, – 2) → A1(6, – 2)

B(–3, – 4) → B1(3, – 4)C(–2, – 1) → C1(2, – 1)

b A(–6, – 2) → A2(2, 6)

B(–3, – 4) → B2(4, 3)C(–2, – 1) → C2(1, 2)

c A(–6, – 2) → A3(–6, 2)

B(–3, – 4) → B3(–3, 4)C(–2, – 1) → C3(–2, 1)


6 a (x, y) → (y, x)

(6, – 2) → (–2, 6) b (x, y) → ( – x, y)

(6, – 2) → (–6, 2) c (x, y) → (x, – y)

(6, – 2) → (6, 2) 7 a (x, y) → (y, x)

(0, – 1) → (–1, 0) b (x, y) → ( – x, y)

(0, – 1) → (0, – 1) c (x, y) → (x, – y)

(0, – 1) → (0, 1) d (x, y) → ( – y, – x)

(0, – 1) → (1, 0)


Exercise 8C Solutions 1 a (x, y) → (x, 3y)

(1, 3) → (1, 3 × 3) = (1, 9) b (x, y) → (2x, y)

(1, 3) → (2 × 1, 3) = (2, 3) c (x, y) → (4x, y)

(1, 3) → (4 × 1, 3) = (4, 3) 2 The factor is 8

2= 4.

(x, y) → (x, 4y) 3 The factor is 9

3= 3.

(x, y) → (3x, y)

4 a (x, y) → (x, 3y)

A(0, 0) → A(0, 0)B(0, 1) → B'(0, 3)C(1, 1) → C'(1, 3)D(1, 0) → D(1, 0)

b (x, y) → (3x, y)

A(0, 0) → A(0, 0)B(0, 1) → B(0, 1)C(1, 1) → C''(3, 1)D(1, 0) → D''(3, 0)

5 a i A (0, 0) → A(0, 0)

B(3, 0) → B(3, 0)C(3, 4) → C'(3, 2)

ii A(0, 0) → A(0, 0)

B(3, 0) → B''⎝⎜⎛3

2, 0⎠⎟⎞

C(3, 4) → C''⎝⎜⎛3

2, 4⎠⎟⎞

b


Exercise 8D Solutions 1 a i ( x, y) → (x + 5, y + 6) ii No invariant points. b i ( x, y) → (x, 4y) ii The x-coordinate is invariant: {( x, 0):x ∈ R} c i (x, y) →

⎝⎜⎛1

3 x, y

⎠⎟⎞

ii The y-coordinate is invariant: {( 0, y):y ∈ R} d i ( x, y) → (x – 2, y + 3) ii No invariant points. e i ( x, y) → (y, x) ii The coordinates will be invariant when x = y: {( x, x):x ∈ R} f i ( x, y) → ( – x, y) ii The y-coordinate is invariant: {( 0, y):y ∈ R} 2 a (2, 3) → (3 – 2, 2 × 3 + 1) = (1, 7) b (x, y) → (3 – x, 2y + 1) = (–6, 12)

3 – x = –6 2y + 1 = 12x = 9 y = 11

2

⎝⎜⎛9, 11

2 ⎠⎟⎞

c (x, y) → (3 – x, 2y + 1) = (x, y)

3 – x = x 2y + 1 = yx = 3

2 y = –1

⎝⎜⎛3

2, – 1

⎠⎟⎞

3 (x, y) → (3 – y, 4 – x) = (x, y)3 – y = x 4 – x = yThese can be arranged as two simultaneous equations: –x – y = –3 –x – y = –4 No solution satisfies both equations, so an invariant point does not exist.

4 a (4, – 1) → (3 – 2 × 4, – 2 × –1 + 1) = (–5, 3) b (x, y) → (3 – 2x, – 2y + 1) = (7, 12)

3 – 2x = 7 – 2y + 1 = 12x = –2 y = – 11

2

⎝⎜⎛–2, – 11

2 ⎠⎟⎞

c (x, y) → (3 – 2x, – 2y + 1) = (x, y)

3 – 2x = x – 2y + 1 = yx = 1 y = 1

3

⎝⎜⎛1, 1

3⎠⎟⎞

5 a (–1, 3) → ( – (1), – 2 × 3) = (1, – 6) b (x, y) → ( – x, – 2y) = (0, 0)

– x = 0 – 2y = 0x = 0 y = 0

(0, 0) c (x, y) → ( – x, – 2y) = (x, y)

– x = x – 2y = yx = 0 y = 0

(0, 0)


Exercise 8E Solutions 1 a i ( x, y) → (x + 2, y + 3)

→ (x – 1, y + 1) ii (

x, y) → (x – 3, y – 2)

→ (x – 1, y + 1) b i ( x, y) → (x, 2y)

→ (2x, 2y) ii ( x, y) → (2x, y)

→ (2x, 2y) c i ( x, y) → (x + 4, y + 5)

→ (x + 4, 3(y + 5)) ii ( x, y) → (x, 3y)

→ (x + 4, 3y + 5) d i ( x, y) → (x – 1, y + 2)

→ ( – (x – 1), y + 2) ii ( x, y) → ( – x, y)

→ ( – x – 1, y + 2) e i ( x, y) → (y, x)

→ ( – x, – y) ii (

x, y) → ( – y, – x)

→ ( – x, – y)

f i Find the rule for reflection in x = 2.

x' = x – 2(x – 2)

= 4 – xy' = y

( x, y) → (4 – x, y)→ (4 – x, – y)

ii ( x, y) → (x, – y)→ (4 – x, – y)

g i (x, y) →⎝⎜⎛1

2 x, y

⎠⎟⎞

→⎝⎜⎛1

2 x – 1, y + 2

⎠⎟⎞

ii (x, y) → (x – 1, y + 2)

→⎝⎜⎛1

2(x – 1), y + 2

⎠⎟⎞

h i (x, y) → (2x, y)

→ (2x + 2, y – 3) ii ( x, y) → (x + 2, y – 3)

→ (2(x + 2), y – 3)


Exercise 8F Solutions 1 (x, y) → (x + 1, y + 4)

x' = x + 1 y' = y + 4x = x' – 1 y = y' – 4

a y = x becomes y' – 4 = x' – 1 y' = x' + 3 The equation of the image is y = x + 3.

b becomes y = x2 y' – 4 = (x' – 1)2

y' = (x' – 1)2 + 4

The equation of the image is y = (x – 1)2 + 4.

c y = 1

x becomes y' – 4 = 1

x' – 1 y' = 1

x' – 1+ 4

The equation of the image is

y = 1x – 1

+ 4.

d y = 1x2 becomes y' – 4 = 1

(x' – 1)2

y' = 1(x' – 1)2 + 4

The equation of the image is

y = 1

(x – 1)2 + 4.

e becomes

x2 + y2 = 1(x' – 1)2 + (y' – 4)2 = 1The equation of the image is

(x – 1)2 + (y – 4)2 = 1.

2 (x, y) → (x, 2y)

x' = x y' = 2yx = x' y = y'

2

a y = x becomes y'2

= x'

y' = 2x'

Image: y = 2x


b becomes y = x2 y'2

= (x')2

y' = 2(x')2

Image: y = 2x2

c y = 1

x becomes y'

2= 1

x'y' = 2

x'

Image: y = 2x

d y = 1

x2 becomes y'2

= 1(x')2

y' = 2(x')2

Image: y = 2x2

e becomes x2 + y2 = 1

(x')2 +⎝⎜⎛ y'

2 ⎠⎟⎞2

= 1

(x')2 + (y')2

4= 1

Image: x2 + y2

4= 1

3 (x, y) → ( – x, y)

→ ( – x + 3, y + 2)x' = x + 3 y' = y + 2x = 3 – x' y = y' – 2

a y = x becomes y' – 2 = 3 – x' y' = 5 – x' Image: y = 5 – x b y = x2 becomes y' – 2 = (3 – x')2

= (x' – 3)2

y' = (x' – 3)2 + 2

Image: y = (x – 3)2 + 2 c y = 1

x becomes y' – 2 = 1

3 – x'y' = 1

3 – x'+ 2

Image: y = 13 – x

+ 2

d y = 1

x2 becomes y' – 2 = 1(3 – x')2

= 1(x' – 3)2

y' = 1(x' – 3)2 + 2

Image: y = 1(x – 3)2 + 2



(3 – x')2 + (y' – 2)2 = 1(x' – 3)2 + (y' – 2)2 = 1

Image: ( x – 3)2 + (y – 2)2 = 1 4 (x, y) → (2x, y)

→ (2x – 3, y + 2)x' = 2x – 3 y' = y + 1x = x' + 3

2y = y' – 1

a y = x becomes y' – 1 = x' + 32

= x'2

+ 32

y' = x'2

+ 52

Image: y = 12 x + 5

2

b becomes y = x2 y' – 1 =⎝⎜⎛ x' + 3

2 ⎠⎟⎞2

= (x' + 3)2

4

y' = (x' + 3)2

4+ 1

Image: y = 14 (x + 3)2 + 1

c y = 1x

becomes y' – 1 = 2x' + 3

y' = 2x' + 3

+ 1

Image: y = 2x + 3

+ 1

d y = 1x2 becomes y' – 1 = 4

(x' + 3)2

y' = 4(x' + 3)2 + 1

Image: y = 4(x + 3)2 + 1


⎝⎜⎛ x' + 3

2 ⎠⎟⎞2

+ (y' – 1)2 = 1

(x' + 3)2

4+ (y' – 1)2 = 1

Image: 14 (x + 3)2 + (y – 1)2 = 1

5 (x, y) → (y, xy) → (y – 4, x + 6)

x' = y – 4 y' = x + 6y = x' + 4 x = y' – 6

y = 2x + 3 becomes x' + 4 = 2(y' – 6)x' + 4 = 2y' – 12 + 3

2y' = x + 4 + 12 – 3 = x' + 13

y' = 12

(x' + 13)

y = 12

(x + 13)

6 a (x, y) → (x + 2, y + 4)

x' = x + 2 y' = y + 4x = x' – 2 y = y' – 4

y = x + 2 becomes y' – 4 = (x' – 2) + 2 y' = x' + 4 {(x, y): y = x + 4}

b (x, y) → (x, – y)

x' = x y' = – yx = x' y = – y'

y = x + 2 becomes –y' = x' + 2 y' = –(x' + 2) {(x, y): y = –(x + 2)}


c (x, y) → (x, 4y)x' = x y' = 4yx = x' y = 1

4 y'

y = x + 2 becomes 14

y' = x' + 2

y' = 4(x' + 2)

{(x, y): y = 4(x + 2)}

d (x, y) → (y, x)

x' = y y' = xy = x' x = y'

y = x + 2 becomes x' = y' + 2 y' = x' – 2 {(x, y): y = x – 2}

e (x, y) → ( – x, y)

x' = – x y' = yx = – x' y = y'

y = x + 2 becomes y' = –x' + 2 {(x, y): y = –x + 2}

7 (x, y) → (x + 1, y + 4)x' = x + 1 y' = y + 4x = x' – 1 y = y' – 4

becomes x2 + y2 = 4 (x' – 1)2 + (y' – 4)2 = 4 {(x, y): (x – 1)2 + (y – 4)2 = 4}

8 a (x, y) → (x, – y)

→ (x, – 2y)x' = x y' = –2yx = x' y = – y'

2

y = becomes x2 – y'2

= (x')2

– y' = 2(x')2

{(x, y): y = –2x2} b (x, y) → (x + 5, y + 2)

→⎝⎜⎛1

2(x + 5), y + 2

⎠⎟⎞

x' = 12 (x + 5) y' = y + 2

x + 5 = 2x' y = y' – 2x = 2x' – 5

y = becomes x2 y' – 2 = (2x' – 5)2

y' = 2(2x' – 5)2 + 2

{(x, y): y = (2x – 5)2 + 2}


c (x, y) →⎝⎜⎛ x, 1

2 y⎠⎟⎞

→⎝⎜⎛ x + 5, 1

2 y + 2

⎠⎟⎞

x' = x + 5 y' = 12

y + 2

x = x' – 5' 2y' = y + 42 y = 2y' – 4 becomes 2y = x2 y' – 4 = (x' – 5)2

2y' = (x' – 5)2 + 4y' = 1

2 (x' – 5)2 + 2

⎩⎨⎧ (x, y): y = 1

2 (x – 5)2 + 2

⎭⎬⎫

d (x, y) → (x + 2, y + 1)

→ ( – (x + 2), y + 1)x' = – x – 2 y' = y + 1x = – x' – 2 y = y' – 1

becomes y = x2 y' – 1 = ( – x' – 2)2

= (x' + 2)2

y' = (x' + 2)2 + 1

{(x, y): y = (x + 2)2 + 1} e (x, y) → (y, x)

→ (y, x + 2)x' = y y' = x + 2y = x' x = y' – 2

becomes y = x2

x' = (y' – 2)2

y' – 2 = ± x 'y' = 2 ± x '

{(x, y): y = 2 ± x } 9 a

b

c

d

e

10 a (x, y) → (x + 1, y)

x' = x + 1 y' = yx = x' – 1 y = y'

y = 2x becomes y' = 2x' – 1

{(x, y): y = 2x – 1}

b

(x, y) → (x, y + 1)x' = x y' = y + 1x = x' y = y' – 1

y = 2x becomes y' – 1 = 2x'

y' = 2x' + 1

{(x, y): y = 2x + 1)}


c (x, y) → (2x, y)

→⎝⎜⎛2x, 1

3 y⎠⎟⎞

x' = 2x y' = 13

y

x = 12

x' y = 3y'

y = 2x becomes 3y' = 2x'2

y' = 13 2

x'2

⎩⎨⎧

(x, y): y = 13

2x2

⎭⎬⎫

d (x, y) →

⎝⎜⎛ x, 1

3 y⎠⎟⎞

→⎝⎜⎛2x, 1

3 y⎠⎟⎞

x' = 2x y' = 13

y

x = 12

x' y = 3y'

y = 2x becomes 3y' = 2x'2

y' = 13 2

x'2

⎩⎨⎧

(x, y): y = 13

2x2

⎭⎬⎫

e (x, y) → (x + 2, y + 4)

→ ( – (x + 2), y + 4)x' = – x – 2 y' = y + 4x = – x' – 2' y = y' – 4

y = 2x becomes y' = 2 – x' – 2

y' = 12x' + 2 + 4

⎩⎨⎧ (x, y): y = 1

2x + 2 + 4⎭⎬⎫

f (x, y) → (x, y + 2

→⎝⎜⎛1

2 x, y + 2

⎠⎟⎞

→⎝⎜⎛1

2 x, – (y + 2)

⎠⎟⎞

x' = 12 x y' = – y – 2

x = 2x' y = – y' – 2

y = 2x becomes – y' – 2 = 22x'

– y' = 22x' + 2

{(x, y): y = (–22x' + 2)}

11 a

b

c

d


e

f


Exercise 8G Solutions 1 a Rearrange the image equation: y' = 2x'2

y' = 2y(x, y) → (x, 2y)

A dilation of factor 2 from the x-axis. Alternatively: y' = 2x'2

y'2

= x'2

x' = y'2

(x, y) →⎝⎜⎛ x

2 , y

⎠⎟⎞

A dilation of factor 12

from the y-axis.

b x' + 2 = 2

x' = x – 2(x, y) → (x – 2, y).

A translation by ⎣⎢⎡ –2

0⎦⎥⎤ .

c (x, y) → (y, x) Reflection in the line y = x. d x' = 2x

x = x'2

(x, y) →⎝⎜⎛1

2 x, y

⎠⎟⎞


from the y-axis.

e Rearrange the image equation:

y'3

= 2x'

y' = 3y(x, y) → (x, 3y)

A dilation of factor 3 from the x-axis. f x' – 3 = x

x' = x + 3(x, y) → (x + 3, y)

A translation by ⎣⎢⎡ 3

0⎦⎥⎤ .

g (x, y) → ( – x, y) Reflection in the y-axis. h (x, y) → (x, – y) Reflection in the x-axis. 2 a Rearrange the image equation: y'

2= (x' – 3)2

x' – 3 = x y'2

= y

x' = x + 3 y' = 2y(x, y) → (x + 3, y)

→ (x + 3, 2y) or(x, y) → (x, 2y)

→ (x + 3, 2y)

A dilation of factor 2 from the x-axis, and a translation by

⎣⎢⎡ 3

0⎦⎥⎤ .

In this case, they can be in either order. b Rearrange the image equation:

y' + 32

= x'2

x' = x y' + 32

= y

x' = x y' = 2y – 3(x, y) → (x, 2y)

→ (x, 2y – 3)

A dilation of factor 2 from the x-axis, followed by a translation by

⎣⎢⎡ 0

–3⎦⎥⎤ .

c Rearrange the image equation:

y' – 12

= (x' – 3)2

x' – 3 = x y' – 12

= y

x' = x + 3 y' = 2y + 1(x, y) → (x, 2y)

→ (x + 3, 2y + 1)


⎣⎢⎡ 3

1⎦⎥⎤ .


d Rearrange the image equation: g Rearrange the image equation: y'

2= 1

x' – 3x' – 3 = x y'

2= y

x' = x + 3 y' = 2y(x, y) → (x + 3, 2y)

→ (x + 3, 2y) or(x, y) → (x, 2y)

→ (x + 3, 2y)

y' – 4 = 2x' – 1

3

x' – 13

= x y' – 4 = y

x' = 3x + 1 y' = y + 4(x, y) → (3x, y)

→ (3x + 1, y + 4) A dilation of factor 3 from the y-axis, and a translation by

⎣⎢⎡ 1

4⎦⎥⎤ .

A dilation of factor 2 from the x-axis, and a translation by in any order.

⎣⎢⎡ 3

0⎦⎥⎤ , h Rearrange the image equation:

2y'3

= 1x' – 2

x' – 2 = x 2y'3

= y

x' = x + 2 y' = 32

y

(x, y) →⎝⎜⎛ x, 3y

2 ⎠⎟⎞

→⎝⎜⎛ x + 2, 3y

2 ⎠⎟⎞


y' + 32

= 1x'

x' = x y' + 32

= y

x' = x y' = 2y – 3(x, y) → (x, 2y)

→ (x, 2y – 3)


from the x-axis

followed by a translation by ⎣⎢⎡ 2

0⎦⎥⎤ .


⎣⎢⎡ 0

–3⎦⎥⎤ .

f Rearrange the image equation:

y' = 1– (x' – 3)

= 1– x' + 3

– x' + 3 = x y' = yx' = – x + 3 y' = y

(x, y) → ( – x, y) → ( – x + 3, y)

Reflection in the y-axis, followed by a translation by

⎣⎢⎡ 3

0⎦⎥⎤ .


Exercise 8H Solutions 1 a (x, y) → (x – 1, y + 3)

x' = x – 1 y' = y + 3x = x' + 1 y = y' – 3

y' – 3 = |x' + 1|y' = |x' + 1| + 3

Image: {(x, y): y = |x + 1| + 3}

b (x, y) → ( – y, – x)

x' = – y y' = – xy = – x' x = – y'

– x' = |y'| Image: {(x, y): x = – |y|}

c (x, y) → (4x, y)

x' = 4x y' = yx = 1

4 x' y = y'

y' =⎪⎪⎪ x'

4 ⎪⎪⎪

Image: ⎩⎨⎧ (x, y): y =

⎪⎪⎪ x

4⎪⎪⎪⎭⎬⎫

d (x, y) → (x, – y)x' = x y' = – yx = x' y = – y'

– y' = |x'| Image: {(x, y): y = – |x|}

e (x, y) → (x – 1, y + 3)

→ (y + 3, x – 1)x' = y + 3 y' = x – 1y = x' – 3 x = y' + 1

x' – 3 = |y' + 1|x' = |y' + 1| + 3

Image: {(x, y): x = |y + 1| + 3}

f (x, y) → (y, x)

→ (y – 1, x + 3)x' = y – 1 y' = x + 3y = x' + 1 x = y' – 3

x' + 1 = |y' – 3|x' = |y' – 3| – 1

Image: {(x, y): x = |y – 3| – 1}


g (x, y) → (x, 2y) → (x, – 2y)

x' = x y' = –2yx = x' – 3 y = 1

2 y'

– 12

y' = |x'|

– y' = 2|x'|

Image: {(x, y): y = –2|x|}

2 a Rearrange the image equation:

y' – 3 = |x'|x' = x y' – 3 = yx' = x y' = y + 3

(x, y) → (x, y + 3)


3⎦⎥⎤ .

b Rearrange the image equation:

y' – 3 = |x' – 3|x' – 3 = x y' – 3 = y

x' = x + 3 y' = y + 3(x, y) → (x + 3, y + 3)


3⎦⎥⎤ .

c 2x' = x y' = y

x' = 12

x y' = y

(x, y) →⎝⎜⎛1

2 x, y

⎠⎟⎞


from the y-axis

d Rearrange the image equation:

y–2

= |x|

x' = x – 12

y' = y

x' = x y' = –2y(x, y) → (x, – y)

→ (x, – 2y) A reflection in the x-axis followed by a dilation of factor 2 from the x-axis.

3 a (x, y) → (2x, y)

x' = 2x y' = yx = 1

2 x' y = y'

y' =⎣⎢⎡ x'

2 ⎦⎥⎤

Image: ⎩⎨⎧ (x, y): y =

⎣⎢⎡ x

2⎦⎥⎤⎭⎬⎫

b (x, y) → (x + 2, y)

x' = x + 2 y' = yx = x' – 2 y = y'y' = [x' – 2]

Image: {(x, y): y = [x – 2]}

c (x, y) → (x, y + 2)

x' = x y' = y + 2x = x' y = y' – 2

y' – 2 = [x']

Image: {(x, y): y = [x] – 2}


g (x, y) → (2x, y) → (2x, – y)

x' = 2x y' = – yx = 1

2 x' y = – y'

– y' =⎣⎢⎡ x'

2 ⎦⎥⎤

d (x, y) → (y, x)x' = y y' = xy = x' x = y'x' = [y']

Image: {(x, y): x = [y]}

Image: ⎩⎨⎧ (x, y): y = –

⎣⎢⎡ x

2⎦⎥⎤⎭⎬⎫

e (x, y) → ( – y, – x)

x' = – y y' = – xy = – x' x = – y'

– x' = [y'] h (x, y) → (x, – y)

→ (x, – y + 2)x' = x y' = – y + 2x = x' y = – y' + 2

– y' + 2 = [x']

Image: {(x, y): x = – [y]}

Image: {(x, y): x = – [x] + 2}

f

(x, y) → (x + 4, y)x' = x + 4 y' = yx = x' – 4 y = y'y' = [x' – 4]

Image: {(x, y): y = [x – 4]}


Exercise 8I Solutions 1 a (x, y) → (x + 2, y + 3)

x' = x + 2 y' = y + 3x = x' – 2 y = y' – 3

y' – 3 = f(x' – 2)

Image: y = f(x – 2) + 3 b (x, y) → (y, x)

x' = y y' = xy = x' x = y'x' = f(y')

Image: x = f(y)

c (x, y) →⎝⎜⎛1

2 x, y

⎠⎟⎞

x' = 12

x y' = xy

x = 2x' y = y'y' = f(2x')

Image: y = f(2x) d (x, y) → (x, 2y)

x' = x y' = 2yx = x' y = 1

2 y'

12

y' = f(x')

Image: y = 2f(x) 2 f(x) = 2x a x = x + 3

2x = 2x + 3

b x = 2x

2x = x2x

= 4x

c x = x

2

2x = xx2

d x = x – 3

4 × 2x = 4 × 2x – 3

= 22 × 2x – 3

= 2x – 1

3 4f⎝⎜⎛ x

2⎠⎟⎞ = 4

⎝⎜⎛ x

2⎠⎟⎞2

= 4⎝⎜⎛ x2

4 ⎠⎟⎞ = x2

y = x2 is mapped to y = x2

No transformation is necessary.

4 f(2x – 3) + 4 =⎝⎜⎛ 1

2x – 3⎠⎟⎞ + 4

y = 1x is mapped to y = 1

2x – 3+ 4

y' – 4 = 12x' – 3

1x

= 12x' – 3

x = 2x' – 3

x' = 12⎝⎜⎛ x + 3

2⎠⎟⎞

y = y' – 4y' = y + 4

The translation is a dilation of factor 12

from

the y-axis, followed by a translation of

⎣⎢⎢⎢⎡

324⎦⎥⎥⎥⎤.

5 –3f(2 – x) = –3(2 – x)2

y = x2 is mapped to y = – 3(2 – x)2

– y'3

= (2 – x')2

x = (2 – x')x' = 2 – xy = – y'

3y' = –3y

The translation is a reflection in the x-axis, followed by a dilation of factor 3 from the x-axis, followed by a

translation of ⎣⎢⎡ 2

0⎦⎥⎤ .

A reflection in the y-axis also occurs, but is irrelevant since the graph of f(x) is symmetrical about the y-axis.


Exercise 8J Solutions 1

a ⎣⎢⎡ 2

5–34⎦⎥⎤⎣⎢⎡ –1

2⎦⎥⎤ =

⎣⎢⎡ 2 × –1 + –3 × 2

5 × –1 + 4 × 2⎦⎥⎤

=⎣⎢⎡ –8

3⎦⎥⎤

b

⎣⎢⎡ –3

–1–1

3⎦⎥⎤⎣⎢⎡ a

b⎦⎥⎤ =

⎣⎢⎡ –3a + –1b

–1a + 3b⎦⎥⎤

=⎣⎢⎡ 3a – b

– a + 3b⎦⎥⎤

2 ⎣⎢⎡ 2

–4–13⎦⎥⎤⎣⎢⎡ 1

0⎦⎥⎤ =

⎣⎢⎡ 2 × 1 + –1 × 0

–4 × 1 + 3 × 0⎦⎥⎤

=⎣⎢⎡ 2

–4⎦⎥⎤

(1, 0) → (2, – 4)

⎣⎢⎡ 2

–4–13⎦⎥⎤⎣⎢⎡ 0

1⎦⎥⎤ =

⎣⎢⎡ 2 × 0 + –1 × 1

–4 × 0 + 3 × 1⎦⎥⎤

=⎣⎢⎡ –1

3⎦⎥⎤

(0, 1) → (–1, 3)

⎣⎢⎡ 2

–4–13⎦⎥⎤⎣⎢⎡ 3

2⎦⎥⎤ =

⎣⎢⎡ 2 × 3 + –1 × 2

–4 × 3 + 3 × 2⎦⎥⎤

=⎣⎢⎡ 4

–6⎦⎥⎤

(3, 2) → (4, – 6) 3

a

⎣⎢⎡ 2

1–11⎦⎥⎤⎣⎢⎡ 1

0⎦⎥⎤ =

⎣⎢⎡ 2 × 1 + –1 × 0

1 × 1 + 1 × 0⎦⎥⎤

=⎣⎢⎡ 2

1⎦⎥⎤

(1, 0) → (2, 1)

⎣⎢⎡ 2

1–1

1⎦⎥⎤⎣⎢⎡ –1

2⎦⎥⎤ =

⎣⎢⎡ 2 × –1 + –1 × 2

1 × –1 + 1 × 2⎦⎥⎤

=⎣⎢⎡ –4

1⎦⎥⎤

(–1, 2) → (–4, 1)

b

⎣⎢⎡ –2

001⎦⎥⎤⎣⎢⎡ 1

0⎦⎥⎤ =

⎣⎢⎡ –2 × 1 + 0 × 0

0 × 1 + 1 × 0⎦⎥⎤

=⎣⎢⎡ –2

0⎦⎥⎤

(1, 0) → (–2, 0)

⎣⎢⎡ –2

001⎦⎥⎤⎣⎢⎡ –1

2⎦⎥⎤ =

⎣⎢⎡ –2 × –1 + 0 × 2

0 × –1 + 1 × 2⎦⎥⎤

=⎣⎢⎡ 2

2⎦⎥⎤

(–1, 2) → (2, 2)

c

⎣⎢⎡ 2

33

–1⎦⎥⎤⎣⎢⎡ 1

0⎦⎥⎤ =

⎣⎢⎡ 2 × 1 + 3 × 0

3 × 1 + –1 × 0⎦⎥⎤

=⎣⎢⎡ 2

3⎦⎥⎤

(1, 0) → (2, 3)

⎣⎢⎡ 2

33

–1⎦⎥⎤⎣⎢⎡ –1

2⎦⎥⎤ =

⎣⎢⎡ 2 × –1 + 3 × 2

3 × –1 + –1 × 2⎦⎥⎤

=⎣⎢⎡ 4

–5⎦⎥⎤

(–1, 2) → (4, – 5) 4 a Let the matrix be

⎣⎢⎡a

cbd⎦⎥⎤ .

⎣⎢⎡ a

cbd⎦⎥⎤⎣⎢⎡ 1

–2⎦⎥⎤ =

⎣⎢⎡ a – 2b

c – 2d⎦⎥⎤

=⎣⎢⎡ –4

5⎦⎥⎤

a – 2b = –4; c – 2d = 52a – 4b = –8; 2c – 4d = 10

⎣⎢⎡ a

cbd⎦⎥⎤⎣⎢⎡ 3

4⎦⎥⎤ =

⎣⎢⎡ 3a + 4b

3c + 4d⎦⎥⎤

=⎣⎢⎡ 18

5⎦⎥⎤

3a + 4b = 18; 3c + 4d = 5 + :

5a = 10 5c = 15a = 2 c = 3

2 – 2b = –4; 3 – 2d = 5b = 3; d = –1

The matrix is ⎣⎢⎡ 2

33

–1⎦⎥⎤ .


b Let the matrix be the points ⎣⎢⎡a

cbd⎦⎥⎤ ,

(x1, y1) and (x2, y2), and the images (m, n) and (p, q).

⎣⎢⎡ a

cbd⎦⎥⎤⎣⎢⎡ x1

y1⎦⎥⎤ =

⎣⎢⎡ x1 a + y1 b

x1 c + y1 d⎦⎥⎤

=⎣⎢⎡m

n⎦⎥⎤

x1 a + y1 b = m; x1 c + y1 d = n

⎣⎢⎡ a

cbd⎦⎥⎤⎣⎢⎡ x2

y2⎦⎥⎤ =

⎣⎢⎡ x2 a + y2 b

x2 c + y2 d⎦⎥⎤

=⎣⎢⎡ p

q⎦⎥⎤

x2 a + y2 b = p; x2 c + y2 d = qThese two simultaneous equations will give a unique solution for a, b, c and d, provided the discriminant is not zero.

Δ = 0 if x1 y2 – y1 x2 = 0x1 y2 = y1 x2

y2

x2= y1

x1

This means the gradient of the line from the origin to (x1, y1) will be the same as the gradient of the line from the origin to (x2, y2). This means the two points and the origin will all lie on the same straight line.

The information will be sufficient, provided the two points are not collinear with the origin.

c Let the matrix be

⎣⎢⎡a

cbd⎦⎥⎤ .

⎣⎢⎡ a

cbd⎦⎥⎤⎣⎢⎡ 1

0⎦⎥⎤ =

⎣⎢⎡ a

c⎦⎥⎤

=⎣⎢⎡ 1

1⎦⎥⎤

a = 1, c = 1

⎣⎢⎡ a

cbd⎦⎥⎤⎣⎢⎡ 0

1⎦⎥⎤ =

⎣⎢⎡ b

d⎦⎥⎤

=⎣⎢⎡ 2

2⎦⎥⎤

b = 2, d = 2

The matrix is ⎣⎢⎡1

122⎦⎥⎤ .

⎣⎢⎡ 1

122⎦⎥⎤⎣⎢⎡ x

y⎦⎥⎤ =

⎣⎢⎡ x + 2y

x + 2y⎦⎥⎤

=⎣⎢⎡ x'

y'⎦⎥⎤

Clearly, all points will be mapped somewhere on the line y = x. The range of the transformation is y = x.

5 Use the fact that the image of (1, 0) gives

the values of a11 and a21 (a and c in the matrix above), and the image of (0, 1) gives the values of a12 and a22 (b and d.)

a (1, 0) → (–1, 0) and (0, 1) → (0, 1)

⎣⎢⎡ –1

001⎦⎥⎤

b (1, 0) → (0, 1) and (0, 1) → (1, 0)

⎣⎢⎡ 0

110⎦⎥⎤

c (1, 0) → (0, – 1) and (0, 1) → (–1, 0)

⎣⎢⎡ 0

–1–10⎦⎥⎤

d (1, 0) → (1, 0) and (0, 1) → (0, 2)

⎣⎢⎡ 1

002⎦⎥⎤

e (1, 0) → (3, 0) and (0, 1) → (0, 3)

⎣⎢⎡ 3

003⎦⎥⎤

f (1, 0) → (3, 0) and (0, 1) → (0, 1)

⎣⎢⎡ 3

001⎦⎥⎤


Solutions to Multiple-choice Questions 1 y = f(x) points: (0, 3) and (–3, 0) y = g(x) points: (0, 3) and (3, 0) (x, y) → ( – x, y) This is reflection in the y-axis. C 2 (x, y) → (x, – y)

x' = x y' = – yx = x' y = – y'

– y' = f(x') Image: g(x) = –f(x) f(x) = –g(x) B 3 Rewrite the equation as

y + 2 = (x – 5)2

x' – 5 = x y' + 2 = yx = x' + 5 y = y' – 2

(x, y) → (x + 5, y – 2)

⎣⎢⎡ 5

–2⎦⎥⎤

D

4 Find the translation that maps g(x)

onto f(x). Rewrite the equation as g(x) – 8 = (x + 2)2

x' + 2 = x y' – 8 = yx = x' – 2 y = y' + 8

(x, y) → (x – 2, y + 8)

⎣⎢⎡ –2

8⎦⎥⎤

The translation that maps f(x) onto g(x) is C

⎣⎢⎡ 2

–8⎦⎥⎤ .

5 y = 3f(x)

= 3x2

This will be a similar curve to y = f(x) but with a larger y-value at every point.

Only D fits this criteria. D

6 (x, y) → ( – x, y) → ( – x + 2, y + 3)

x' = – x + 2 y' = y + 3x = – x' + 2 y = y' – 3

y' – 3 = – x' + 2

Image: y = – x + 2 + 3 E 7 (x, y) → (x, – y)

→ (x, – 2y)x' = x y' = –2yx = x' y = – y'

2– y'

2= 1

x2

Image: y = – 2x2 A

8 (x, y) →

⎝⎜⎛1

3 x, y

⎠⎟⎞

→⎝⎜⎛1

3 x – 5, y + 2

⎠⎟⎞

x' = 13 x – 5 y' = y + 2

x = 3x' + 15 y = y' – 2y' – 2 = |3x' + 15|

Image: y = |3x + 15| + 2 = 3|x + 5| + 2 B 9 [–4.6] + [7.2] + [8.7] = –5 + 7 + 8 = 10 B 10 (x, y) → (x, – y)

→ (x + 5, – y – 4)x' = x + 5 y' = – y – 4x = x' – 5 y = – y' – 4

– y' – 4 = [x' – 5] Image: y = – [x – 5] – 4 B


Solutions to Short-answer Questions 1 a (x, y) → (3x, y)

(3, – 1) → (9, – 1) b (x, y) → (x, 2y)

(3, – 1) → (3, – 2) c (x, y) → (x – 3, y + 2)

(3, – 1) → (0, 1) d (x, y) → ( – x, y)

(3, – 1) → (–3, – 1) e (x, y) → (x, – y)

(3, – 1) → (3, 1) f (x, y) → (y, x)

(3, – 1) → (–1, 3) 2 a (x, y) → (x – 2, y + 3)

x' = x – 2 y' = y + 3x = x' + 2 y = y' – 3

y' – 3 = (x' + 2)2

Image: y = (x + 2)2 + 3 b (x, y) → (y, x)

x' = y y' = xy = x' x = y'x' = (y')2

Image: x = y2 or y = ± x c (x, y) → (x, – y)

x' = x y' = – yx = x' y = – y'

– y' = (x')2

Image: y = – x2

3 a (x, y) → (y, x)

→ (y – 2, x + 3) b

(x, y) → (x, 5y)

→ (x, – 5y)

c (x, y) → (4x, y) → (4x – 2, y + 3)

d (x, y) → (x – 2, y + 3)

→ (x – 2, 4(y + 3)) 4 a (x, y) → (3x, y)

x' = 3x y' = yx = 1

3 x' y = y'

y = 23 x' – 1

Image: y = 23 x – 1

b (x, y) → (x, 2y)

x' = x y' = 2yx = x' y = 1

2 y'

12

y' = 2x' – 1

Image: y = 4x – 2 c (x, y) → (x – 3, y + 2)

x' = x – 3 y' = y + 2x = x' + 3 y = y' – 2

y' – 2 = 2(x' + 3) – 1 = 2x' + 6 – 1

Image: y = 2x + 7 d (x, y) → ( – x, y)

x' = – x y' = yx = – x' y = y'y' = 2 × – x' – 1

Image: y = –2x – 1 or2x + y + 1 = 0

e ( x, y) → (x, – y)

x' = x y' = – yx = x' y = – y'

– y' = 2x' – 1 Image: y = –2x + 1 or

2x + y = 1


f (x, y) → (y, x)x' = y y' = xy = x' x = y'x' = 2y' – 1

Image: y = 12 (x + 1)

5 a x' = y – 2 y' = x + 3

y = x' + 2 x = y' – 3x' + 2 = 2 – (y' – 3)2

Image: – x = (y – 3)2

y – 3 = ± – xy = 3 ± – x

b (x, y) → (x, – 5y)

x' = x y' = –5yx = x' y = – 1

5 y'

–1.5 y' = 2 – (x')2

Image: y = 5x2 – 10 = 5(x2 – 2)

c (x, y) → (4x – 2, y + 3)

x' = 4x – 2 y' = y + 3x = x' + 2

4y = y' – 3

y' – 3 = 2 –⎝⎜⎛ x' + 2

4 ⎠⎟⎞2

Image: y = 5 – 116

(x + 2)2

d (x, y) → (x – 2, 4(y + 3))

x' = x – 2 y' = 4(y + 3)x = x' + 2 y = y'

4– 3

y'4

– 3 = 2 – (x' + 2)2

y'4

= 5 – (x' + 2)2

Image: y = 20 – 4(x + 2)2

6 a y = – |x|

b y = |2x| + 3

c y = 4 – |2x|

7 a Rearrange the image equation: y' – 4

2= 1

x' – 3x' – 3 = x y' – 4

2= y

x' = x + 3 y' = 2y + 4(x, y) → (x + 3, 2y + 4)

b Rearrange the image equation: y' – 3

2= (x' – 4)2

x' – 4 = x y' – 32

= y

x' = x + 4 y' = 2y + 3(x, y) → (x + 4, 2y + 3)

c Rearrange the image equation: y' + 4 = |3x'|

3x' = x y' + 4 = 4x' = x

3 y' = y – 4

(x, y) →⎝⎜⎛ x

3 , y – 4

⎠⎟⎞


d Rearrange the image equation: y = x + 1

x – 1= x – 1 + 2

x – 1 = 1 + 2

x – 1y – 1 = 2

x – 1y – 1

2= 1

x – 1x' – 1 = x y' – 1

2= y

x' = x + 1 y' = 2y + 1(x, y) → (x + 1, 2y + 1)

e Rearrange the image equation: – y + 3 = (x – 2)2

x' – 2 = x – y' + 3 = yx' = x + 3 y' = – y + 3

(x, y) → (x + 3, – y + 3) → (x + 3, 3 – y)

8 a x' = x y' – 1 = y

x' = x y' = y + 1(x, y) → (x, y + 1)

Translate the graph of one unit up. y = x2

b x' – 1 = x y' + 2 = y

x' = x + 1 y' = y – 2(x, y) → (x + 1, y – 2)

Translate the graph of one unit right and two units down.

y = x2

c x' – 2 = x 3(y' + 2) = yx' = x + 2 y' = y

3– 2

(x, y) →⎝⎜⎛ x, y

3⎠⎟⎞

→⎝⎜⎛ x + 2, y

3– 2

⎠⎟⎞

Dilate the graph of a factor y = x2 13

from the x-axis then translate it two units right and two units down.

d 4x' = x y' = yx' = x

4y' = y

(x, y) →⎝⎜⎛ x

4, y⎠⎟⎞

Dilate the graph of y = [x] a factor 14

from the y-axis.



y' = –6⎪⎪⎪ x' – 1

2⎪⎪⎪

– y'6

=⎪⎪⎪ x' = 1

2⎪⎪⎪

x' – 12

= x – y'6

= y

x' = x + 12

y' = –6y

(x, y) → (x, 6y) → (x, – 6y)

→⎝⎜⎛ x + 1

2, 2y

⎠⎟⎞

Dilate y = |x| by a factor 6 from the x-axis, reflect in the x-axis and translate the result 1

2 a unit right.

f Rearrange the image equation: y' – 2

–3= |x' – 2|

x' = x – 2 y' – 2–3

= y

x' = x + 2 y' = –3y + 2(x, y) → (x, 3y)

→ (x, – 3y) → (x + 2, – 3y + 2)

Dilate y = |x| by a factor 3 from the x-axis, reflect in the x-axis and translate two units right and two units up.


Chapter 9 – Ratios and similarity Exercise 9A Solutions 1 One part = 9000 ÷ 9 = 1000 Two parts = 1000 × 2 = 2000 Seven parts = 1000 × 7 = 7000 2 One part = 15 000 ÷ 5 = 3000 Two parts = 3000 × 2 = 6000

3 x6

= 915

x = 9 × 615

= 3.6

4 144p

= 611

p144

= 116

p = 11 × 1446

= 264

5 x

3= 15

2

x = 15 × 32

= 22.5

6 6 : 5 : 7 = 180° One part = 180° ÷ 18 = 10° Six parts = 10° × 6 = 60° Five parts = 10° × 5 = 50° Seven parts = 10° × 7 = 70° 7 Suppose they receive $x, $y and $z

respectively.

x + 2x

= 32

2(x + 2) = 3x2x + 4 = 3x

x = 4 X receives $4 and Y receives $6. Two parts = $4 One part = $2 Seven parts = $14 Z receives $14. 8 One part = 10 g Three parts = 10 g × 3 = 30 g (zinc) Four parts = 10 g × 4 = 40 g (tin)

9 Seven parts = 56 One part = 56 ÷ 7 = 8 green beads Two parts = 8 × 2 = 16 white beads 10 One part = 45 mm 125 000 parts = 45 mm × 125 000 = 5 625 000 mm = 5.625 km 11 One part = $5200 ÷ 13 = $400 Eight parts = $400 × 8 = $3200 (mother) Five parts = $400 × 5 = $2000 (daughter) Difference = $1200 12 If BC is one part, AB and CD are each

two parts. AD is 5 parts and BD is 3 parts, so BD = 3

5 AD.

13 The ratio will be π : 1, as for any circle. 14 One part = 30 ÷ 5 = 6 Two parts = 6 × 2 = 12 (boys) Three parts = 6 × 3 = 18 (girls) After six boys join the class, there are 18

boys and 18 girls, so the ratio is 1 : 1. 15 b

a= 4

3 and b + c

a= 5

2b + c

a= 5

2ba

+ ca

= 52

43

+ ca

= 52

ca

= 52

– 43

= 15 – 86

= 76

∴a : c = 6 : 7

16 One part = 3.5 cm 250 000 parts = 3.5 cm × 250 000 = 875 000 cm = 8.75 km


Exercise 9B Solutions 1 a AAA

x5

= 94

x = 9 × 54

= 11.25 cm

b AAA

Note that E corresponds with B, so x corresponds with 14 cm. x

14= 10

12

x = 10 × 1412

= 11 23

cm

c AAA x

2= 6

4

x = 6 × 24

= 3 cm

d AAA

Note that Q corresponds with B and R corresponds with C, so x corresponds with 6 cm.

x6

= 108

x = 10 × 68

= 7.5 cm

2 a AAA

x + 1212

= 2416

= 32

2x + 24 = 362x = 12

x = 6 cm b AAA

x + 22

= 53

3x + 6 = 103x = 4

x = 1 13

cm

c AAA x + 8

x= 8

2= 4

x + 8 = 4x3x = 8

x = 2 23

cm

d AAA x + 1.5

x= 12

10= 6

55x + 7.5 = 6x

x = 7.5 cm

3 AC

14= 15

12= 5

4

AC = 5 × 144

= 17.5 cm

AE + 4AE

= 54

4AE + 16 = 5AEAE = 16 cmAB = AE + EB

= 20 cm

4 tree

33= 30

224= 15

112

Tree height = 15 × 33112

= 4.42 m

Note: It is valid to leave the measurements of the stick and its shadow in cm, as you are comparing the ratio of measurements with the same units.

5 h

15= 20

40= 1

2h = 15

2= 7.5 m high

6 h

300= 1

20h = 300

20= 15 m high

7 CY

45= 15

30= 1

2CY = 45

2= 22.5 m


8 h32

= 26.2

h = 646.2

= 10 1031

m high

9 x

4= 20 – x

82x8

= 20 – x8

2x = 20 – x3x = 20

x = 203

= 6 23 cm high

10 Let x be the height of A above the 80 cm

leg of the table.

x30

= 12100

h = 12 × 30100

= 3.6

Height = 80 cm + 3.6 cm = 83.6 cm 11 x

1.3 – x= 1.5

0.8= 15

88x = 19.5x – 15x

23x = 19.5x = 19.5

23= 39

46 m

12

x

100= 1.3

3.5

x = 1.3 × 1003.5

= 13003.5

= 2607

≈ 37.1

Height ≈ 37.1 m + 1.7 m = 38.6 m

13

h

8= 9

10h = 72

10= 7.2 m high

14 Taking the heights above the spotlight, h – 0.6

8= 0.5

3= 1

6h – 6

10= 8

6= 4

3h = 4

3+ 3

5 = 20 + 9

15 = 1 14

15 m high

15 a Vertically opposite angles at C are equal: ∠B = ∠D = 90° The third angles in the triangle must be

equal: ∠A = ∠E ∴ ΔABC ~ ΔEDC b x

4= 5

2x = 20

2= 10

c y2 = 22 + 42

= 4 + 16 = 20y = 20 = 4 × 5 = 2 5

z2 = 102 + 52

= 100 + 25 = 125z = 125 = 25 × 5 = 5 5


d y : z = 2 5 : 5 5 = 2 : 5

ED : AB = 2 : 5∴ y : z = ED : AB

16 a + 12

12= 10

7a + 12 = 120

7a = 120

7– 12

= 367

= 5 17

17 h

3= 1.8

0.76

h = 1.8 × 30.76

≈ 7.11 m

18 In ΔTRN, ∠TRN = 90° – ∠T

In ΔRST, ∠S = 90° – ∠T∠TRN = ∠S∠SRN = ∠T

ΔTRN ~ ΔTSR NT

RT= RT

ST = 4

10= 2

5NT4

= 25

NT = 2 × 45

= 1.6 m

19

In ΔAPQ and ΔACB, AQ

AB= 3

14APAC

= 1.57

= 314

∴ AQAB

= APAC

∠A is common to both triangles. ΔAPQ ~ ΔACB

PQBC

= AQAB

PQ10

= 314

PQ = 3014

= 2 17

m

20 Note that the three triangles are all

similar, as shown in Q.18. x + 4

6= 6

4= 3

2

x + 4 = 3 × 62

= 9

∴ x = 5yx

= 4y

y2 = 4x = 4 × 5∴ y = 2 5

ay

= 64

= 32

∴ a = 3y2

= 3 5


Exercise 9C Solutions 1 a 2 : 4 : 6 : 8 = 1 : 2 : 3 : 4 b 2 : 8 : 18 : 32 = 1 : 4 : 9 : 16 c The second ratio is the square of the

first. 2 a 2 : 4 : 6 : 8 = 1 : 2 : 3 : 4 b 1 : 4 : 9 : 16 c The second ratio is the square of the

first. 3 A'B'

AB= 5

3

Area A'B'C'D' = 7 ×⎝⎜⎛ 5

3⎠⎟⎞2

= 7 × 259

= 19 49 cm2

4 20 × 2.12 = 20 × 4.41

= 88.2 cm2

5 a F is the midpoint of AC, so AF = 1 cm. BF 2 = BA2 – AF 2

= 22 – 12 = 3BF = 3 cm

b A'C'

AC= B'F'

BFa2

= 23

a = 43

= 4 33

c Area A'B'C'Area ABC

=⎝⎜⎛ B'F '

BF ⎠⎟⎞2

=⎝⎜⎛ 2

3 ⎠⎟⎞2

= 43

6 Area ratio = 16 : 25

Side ratio = 1625

= 42

52 = 4 : 5

7 30 × 9

12= 22.5 cm

8 a 1 : 2 : 3 b 1 : 2 : 3 c 1 : 8 : 27 d The third ratio is the cube of the first. 9 a i 8 : 12 = 2 : 3 ii 4 : 6 = 2 : 3 iii 3 : 4 1

2= 2 : 3

b 8 × 4 × 3 : 12 × 6 × 4 1

2= 96 : 324

= 8 : 27

c The ratio in b is the cube of the ratios in a. 10 a 3 : 2 : 5 b Sphere 1: V = 4

3× π × 33 = 36π

Sphere 2: V = 43

× π × 23 = 32π3

Sphere 3: V = 43

× π × 53 = 500π3

36 : 323

: 5003

= 108 : 32 : 500

= 27 : 8 : 125

c The second ratio is the cube of the first. 11 (2 : 1)3 = 23 : 13

= 8 : 1


12 (3 : 4)3 = 33 : 43

= 27 : 64 13 3 8 : 27 = 3 8 : 3 27

= 2 : 3

14 Volume ratio = 64 : 27 a Height ratio = 3 64 : 27

= 4 : 3

b Radius ratio = 3 64 : 27

= 4 : 3

15 Height ratio = 2 : 1 a Area ratio = (2 : 1)2

= 4 : 1 b Capacity ratio = (2 : 1)3

= 8 : 1 16 a (1 : 10)2 = 1 : 100 b (1 : 10)3 = 1 : 1000 c (1 : 10)1 = 1 : 10 d Both models will have the same number

of wheels, so 1 : 1.

17 12

×⎝⎜⎛ 12

8 ⎠⎟⎞3

= 12

×⎝⎜⎛ 3

2⎠⎟⎞3

= 2716

litres

12

×⎝⎜⎛ 16

8 ⎠⎟⎞3

= 12

× 23

= 4 litres

18 343 ×⎝⎜⎛ 7.5

10.5⎠⎟⎞3

= 343 ×⎝⎜⎛ 5

7⎠⎟⎞3

= 125 mL

343 ×⎝⎜⎛ 9

10.5⎠⎟⎞3

= 343 ×⎝⎜⎛ 6

7⎠⎟⎞3

= 216 mL

19 a Length ratio = 1 : 2500

= 1 : 50

b Capacity ratio = (area ratio) 3

= (1 : 50)3

= 1 : 125 000 c Width = 150 × 1

50= 3 cm

d Area = 3 ÷ 1

2500 = 3 × 2500 = 7500 cm2

20 a Height ratio = 144 : 169

= 12 : 13

b Capacity ratio = (12 : 13)3

= 1728 : 2197 21 a Ratio of sides = 1 : 2 Ratio of areas = 12 : 22 = 1 : 4 Four times b Area ΔAKM = 15

4= 3.75

22 ΔBDE ~ ΔCAF and AB = AC = 2AD ∴ BD2 = BA2 – AD2

= (2AD)2 – AD2

= 3AD2

Ratio of areas = BD2

AC2

= 3AD2

(2AD)2

= 3AD2

4AD2 = 34

Note: It is easier to express lengths in terms of AD as fractions are avoided.


Exercise 9D Solutions 1 Construct a line OB of length 4 units,

and draw a circle of radius 2 units from the centre of OB (2 units from either end). Mark point A 3 units from O. Construct a line through A perpendicular to OB which cuts the circle at D. The line AD will be 3 units long.

2 Construct a line OB of length 6 units,

and draw a circle of radius 3 units from the centre of OB (3 units from either end). Mark point A 5 units from O. Construct a line through A perpendicular to OB which cuts the circle at D. The line AD will be 5 units long.

3 Draw a line AB 10 cm long. Draw any

line AX. Mark any point C on AX and replicate the line segment AC four times as shown below. Join DB and draw lines through C, C', C'' parallel to DB. These will divide AB into four equal line segments.

4 Draw a line AB 20 cm long. Draw any line AX. Mark any point C on AX and replicate the line segment AC nine times. Join DB and draw lines through C, C', C'' parallel to DB. These will divide AB into nine equal line segments.

The diagram will be the same as in Q.3, except that there will be eight circles instead of three.

5 Take 1 cm to be a unit.

Draw OB 14 cm long and OA 4 cm long. Join AB. Mark OD 1 cm along OB from O and draw a line through D parallel to BA. OC will be 4

14= 2

7 units (cm) long.

6 Take 1 cm to be a unit.

Draw OB 13 cm long and OA 9 cm long. Join AB. Mark OD 1 cm along OB from O and draw a line through D parallel to BA. OC will be 9

13 units (cm) long.

The diagram will be like the one for Q.5, except that OA will be longer and OB will be slightly shorter.

7 Draw OB 3 units long and OA 10 units long. Join AB. Mark OD 1 unit along OB and draw a line through D parallel to BA. OC will be 10

3 units long.

The diagram will be like the one for Q.5, except that OA will be just over three times as long as OB.

8 Make three copies of the 4 unit length

(below) or four copies of the 3 unit length. AB will be 3 × 4 units long.


Exercise 9E Solutions 1

a φ – 1 = 1 + 52

– 1

= 1 + 5 – 22

= 5 – 12

∴ 1φ

= φ – 1

b φ3 = (1 + 5 )2(1 + 5 )8

= (1 + 2 5 + 5)(1 + 5 )8

= (6 + 2 5 )(1 + 5 )8

= 6 + 8 5 + 108

= 16 + 8 58

= 2 + 5

2φ + 1 = 1 + 5 + 1 = 2 + 5

∴ φ3 = 2φ + 1

c As shown above, φ – 1 = 1

φ.

∴ (φ – 1)2 = 1φ2

2 – φ = 2 – 1 + 52

= 4 – 1 – 52

= 3 – 52

(φ – 1)2 =⎝⎜⎛ 1 + 5 – 2

2 ⎠⎟⎞2

= ( 5 – 1)2

4

= 5 – 2 5 + 14

= 3 – 52

= 2 – φ

∴ 2 – φ = (φ – 1)2 = 1φ2

2 a In ΔACX, ∠ACX = 90° – ∠BCX

In ΔCBX, ∠B = 90° – ∠BCX

∠ACX = ∠B∠A = ∠BCX

ΔACX ~ ΔCBX∴ AX

CX= CX

BX b Multiply both sides of the above

equation by CX·BX i CX 2 = AX⋅BX

= 2 × 8 = 16CX = 4

ii CX2 = AX⋅BX = 1 × 10 = 10

CX = 10 3 Join AB and BC. This will produce a right-

angled triangle with an altitude. In Q.2 we proved that the altitude was the geometric mean of the two segments that divided the base. Therefore, as in Q.2:

ADBD

= BDCD

ECDE

= DEDE + EC

Since BD = DE, AD = EC and CD = DE + EC DEEC

= DE + ECDE

= 1 + ECDE

x = DEEC

= 1 + 1x

∴ x2 – x – 1 = 0 Using the quadratic formula:

x = –1 + 1 – 4 × 1 × –12

= –1 + 52

= φ

(Rejecting the negative root as x > 0)


ECDE

= 1φ

= φ – 1

ADBD

= ECDE

= φ – 1

∴ ADBD

= BDCD

= φ – 1

4 a i ∠AOB = 360

10= 36°

ii ∠OAB = 180 – 362

= 72°

b i ∠XAB = 72

2= 36°

∠ABO = ∠OAB = 72°∠AXB = 180 – 36 – 72

= 72°∠ABO = ∠AXB

∴ AX = AB

ii ∠XAO = 722

= 36° = ∠AOX∴ AX = OX

iii Corresponding angles are equal, so the triangles must be similar. c ΔAOB ~ ΔXAB

OBAB

= ABXB

OX + XBAB

= ABXB

OX = XA = ABAB + XB

AB= AB

XB1 + XB

AB= AB

XBx = XB

AB = 1 + 1

x ∴ x2 – x – 1 = 0

Using the quadratic formula:

x = –1 + 1 – 4 × 1 × –12

= –1 + 52

= φ

(Rejecting the negative root as x > 0)

XBAB

= 1φ

= φ – 1

= –1 + 52

(Refer to Q.1 part a.) XB

AB= AB

OB = AB = φ – 1 since OB = 1

AB = –1 + 52

≈ 0.62

d i Draw a circle of radius 1 unit.

Use the construction in section 9.5 of the textbook to find φ, then cut off a length of 1 unit to obtain a length of φ – 1. Mark off this length around the circumference of the circle to divide the circumference into ten equal parts. Join these points to produce a regular decagon.

ii Repeat i but join every second point. 5 φ0 = 1

φ1 = φ = 1 + 52

φ – 1 = 1φ

∴ φ = 1φ

+ 1

φ2 = φ⎝⎜⎛ 1

φ+ 1

⎠⎟⎞

= 1 + φ = 3 + 52

φ3 = φ(1 + φ) = φ2 + φ = (1 + φ) + φ = 1 + 2φ

= 4 + 2 52

= 2 + 5

φ4 = φ(1 + 2φ) = φ + 2φ2

= φ + 2(1 + φ) = 2 + 3φ

= 4 + 3(1 + 5 )2

= 7 + 3 52


φ–1 = 1φ

= φ – 1

= 1 + 5 – 22

= –1 + 52

φ–2 = 1φ

(φ – 1)

= 1 – (φ – 1) = 2 – φ

= 4 – (1 + 5 )2

= 3 – 52

φ–3 = 1φ

(2 – φ)

= 2⎝⎜⎛ 1

φ⎠⎟⎞ – 1

= 2(φ – 1) – 1 = 2φ – 3

= 2 + 2 5 – 62

= 5 – 2

φ–4 = 1φ

(2φ – 3)

= 2 – 3φ

= 2 – 3(φ – 1) = 5 – 3φ

= 10 – 3 – 3 52

= 7 – 3 52

Alternatively, the surd expressions can be multiplied and simplified, for the same answers:

φ – 1 = 1φ

φ = 1 + 1φ

φn + 1 = φ × φn

=⎝⎜⎛ 1 + 1

φ⎠⎟⎞ × φ

b

= φn + φn – 1

6 tn > tn – 1

tn + 1

tn= 1 + tn – 1

tn

Since the Fibonacci sequence is increasing, 1 < tn + 1

tn< 2.

This means the sequence is not diverging to infinity, and has a limit between 1 and 2.

If there is a limit, then when n is large,

tn + 1

tn≈ tn – 1

tn

= 1 + tn – 1

tn

= 1 +1

tn – 1

tn

x = tn + 1

tn

≈ tn – 1

tn

= 1 + 1x

∴ x2 – x – 1 = 0

Using the quadratic formula:

x = –1 + 1 – 4 × 1 × –12

= –1 + 52

= φ

(Rejecting the negative root as x > 0.) Thus the sequence will approach φ as n → ∞.


Solutions to Multiple-choice Questions 1 x

7= 3

5

x = 3 × 75

= 215

D

2 100 parts = 400 kg One part = 4 kg 85 parts = 85 × 4 = 340 kg (copper) B 3 Cost of one article is Q

P.

Cost of R articles = QP

× R

= QRP

D

4 100 parts = 3.2 m 1 part = 3.2

100

= 0.032 m = 3.2 cm C 5 75 parts = 9 seconds 1 part = 9

75= 3

25 seconds

100 parts = 325

× 100

= 12 seconds B

6 10 parts = 50 One part = 5 Largest part is 6 parts = 30 D 7 Ratio of lengths = 10 : 30 = 1 : 3 Ratio of volumes = 1 3 : 33

= 1 : 27 C 8 Ratio of lengths = 4 : 5 Ratio of volumes = 4 3 : 53

= 64 : 125 E 9 XY

3= 12

10= 6

5

XY = 6 × 35

= 3.6 cm E 10 XY ' = 2

3 XY

Area of triangle XY 'Z ' = 4

9 area of triangle XYZ

= 49

× 60 = 803

cm2 E


Solutions to Short-answer Questions 1 a

Both triangles share a common angle X.

∠XPQ = ∠XYZ∠XQP = ∠XYZ

(alternate angles on parallel lines) (AAA) ∴ ΔXPQ ~ ΔXYZ

b i XQXZ

= ZPXY

XQ30

= 2436

= 23

XQ = 20 cm

ii QZ = XZ – XQQZ = 30 – 20

= 10 cm c XP : PY = 24 : 12 = 2 : 1 PQ : YZ = 2 : 3 2 a Ratio of areas ABC : DEF = 12.5 : 4.5 = 25 : 9 AB : DE = 5 : 3 DE = 3 cm b AC : DF = 5 : 3 c EF : BC = 3 : 5

3 h21

= 12.3

h = 212.3

= 21023

m

4 BC = 5 (3-4-5 triangle) So YB = 2.5 ΔBAC ~ ΔBYX

XYYB

= CAAB

XY2.5

= 34

XY = 34

× 2.5 = 158

5 The triangles are similar (AAA).

x – 77

= 34

4x – 28 = 214x = 49

x = 12.25 6

If the two sloping lines were extended to

form a triangle, then the left side of the top triangle would be given by:

xx + 8.8

= 7.226.4

= 72264

= 311

11x = 3x + 26.48x = 26.4

x = 3.3 Now compare the top two triangles: y

7.2= 5.5

3.3= 5

3

y = 5 × 7.23

= 12

7 a Volume of block = 64 cm3

8 parts = 64 cm3

1 part = 8 cm3

5 parts = 40 cm3

3 parts = 24 cm3

Mass of X = 40 × 85

= 64 g

Mass of Y = 24 × 43

= 32 g

Total mass = 96 g b X : Y = 64 : 32 = 2 : 1 (by mass)


c Volume (cm3) : mass (g) = 64 : 96 = 2 : 3 = 1000 : 1500

Volume of 1500 g block is 1000 cm3.

d 3 1000 = 10 cm 8 a Consider ΔBMA and ΔPAD.

∠B = ∠P = 90°∠BAM = ∠PDA

= 90° – ∠PAD∠BMA = ∠PAD

= 90° – ∠BAM ΔBMA ~ ΔPAD (AAA) b BM = 30 cm AM = 50 cm (3-4-5 triangle) Comparing corresponding sides AM and AD: AM : AD = 50 : 60 = 5 : 6 Ratio of areas = 5 2 : 62

= 25 : 36

c PDBA

= ADMA

PD40

= 6050

= 65

PD = 6 × 405

= 48 cm

9 a The same units (cm) must be used to

compare these quantities. 200 : 30 = 20 : 3

b A360

= 202

32 = 4009

A = 4009

× 360

= 16 000 cm2 = 1.6 m2

c V1000

= 203

33 = 800027

V = 800027

× 1000

= 8 000 00027

cm3

= 827

m3

10 a Ratio of radii = 101 : 100 = 1.01 : 1 Ratio of areas = 1 .012 : 1

= 1.0201 : 1= 102.01 : 100

Percentage increase = 2.01% ≈ 2% b Ratio of volumes = 1 .013 : 1

= 1.030301 : 1= 103.0301 : 100

Percentage increase ≈ 3% 11 a XY

BC= AX

AB = 3

9= 1

3

b AY

AC= AX

AB = 3

9= 1

3

c CY

AC= 2

3

d YZ

AD= CY

AC = 2

3

e area AXYarea ABC

= 12

32

= 19

f area CYZarea ACD

= 22

32

= 49


12

Consider ΔAOB and ΔCOD ∠AOB = ∠COD (vertically opposite angles) ∠ABO = ∠CDO (alternate angles on parallel lines) ∠OAB = ∠OCD (alternate angles on parallel lines) ΔAOB ~ ΔCOD (AAA)

COAO

= CDAB

= 31

= 3

CO = 3AOCO + AO = 4AO

AC = $AOAO = 1

4 AC

13 a

PQAB

= YQXB

(corresponding sides of similar triangles) ∠B = ∠Q (corresponding angles of similar triangles) ∴ ΔABX ~ ΔPQY (PAP) b AX

PY= AB

PQ

(similar triangles proven above)

ABPQ

= BCQR

(ABC and PQR are similar) ∴ AX

PY= BC

QR


Chapter 10 – Circular functions I Exercise 10A Solutions 1

a 60° = 60 × π180

= π3

b 144° = 144 × π180

= 4π5

c 240° = 240 × π180

= 4π3

d 330° = 330 × π180

= 11π6

e 420° = 420 × π180

= 7π3

f 480° = 480 × π180

= 8π3

2

a 2π3

= 2π × 1803 × π

= 120°

b 5π6

= 5π × 1806 × π

= 150°

c 7π6

= 7π × 1806 × π

= 210°

d 0.9π = 9π × 18010 × π

= 162°

e 5π9

= 5π × 1809 × π

= 100°

f 9π5

= 9π × 1805 × π

= 324°

g 11π9

= 11π × 1809 × π

= 220°

h 1.8π = 18π × 18010 × π

= 324°

3 Methods will vary depending on the

calculator used. a 0.6 = 34.38° b 1.89 = 108.29° c 2.9 = 166.16° d 4.31 = 246.94° e 3.72 = 213.14° f 5.18 = 296.79° g 4.73 = 271.01° h 6.00 = 343.77°


4 Methods will vary depending on the calculator used.

a 38° = 0.66 b 73° = 1.27 c 107° = 1.87 d 161° = 2.81 e 84°10' = 1.47 f 228° = 3.98 g 136°40' = 2.39 h 329° = 5.74 5

a – π3

= –1π × 1803 × π

= –60°

b –4π = –4π × 180π

= –720°

c –3π = –3π × 180π

= –540°

d – π = –1π × 180π

= –180°

e 5π3

= 5π × 1803 × π

= 300°

f – 11π6

= –11π × 1806 × π

= –330°

g 23π6

= 23π × 1806 × π

= 690°

h – 236

= –23π × 1806 × π

= –690°

6

a –360° = –360 × π180

= –2π

b –540° = –540 × π180

= –3π

c –240° = –240 × π180

= – 4π3

d –720° = –720 × π180

= –4π

e –330° = –330 × π180

= – 11π6

f –210° = –210 × π180

= – 7π6


Exercise 10B Solutions 1 a sin 0 = y-coordinate of (1, 0) = 0 cos 0 = x-coordinate of (1, 0) = 1

b sin 3π2

= –1

cos 3π2

= 0

c sin – 3π2

= 1

cos – 3π2

= 0

d sin 5π2

= sin ⎝⎜⎛ 2π + π

2 ⎠⎟⎞

= sin π2

= 1

cos 5π2

= cos ⎝⎜⎛ 2π + π

2 ⎠⎟⎞

= cos π2

= 0

e sin – 3π = sin (–2π – π)

= sin – π = 0cos – 3π = cos (–2π – π)

= cos – π = –1

f sin 9π2

= sin ⎝⎜⎛ 4π + π

2 ⎠⎟⎞

= sin π2

= 1

cos 9π2

= cos ⎝⎜⎛ 4π + π

2 ⎠⎟⎞

= cos π2

= 0

g sin 7π2

= sin ⎝⎜⎛ 2π + 3π

2 ⎠⎟⎞

= sin 3π2

= –1

cos 7π2

= cos ⎝⎜⎛ 2π + 3π

2 ⎠⎟⎞

= cos 3π2

= 0

h sin 4π = sin 0 = 1

cos 4π = cos 0 = 1 2 Methods will vary depending on the

calculator used. a sin 1.9 = 0.95 b sin 2.3 = 0.75 c sin 4.1 = –0.82 d cos 0.3 = 0.96 e cos 2.1 = –0.50 f cos (–1.6) = –0.03 g sin (–2.1) = –0.86 h sin (–3.8) = 0.61 3 a sin 27π = sin (26π + π)

= sin π = 0cos 27π = cos (26π + π)

= cos π = –1

b sin – 5π2

= sin ⎝⎜⎛ –2π – π

2 ⎠⎟⎞

= sin – π2

= –1

cos – 5π2

= cos ⎝⎜⎛ –2π – π

2 ⎠⎟⎞

= cos – π2

= 0

c sin 27π2

= sin ⎝⎜⎛ 12π + 3π

2 ⎠⎟⎞

= sin 3π2

= –1

cos 27π2

= cos ⎝⎜⎛ 12π + 3π

2 ⎠⎟⎞

= cos 3π2

= 0


d sin – 9π2

= sin ⎝⎜⎛ –4π – π

2 ⎠⎟⎞

= sin – π2

= –1

cos – 9π2

= cos ⎝⎜⎛ –4π – π

2 ⎠⎟⎞

= cos – π2

= 0

e sin 11π2

= sin ⎝⎜⎛ 4π + 3π

2 ⎠⎟⎞

= sin 3π2

= –1

cos 22π2

= cos ⎝⎜⎛ 4π + 3π

2 ⎠⎟⎞

= cos 3π2

= 0

f sin 57π = sin (56π + π)

= sin π = 0cos 57π = cos (56π + π)

= cos π = –1 g sin 211π = sin (210π + π)

= sin π = 0cos 211π = cos (210π + π)

= cos π = –1 h sin – 53π = sin (–52π – π)

= sin – π = 0cos – 53π = cos (–52π – π)

= cos – π = –1


Exercise 10C Solutions 1 Methods will vary depending on the

calculator used. a tan 1.6 = –34.23 b tan (–1.2) = –2.57 c tan 136° = –0.97 d tan (–54°) = –1.38 e tan 3.9 = 0.95 f tan (–2.5) = 0.75 g tan 239° = 1.66 2

a tan π = sin πcos π

= 0–1

= 0

b tan – π = sin – πcos – π

= 0–1

= 0

c tan 7π2

=sin 7π

2

cos 7π2

=sin 3π

2

cos 3π2

= – 10

= undefined

d tan (–2π) = sin (–2π)cos (–2π)

= 0–1

= 0

e tan 5π2

=sin 5π

2

cos 5π2

=sin π

2

cos π2

= 10

= undefined

f tan ⎝⎜⎛ – π

2 ⎠⎟⎞ =

sin ⎝⎜⎛ – π

2 ⎠⎟⎞

cos ⎝⎜⎛ – π

2 ⎠⎟⎞

= – 10

= undefined

3 a tan 180 = sin 180

cos 180 = 0

–1= 0

b tan 360 = sin 360cos 360

= 01

= 0

c tan 0 = sin 0cos 0

= 01

= 0

d tan – 180 = sin – 180cos – 180

= 0–1

= 0

e tan – 540 = sin – 540cos – 540

= 0–1

= 0

f tan 720 = sin 720

cos 720 = 0

1= 0


Exercise 10D Solutions 1 a cos θ = 3

8= 0.375

θ = 67°59'

b x

5= cos 25°

x = 5 × 0.9063. . . ≈ 4.5315

c x

6= sin 25°

x = 6 × 0.4226. . . ≈ 2.5357

d x

10= cos 50°

x = 10 × 0.642787. . . ≈ 6.4279

e tan θ = 6

5= 1.2

θ = 50°12'

f x

10= sin 20°

x = 10 × 0.342020. . . ≈ 3.4202

g 5

x= tan 65°

x = 5tan 65

= 52.1445. . .

≈ 2.3315

h x

7= sin 70°

x = 7 × 0.9396. . . ≈ 6.5778

i 5

x= cos 40°

x = 5cos 40°

= 50.7660. . .

≈ 6.5270

2 a a = cos 40° = 0.7660

b = sin 40° = 0.6428 b c = –a = –0.7660 d = b = 0.6428 c i cos 140° = –0.7660

sin 140° = 0.6428 ii cos 40° = 0.7660

∴cos 140° = – cos 40°


Exercise 10E Solutions 1 a sin (π + θ) = – sin θ

= –0.42 b cos (π – x) = – cos x

= –0.7 c sin (2π – θ) = – sin θ

= –0.42 d tan (π – α) = – tan α

= –0.38 e sin (π – θ) = sin θ

= 0.42 f tan (2π – α) = – tan α

= –0.38 g cos (π + x) = – cos x

= –0.7 h cos (2π – x) = cos x

= 0.7 2 a sin (180 + x)° = – sin x°

= –0.7

b cos (180 + θ)° = – cos θ° = –0.6

c tan (360 – α)° = – tan α° = –0.4

d cos (180 – θ)° = – cos θ° = –0.6

e sin (360 – x)° = – sin x° = –0.7

f sin ( – x)° = – sin x° = –0.7

g tan (360 + α)° = tan α° = 0.4

h cos ( – θ)° = cos θ° = –0.6

3 a a = cos (π – θ) = – cos θ

= – 12

b b = sin (π – θ) = sin θ

= 32

c c = cos ( – θ) = cos θ

= 12

d d = sin ( – θ) = – sin θ

= – 32

e tan (π – θ) = – tan θ

= –

⎝⎜⎜⎜⎛ 3

212 ⎠⎟⎟⎟⎞

= – 3

f tan ( – θ) = – tan θ

= – 3

4 a d = sin (π + θ) = – sin θ

= – 32

b c = cos (π + θ) = – cos θ

= 12

c tan (π + θ) = tan θ

=

32

– 12

= – 3

d sin (2π – θ) = – sin θ

= – 32

e cos (2π – θ) = cos θ

= – 12


Exercise 10F Solutions 1 Use symmetry to evaluate each value.

a sin 2π3

= sin π3

= 32

b cos 3π4

= – cos π4

= – 12

c tan 5π6

= – tan π6

= – 13

d sin 7π6

= – sin π6

= – 12

e cos ⎝⎜⎛5π

4 ⎠⎟⎞ = – cos π

4= – 1

2

f tan 4π3

= tan π3

= 3

g sin 5π3

= – sin π3

= – 32

h cos 7π4

= cos π4

= 12

i tan 11π6

= – tan π6

= – 13

2

a 120° = 2π3

sin 2π3

= sin π3

= 32

cos 2π3

= – cos π3

= – 12

tan 2π3

= – tan π3

= – 3

b 135° = 3π4

sin 3π4

= sin π4

= 12

cos 3π4

= – cos π4

= – 12

tan 3π4

= – tan π4

= –1

c 210° = 7π6

sin 7π6

= – sin π6

= – 12

cos 7π6

= – cos π6

= – 32

tan 7π6

= tan π6

= 13

d 240° = 4π3

sin 4π3

= – sin π3

= – 32

cos 4π3

= – cos π3

= – 12

tan 4π3

= tan π3

= 3

e 135° = 7π4

sin 7π4

= – sin π4

= – 12

cos 7π4

= cos π4

= 12

tan 7π4

= – tan π4

= –1


f 390° = 13π6

= 2π + π6

sin 13π6

= sin π6

= 12

cos 13π6

= cos π6

= 32

tan 13π6

= tan π6

= 13

g 420° = 7π3

= 2π + π3

sin 7π3

= sin π3

= 32

cos 7π3

= cos π3

= 12

tan 7π3

= tan π3

= 3

h –135° = – 7π4

sin – 7π4

= – sin π4

= – 12

cos – 7π4

= – cos π4

= – 12

tan – 7π4

= tan π4

= 1

i –300° = – 5π3

sin – 5π3

= sin π3

= 32

cos – 5π3

= cos π3

= 12

tan – 5π3

= tan π3

= 3

j –60° = – π3

sin – π3

= – sin π3

= – 32

cos – π3

= cos π3

= 12

tan – π3

= – tan π3

= – 3

3

a sin – 2π3

= – sin π3

= – 32

b cos 11π4

= – cos π4

= – 12

c tan 13π6

= tan π6

= 13

d tan 15π6

= tan 3π6

= tan π2

= undefined

e cos 14π4

= cos 2π4

= cos π2

= 0

f cos 3π4

= – cos π4

= – 12

g sin 11π4

= sin π4

= 12

h cos – 21π3

= cos – 7π

= cos π = –1


Exercise 10G Solutions 1

a i Period = 2πn

= 2π1

= 2π

ii Amplitude = a = 2

b i Period = 2πn

= 2π2

= π


c i Period = 2πn

= 2π3


d i Period = 2πn

=2π12

= 4π


e i Period = 2πn

= 2π3

ii Amplitude = a = 4 f i Period = 2π

n= 2π

4= π

2


g i Period = 2πn

=2π12

= 4π

ii Amplitude = a = 2 2

a Period = 2π2

= π

Amplitude = 3

b Period = 2π3

Amplitude = 2

c Period =2π12

= 4π

Amplitude = 4

d Period = 2π3

Amplitude = 12

e Period = 2π3

Amplitude = 4


f Period = 2π2

= π

Amplitude = 5

g Period =2π12

= 4π

Amplitude = 3

h Period = 2π4

= π2

Amplitude = 2

i Period =2π13

= 6π

Amplitude = 2

3

a Period = 2π2

= π

Amplitude = 1

b Period =2π13

= 6π

Amplitude = 2

c Period = 2π3

Amplitude = 2

d Period = 2π3

Amplitude = 2


4 Period =2π23

= 6π2

= 3π

Amplitude = 52

f(0) = 52

cos 0

= 52

f(2π) = 52

cos 4π3

= – 54

5

a Period (f) = 2π1

= 2π

Amplitude (f) = 1

Period (g) = 2π1

= 2π

Amplitude (g) = 1

b The lines intersect at π4 and 5π4 .


Exercise 10H Solutions 1

a Period = 2π2

= π

Amplitude = 4 Greatest/least y = ± 4

b Period = 2π2

= π


c Period = 2π3


2 a Start with y = 3 sin θ

Period = 2π

1= 2π


Translate graph π2 units to the right.

b Start with y = sin 2θ

Period = 2π2

= π

Amplitude = 1 Greatest/least y = ± 1 Translate graph π units to the left.

c Start with y = 2 sin 3θ

Period = 2π3


Translate graph π4 units to the left.


d Start with y = 3 sin 2θ

Period = 2π2

= π



e Start with y = 3 sin 2x

Period = 2π2

= π


No translation.

f Start with y = 2 cos 3θ

Period = 2π3



g Start with y = 2 cos 2θ

Period = 2π2

= π



h Start with y = –3 sin 2x

Period = 2π2

= π


Reflect graph in the x-axis.

i Start with y = –3 cos 2θ

Period = 2π2

= π


Reflect graph in the x-axis and translate π2 units to the left.


3

a f(0) = cos⎝⎜⎛ – π

3 ⎠⎟⎞

= cos π3

= 12

f(2π) = cos⎝⎜⎛ 2π – π

3 ⎠⎟⎞

= cos π3

= 12

b Start with y = cos x

Period = 2π1

= 2π



4

a f(0) = sin 2 ⎝⎜⎛ – π

3 ⎠⎟⎞

= sin – 2π3

= – 32

f(2π) = sin 2 ⎝⎜⎛2π – π

3 ⎠⎟⎞

= sin – 2π3

= – 32

b Start with y = sin 2x

Period = 2π2

= π



5

a f( – π) = sin 3 ⎝⎜⎛ – π + π

4 ⎠⎟⎞

= sin – 3π4

= – 12

f(π) = sin 3 ⎝⎜⎛ π + π

4 ⎠⎟⎞

= sin 5π4

= – 12

b Start with y = sin 3x

Period = 2π3




Exercise 10I Solutions 1 a sin x = – 1

2

When x = π4

, sin x = 12

∴ x = π + π4

and 2π – π4

= 5π4

and 7π4

b cos x = 12

x = π4

and 2π – π4

= π4

and 7π4

2 a cos x = – 1

2

When x = π4

, cos x = 12

∴ x = π – π4

and – π + π4

= 3π4

and – 3π4

b sin x = 32

x = π3

and π – π3

= π3

and 2π3

c cos x = – 12

When x = π3

, cos x = 12

∴ x = π – π3

and – π + π3

= 2π3

and – 2π3

3

a cos θ = – 32

When θ = 30°, cos θ = 32

∴ θ = 180 – 30 and 180 + 30 = 150° and 210°

b sin θ = 1

2θ = 30 and 180 – 30 = 30° and 150°

c cos θ = – 1

2

When θ = 60°, cos θ = 12

∴ θ = 180 – 60 and 180 + 60 = 120° and 240°

d 2 cos θ = –1cos θ = – 1

2

When θ = 60°, cos θ = 12

∴ θ = 180 – 60 and 180 + 60 = 120° and 240°

e 2 sin θ = 3

sin θ = 32

θ = 60 and 180 – 60 = 60° and 120°

f 2 sin θ = 1

sin θ = 12

θ = 45 and 180 – 45 = 45° and 135°


4 Make sure your calculator is set to radians for these questions.

a sin x = 0.8 x ≈ 0.927 and π – 0.927 ≈ 0.93 and 2.21

b cos x = –0.4 When x = 0.4, cos x = 1.159. . .

∴ x ≈ π – 1.159 and π + 1.159

≈ 4.30 and 1.98 c sin x = –0.35 When x = 0.35, sin x ≈ 0.357

∴ x ≈ π + 0.357 and 2π – 0.357

≈ 3.50 and 5.93 d sin x = 0.4

x ≈ 0.411 and π – 0.411 ≈ 0.41 and 2.73

e cos x = –0.7 When x = 0.7, cos x ≈ 0.795

∴ x ≈ π – 0.795 and π + 0.795

≈ 2.35 and 3.95 f cos x = –0.2 When x = 0.2, cos x ≈ 1.369

∴ x ≈ π – 1.369 and π + 1.369

≈ 1.77 and 4.51 5 a sin x = 0.6

x ≈ 0.643, π – 0.643, 2π + 0.643 and 4π – 0.643

≈ 0.64, 2.498, 6.73 and 8.871

b sin x = – 1

2

When x = π4

, sin x = 12

∴ x = π + π4

, 2π – π4

, 3π + π4

and 4π – π4

= 5π4

, 7π4

, 13π4

and 15π4

c sin x = 32

x = π3

,π – π3

, 2π + π3

and 3π – π3

= π3

, 2π3

, 7π3

and 8π3

6 a 2θ ∈ [0, 4π]

sin 2θ = – 12

When sin 2θ = 1

2 , 2θ = π

6

∴ 2θ = π + π6

, 2π – π6

, 3π + π6

and 4π – π6

= 7π6

, 11π6

, 19π6

and 23π6

θ = 7π12

, 11π12

, 19π12

and 23π12

b 2θ ∈ [0, 4π]

cos 2θ = 32

2θ = π6

, 2π – π6

, 2π + π6

and 4π – π6

= π6

, 11π6

, 13π6

and 23π6

θ = π12

, 11π12

, 13π12

and 23π12

c 2θ ∈ [0, 4π]

sin 2θ = 12

2θ = π6

,π – π6

, 2π + π6

and 3π – π6

= π6

, 5π6

, 13π6

and 17π6

θ = π12

, 5π12

, 13π12

and 17π12


d 3θ ∈ [0, 4π]sin 3θ = – 1

2

When sin 3θ = 12

, 3θ = π4

∴ 3θ = π + π4

, 2π – π4

, 3π + π4

, 4π – π4

,

5π + π4

and 6π – π4

= 5π4

, 7π4

, 13π4

, 15π4

, 21π4

and 23π4

θ = 5π12

, 7π12

, 13π12

, 15π12

, 21π12

and 23π12

e 2θ ∈ [0, 4π]

cos 2θ = – 32

When cos 2θ = 32

, 2θ = π6

∴ 2θ = π – π6

,π + π6

, 3π – π6

and 3π + π6

= 5π6

, 7π6

, 17π6

and 19π6

θ = 5π12

, 7π12

, 17π12

and 19π12

f 2θ ∈ [0, 4π]

sin 2θ = – 12

When sin 2θ = 12

, 2θ = π4

∴ 2θ = π + π4

, 2π – π4

, 3π + π4

and 4π – π4

= 5π4

, 7π4

, 13π4

and 15π4

θ = 5π8

, 7π8

, 13π8

and 15π8

7 a

2θ ∈ [0, 4π]sin 2θ = –0.8

When sin 2θ = 0.8, 2θ ≈ 0.927

∴ 2θ ≈ π + 0.927, 2π – 0.927, 3π + 0.927 and 4π – 0.927

≈ 4.067, 5.353, 10.347 and 11.633θ ≈ 2.034, 2.678, 5.176 and 5.820

b

2θ ∈ [0, 4π]sin 2θ = –0.6

When sin 2θ = 0.6, 2θ ≈ 0.643

∴ 2θ ≈ π + 0.643, 2π – 0.643, 3π + 0.643 and 4π – 0.643

≈ 3.783, 5.639, 10.068 and 11.922θ ≈ 1.892, 2.820, 5.034 and 5.961

c

2θ ∈ [0, 4π]cos 2θ = 0.4

2θ ≈ 1.159, 2π – 1.159, 2π + 1.159 and 4π – 1.159 ≈ 1.159, 5.124, 7.443 and 11.408θ ≈ 0.580, 2.562, 3.721 and 5.704

d 3θ ∈ [0, 3π]

cos 3θ = 0.63θ ≈ 0.927, 2π – 0.927, 2π + 0.927,

4π – 0.927, 4π + 0.927 and 6π – 0.927 ≈ .927, 5.354, 7.210, 11.641, 13.494 and 17.922θ ≈ 0.309, 1.785, 2.403, 3.880, 4.498 and 5.974

8

a Period = 2π1

= 2π

Amplitude = 1

b Solve cos x = 1

2 for x ∈ [–2π, 2π].

cos x = 12

x = π3

, 2π – π3

, – π3

and – 2π + π3

= – 5π3

, – π3

,π3

and 5π3

Mark the points

⎝⎜⎛ – 5π

3, 1

2⎠⎟⎞ ,⎝⎜⎛ – π

3, 12⎠⎟⎞ ,⎝⎜⎛ π

3, 12⎠⎟⎞ and

⎝⎜⎛ 5π

3, 12⎠⎟⎞ .


c Solve cos x = – 12

for x ∈ [–2π, 2π].

x = π – π3

,π + π3

, – π + π3

and – π – π3

= – 4π3

, – 2π3

, 2π3

and 4π3

Mark the points

⎝⎜⎛ – 4π

3, – 1

2⎠⎟⎞ ,⎝⎜⎛ – 2π

3, – 1

2⎠⎟⎞ ,⎝⎜⎛ 2π

3, – 1

2⎠⎟⎞

and ⎝⎜⎛ 4π

3, – 1

2⎠⎟⎞ .


Exercise 10J Solutions 1 x-axis intercepts occur when y = 0 a The graph is y = 2 sin x raised by 1 unit. sin x = – 1

2

When sin x = 12 , x = π

6

∴ x = π + π6

, 2π – π6

= 7π6

and 11π6

b The graph is y = 2 sin 2x (amplitude 2,

period π) lowered by 3 units.

2x ∈ [0, 4π]

sin 2x = 32

2x = π3

,π – π3

, 2π + π3

and 3π – π3

= π3

, 2π3

, 7π3

and 8π3

x = π6

, π3

, 7π6

and 4π3

c The graph is y = 2 cos x raised by 1 unit.

cos x = – 12

When cos x = 12

, x = π4

∴ x = π – π4

,π + π4

= 3π4

and 5π4

d The graph is y = 2 sin 2x (amplitude 2,

period π) moved down 2 units. 2x ∈ [0, 4π]

2 sin 2x = 2sin 2x = 1

2x = π2

and 2π + π2

x = π4

and 5π4


e The graph is y = 2 sin x raised by

1 unit and moved π4 units right.

sin ⎝⎜⎛ x = π

4 ⎠⎟⎞ = – 1

2 for x ∈

⎣⎢⎡ – π

4, 7π

4 ⎦⎥⎤

When sin ⎝⎜⎛ x – π

4 ⎠⎟⎞ = 1

2, x = π

4

∴ x – π4

= – π4

, π + π4

, 2π – π4

x = 0, 3π2

and 2π

2

a The graph of y = 2 sin 3x (period 2π3 , amplitude 2) translated 2 units down.

b The graph of y = 2 cos 3x (period 2π3 , amplitude 2) translated π3 units to the left.


c The graph of y = 2 sin 2x (period π, amplitude 2) translated 3 units down.

d The graph of y = 2 cos 2x (period π, amplitude 2) translated 1 unit up.

e The graph of y = 2 cos 2x (period π, amplitude 2) translated π3 units to the right and 1 unit

down.

f The graph of y = 2 sin 2x (period π, amplitude 2) translated π6 units to the left and 1 unit up.


3 a The graph of y = 2 sin 2x (period π,

amplitude 2) translated π3 units to the

left and 1 unit up.

b The graph of y = 2 sin 2x (period π,

amplitude 2) reflected in the x-axis and

then translated π6 units to the left and

1 unit up.

c The graph of y = 2 cos 2x (period π,

amplitude 2) translated π4 units to the

left and 3 units up.


Exercise 10K Solutions 1

a A =

⎣⎢⎢⎢⎡cos π

6

sin π6

– sin π6

cos π6⎦⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡ 3

212

– 12

32 ⎦⎥⎥⎥⎤

⎣⎢⎢⎢⎡ 3

212

– 12

32 ⎦⎥⎥⎥⎤

⎣⎢⎡ 1

0⎦⎥⎤ =

⎣⎢⎢⎢⎡ 3

212 ⎦⎥⎥⎥⎤

⎝⎜⎛ 3

2, 1

2⎠⎟⎞

b A =

⎣⎢⎢⎢⎡ cos – π

4

sin – π4

– sin – π4

cos – π4 ⎦⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡ 1

2

– 12

1212 ⎦⎥⎥⎥⎤

c A =


2

sin π2

– sin π2

cos π2⎦⎥⎥⎥⎤

=⎣⎢⎡ 0

1–1

0⎦⎥⎤

⎣⎢⎡ 0

1–10⎦⎥⎤⎣⎢⎡ –1

0⎦⎥⎤ =

⎣⎢⎡ 0

–1⎦⎥⎤

(0, – 1)

d A =

⎣⎢⎢⎢⎡ cos – 2π

3

sin – 2π3

– sin 2π3

cos – 2π3 ⎦⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡ – 1

2

– 32

32

– 12⎦⎥⎥⎥⎤

⎣⎢⎢⎢⎡ – 1

2

– 32

32

– 12⎦⎥⎥⎥⎤

⎣⎢⎡ 0

–1⎦⎥⎤ =

⎣⎢⎢⎢⎡ – 3

212 ⎦

⎥⎥⎥⎤

⎝⎜⎛ – 3

2, 12⎠⎟⎞

e A =

⎣⎢⎢⎢⎡cos 5π

4

sin 5π4

– sin 5π4

cos 5π4 ⎦⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡ – 1

2

– 12

12

– 12 ⎦⎥⎥⎥⎤

⎣⎢⎢⎢⎡ – 1

2

– 12

12

– 12 ⎦⎥⎥⎥⎤

⎣⎢⎢⎢⎡ 1

212 ⎦⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡ – 1

2+ 1

2

– 12

– 12 ⎦⎥⎥⎥⎤

=⎣⎢⎡ 0

–1⎦⎥⎤

(0, – 1)


f A =

⎣⎢⎢⎢⎡ cos – 5π

6

sin – 5π6

– sin – 5π6

cos – 5π6 ⎦⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡ – 3

2

– 12

12

– 32 ⎦⎥⎥⎥⎤

⎣⎢⎢⎢⎡ – 3

2

– 12

12

– 32 ⎦⎥⎥⎥⎤

⎣⎢⎢⎢⎡ 3

212 ⎦⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡ – 3

4+ 1

4

– 34

– 34 ⎦⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡ – 1

2

– 32 ⎦⎥⎥⎥⎤

⎝⎜⎛ – 1

2, – 3

2 ⎠⎟⎞

2

a

⎣⎢⎢⎢⎡ 3

212

– 12

32 ⎦⎥⎥⎥⎤

=


6

sin π6

– sin π6

cos π6 ⎦⎥⎥⎥⎤

π6 anticlockwise. b Both the cosine and sine of the angle are

negative, so the angle is in the third quadrant.

cos θ = – 12

θ = π + π3

= 4π3

⎣⎢⎢⎢⎡ – 1

2

– 32

32

– 12⎦⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡cos 4π

3

sin 4π3

– sin 4π3

cos 4π3 ⎦⎥⎥⎥⎤

4π3 anticlockwise.

c The angle is in the first quadrant and cos θ = 4

5.

cos–1 45

anticlockwise.

3

a Rπ4

=


4

sin π4

– sin π4

cos π4 ⎦⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡ 1

212

– 12

12 ⎦⎥⎥⎥⎤

b

⎣⎢⎢⎢⎡ 1

212

– 12

12 ⎦⎥⎥⎥⎤

⎣⎢⎡ 2

1⎦⎥⎤ =

⎣⎢⎢⎢⎡ 2

2– 1

222

+ 12 ⎦⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡ 1

232 ⎦⎥⎥⎥⎤

⎝⎜⎛ 1

2, 3

2 ⎠⎟⎞

c i R3π4

=

⎣⎢⎢⎢⎡cos 3π

4

sin 3π4

– sin 3π4

cos 3π4 ⎦⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡ – 1

212

– 12

– 12 ⎦⎥⎥⎥⎤

ii Rπ2

=


2

sin π2

– sin π2

cos π2 ⎦⎥⎥⎥⎤

=⎣⎢⎡ 0

1–10 ⎦⎥⎤


iii R5π6

=

⎣⎢⎢⎢⎡cos 5π

6

sin 5π6

– sin 5π6

cos 5π6 ⎦⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡ – 3

2

12

– 12

– 32 ⎦⎥⎥⎥⎤

iv R – π4

=

⎣⎢⎢⎢⎡ cos – π

4

sin – π4

– sin – π4

cos – π4 ⎦⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡ 1

2

– 12

1212 ⎦⎥⎥⎥⎤

d Use Pythagoras’ theorem for the unit

circle triangle with hypotenuse 1 and sides cos θ and sin θ.

sin2θ +⎝⎜⎛4

5⎠⎟⎞2

= 12

sin2θ = 1 – 1625

= 925

sin θ = 35

(since θ is in the first quadrant)

Rθ =⎣⎢⎡ cos θ

sin θ– sin θ

cos θ⎦⎥⎤

=

⎣⎢⎢⎢⎡ 4

535

– 3545⎦⎥⎥⎥⎤


Exercise 10L Solutions 1

a i d = 12 + 12 cos 16 π (5.7 + 0.3)

= 12 + 12 cos π = 0

ii d = 12 + 12 cos 16

π(2.7 + 0.3)

= 12 + 12 cos π2

= 6

b 5 = 12 + 12 cos 16

π⎝⎜⎛ t + 1

3⎠⎟⎞

cos 16 π⎝⎜⎛ t + 1

3⎠⎟⎞ = – 7

1216 π⎝⎜⎛ t + 1

3⎠⎟⎞ = 2.1936 or 4.0896

t + 13

= 4.1895 or 7.8106

t = 3.856 or 7.477

3.856 × 30 = 116 days, 26th April 7.477 × 30 = 224 days, 14th August 2

a Sketch the graph of y = 3 sin πt6

(period 12, amplitude 3) translated 10 units up.

b Solve 10 + 3 sin πt6

= 8.5 for

0 ≤ πt6≤ 24 × πt

6= 0 ≤ πt

6≤ 4πt

sin πt6

= 8.5 – 103

= – 12

πt6

= π + π6

, 2π – π6

, 3π + π6

, 4π – π6

t = 7, 11, 19, 23

From the graph: {t: 0 ≤ t ≤ 7}∪{t: 11 ≤ t ≤ 19}∪

{t: 23 ≤ t ≤ 24}

c This would occur half an hour each side of the maximum.

The maximum occurs when t = 3.

w = 10 + 3 sin 2.5π6

= 12.9

3 a The period is the time from high tide to

low tide to high tide = 6 + 6 = 12 hours. 360

r= 12

r = 36012

= 30

Amplitude = 7 – 32

= 2

q = 2 At high tide, D = p + q × 1

7 = p + 2 p = 5

b The graph is of It

has amplitude 2, period 12, and is the cosine curve raised 5 units.

D = 5 + 2 cos(30t°).

c t = 0 is a maximum, therefore a high

tide. The low tide occurs when t = 6. Solve for D = 4, then find the first

solution where t > 6. 4 = 5 + 2 cos 30t°

cos 30t° = – 12

30t° = 150° or 210°t = 5 or 7

Since low tide occurs at t = 6, the ship can enter one hour after low tide.


4 a This occurs when sin 3t = 1 x = 3 + 2 = 5 m b This occurs when sin 3t = –1 x = 3 – 2 = 1 m c sin 3t = 1 for 0 ≤ 3t ≤ 15

3t = π2

, 5π2

, 9π2

t = π6

, 5π6

, 9π6

= 0.524, 2.618, 4.712 s

d sin 3t = 0 for 0 ≤ 3t ≤ 9 3t = 0,π, 2π

t = 0,π3

, 2π3

= 0, 1.047, 2.094 s

5 a t = 4

A°C = 21 – 3 cos π3

= 19.5°C

b D = A – B

= –1 + 2 cos πt12

c The graph is of y = –1 + 2 cos πt12

.

It has amplitude 2, period 24, and is the cosine curve moved 1 unit down.

d –1 + 2 cos πt12

< 0

cos πt12

< 12

π3

< πt12

< 5π3

4 < t < 20

From 4 am to 8 pm.

6

a π6

t – π3

= 0

π6

t = π3

t = 2 am

b 6 + 4 cos ⎝⎜⎛ π

6 t – π

3 ⎠⎟⎞ = 2

cos ⎝⎜⎛ π

6 – π

3 ⎠⎟⎞ = –1

π6

t – π3

= π, 3π

t – 2 = 6, 18t = 8 or t = 20

8 am and 8 pm 7 a i Amplitude = 5 – 2

2 = 1.5 m

ii Period = 12 hours iii From graph, the shape is a cosine curve reflected in the x-axis. Graph will be of the form d(t) = –1.5 cos kt + 3.5

Period = 2πk

= 12

k = 2π12

= π6

d(t) = –1.5 cos πt6

+ 3.5

iv 1.5 m b 3.5 is the middle of the hour hand’s path.

From the graph, this will occur from 3 am to 9 am and from 3 pm to 9 pm.

8 a Use the information given, starting at

noon.


b In this case, it is easiest to make t = 0 at noon, which is the reference point.

D = 4 + cos kt

Period = 2πk

= 16

k = 2π16

= π8

D = 4 + cos πt8

D = 4 ⇒ cos πt8

= 0

πt8

= π2

or – π2

t = 4 or – 4

It can enter after 8 am and must leave by 4 pm.

c D = 4 + cos πt8

d = 3.5 ⇒ cos πt8

= –0.5

πt8

= 2π3

or – 2π3

t = 163

or – 163

= 5 13

or – 5 13

It can enter after 6:40 am and must leave by 5:20 pm.

9

a i 2ππ26

= 2π × 26π

= 52 weeks (1 year)

ii 3000 iii 4000 ± 3000 = [1000, 7000]

b i N(0) = 3000 sin –10π26

+ 4000

= 1194.95

(1195 ants, more or less)

N(100) = 3000 sin 90π26

+ 4000

= 1021.87 (1022 ants, more or less)

ii

c i 3000 sin π(t – 10)26

+ 4000 = 7000

sin π(t – 10)26

= 1

π(t – 10)26

= π2

t = 13 + 10 = 23

t = 23 and t = 75, since the period is 52 weeks.

ii 3000 sin π(t – 10)26

+ 4000 = 1000

sin π(t – 10)26

= –1

π(t – 10)26

= 3π2

t = 39 + 10 = 49

This is the only value since the period is 52 weeks.

d 3000 sin π(t – 10)26

+ 4000 > 5500

sin π(t – 10)26

> 12

π6

< π(t – 10)26

< 5π6

and

13π6

< π(t – 10)26

< 17π6

133

< t – 10 < 653

and

1693

< t – 10 < 2213

⎝⎜⎛ 14 1

3, 31 2

3⎠⎟⎞ ∪

⎝⎜⎛ 66 1

3, 83 2

3⎠⎟⎞


e The given population varies between 10 000 and 40 000, a = 15 000 and d = 25 000.

Maximum to minimum is half a period, so the period = 20.

2π ÷ πb

= 20

2π = 20πb

b = 20π2π

= 10

Maximum at t = 10 means

π(10 – c)10

= π2

10 – c = 5c = 5


Solutions to Multiple-choice Questions

1 Pythagoras’ rule gives AC = a2 + b2

cos x = ABAC

= b2

a2 + b2 B

2 y = 2 sin (3x – π)

= 2 sin 3(x – π3 ⎠⎟⎞

Period = 2π3

A

3 y = –5 cos 5x + 3 Amplitude = |–5| = 5 D 4 5 sin (2x – π) + 2 = 0

Period = 2π2

= π

Amplitude = 5

Maximum and minimum y values are y = 7 and y = –3. Since y = 0 is within the range of the function, the function will pass through the line twice each cycle. The function covers two cycles over the given domain, and so will pass through y = 0 four times. There are four solutions. D

5 3π11

= 3 × π × 180π × 11

= 54011

= 49.09 C

6 x ∈ ⎝⎜⎛5π

12, 23π

12 ⎠⎟⎞

3x ∈ ⎝⎜⎛ 5π

4 , 23π

4 ⎠⎟⎞

sin 3x = – 22

= – 12

3x = π + π4

, 2π – π4

, 3π + π4

,

4π – π4

, 5π + π4

, 6π – π4

Omit the first and last as they are not in the interval.

3x = 7π4

, 13π4

, 15π4

, 21π4

x = 7π

12, 13π

12, 5π

4, 7π

4 D

7 cos ⎝⎜⎛ – 13π

6 ⎠⎟⎞ = cos

⎝⎜⎛–2π – π

6 ⎠⎟⎞

= cos – π6

= sin 2π3

E

8 tan (180 – θ) = sin (180 – θ)cos (180 – θ)

= sin θ– cos θ

= – cos (90 + θ)– sin (90 + θ)

= cos (90 + θ)sin (90 + θ)

E

9 Period = 2π4π – 3π

= 2ππ

= 2 B 10 Both cos θ and sin θ are positive,

so the angle is in the first quadrant. cos θ = 1

2θ = 60°

An anticlockwise rotation of 60° about the origin. B


Solutions to Short-answer Questions 1

a 330° = 330 × π180

= 11π6

b 810° = 810 × π180

= 9π2

c 1080° = 1080 × π180

= 6π

d 1035° = 1035 × π180

= 23π4

e 135° = 135 × π180

= 3π4

f 405° = 405 × π180

= 9π4

g 390° = 390 × π180

= 13π6

h 420° = 420 × π180

= 7π3

i 80° = 80 × π180

= 4π9

2

a 5π6

= 5π × 1806π

= 150°

b 7π4

= 7π × 1804π

= 315°

c 11π4

= 11π × 1804π

= 495°

d 3π12

= 3π × 18012π

= 45°

e 15π2

= 15π × 1802π

= 1350°

f – 3π4

= –3π × 1804π

= –135°

g – π4

= – π × 1804π

= –45°

h – 11π4

= –11π × 1804π

= –495°

i – 23π4

= –23π × 1804π

= –1035°

3

a sin 11π4

= sin ⎝⎜⎛ 2π + 3π

4 ⎠⎟⎞

= sin 3π4

= 12

b cos – 7π4

= cos ⎝⎜⎛–2π + π

4 ⎠⎟⎞

= cos π4

= 12

c sin 11π6

= sin ⎝⎜⎛ 2π – π

6 ⎠⎟⎞

= sin – π6

= – 12

d cos – 7π6

= cos ⎝⎜⎛–2π + 5π

6 ⎠⎟⎞

= cos 5π6

= – 32

e cos 13π6

= cos ⎝⎜⎛ 2π + π

6 ⎠⎟⎞

= cos π6

= 32

f sin 23π6

= sin ⎝⎜⎛ 4π – π

6 ⎠⎟⎞

= sin – π6

= – 12

g cos – 23π3

= cos ⎝⎜⎛–6π + π

3 ⎠⎟⎞

= cos π3

= 12


h sin – 17π4

= cos ⎝⎜⎛ –4π – π

4 ⎠⎟⎞

= cos – π4

= – 12

4

a Period =2π12

= 4π

Amplitude = 2

b Period = 2π4

= π2

Amplitude = 3

c Period = 2π3

Amplitude = 12

d Period = 2π2

= π

Amplitude = 3

e Period =2π13

= 6π

Amplitude = 4

f Period =2π23

= 3π

Amplitude = 23

5 a The graph of y = 2 sin 4x has period

2π4

= π2 and amplitude 2.

b The graph of y = –3 cos x3

has period

2π13

= 6π and amplitude 3. It is

y = 3 cos x3

reflected in the x-axis.

c The graph of y = 2 sin 3x

(period 2π3 , amplitude 2), reflected in

the x-axis.

d The graph of y = 2 cos x

3 has period

2π13

= 6π and amplitude 3.

e The graph of y = sin x, translated

π4 units to the right.


f The graph of y = sin x, translated

2π3 units to the left.

g The graph of y = 2 cos x, translated

5π6 units to the right.

h The graph of y = 3 cos x, reflected in the

x-axis and translated π6 units to the right.

6

a sin θ = – 32

sin θ = 32

⇒ θ = π3

θ = – π3

and – π + π3

= – π3

and – 2π3

b 2θ ∈ [–2π, 2π]

sin 2θ = 32

⇒ 2θ = π3

2θ = – π3

, – π + π3

,π + π3

and 2π – π3

= – π3

, – 2π3

, 4π3

and 5π3

θ = – π6

, – π3

, 2π3

and 5π6

c θ – π3

∈ ⎣⎢⎡ – π

3, 5π

3 ⎦⎥⎤

sin ⎝⎜⎛ θ – π

3 ⎠⎟⎞ = 1

2⇒ θ – π

3= π

6

θ – π3

= – π6

and π + π6

= – π6

and 7π6

θ = π6

and 3π2

d θ + π3

∈ ⎣⎢⎡ π

3 , 7π

3 ⎦⎥⎤

sin ⎝⎜⎛ θ + π

3 ⎠⎟⎞ = –1

θ + π3

= 3π2

θ = 3π2

– π3

= 7π6

e θ = 0 ⇒ π3

– θ = π3

θ = 2π ⇒ π – θ = – 5π3

∴ π3

– π ∈ ⎣⎢⎡ – 5π

3 ,π

3 ⎦⎥⎤

sin ⎝⎜⎛ π

3– θ

⎠⎟⎞ = 1

2⇒ π

3– θ = π

6θ3

– θ = – π6

and – π + π6

= – π6

and – 5π6

– θ = – π2

and – 7π6

θ = π2

and 7π6


7 a The graph of y = 2 sin 4x (period π ,

amplitude 2) translated 1 unit up.

b The graph of y = 2 cos x reflected in the

x-axis and translated 1 unit up.

c The graph of y = 3 cos x translated

π3 units to the left.

d The graph of y = cos x reflected in the

x-axis and translated π3 units to the left

and 2 units up.

e The graph of y = 2 sin 4x (period 2π3 ,

amplitude 2) reflected in the x-axis and translated 1 unit up.


Chapter 11 – Circular functions II Exercise 11A Solutions 1 a cos ( – α) = cos α

= 0.6

b sin ⎝⎜⎛ π

2+ α

⎠⎟⎞ = cos α

= 0.6

c tan ( – θ) = – tan θ

= –0.7

d cos ⎝⎜⎛ π

2– x

⎠⎟⎞ = sin x

= 0.3

e sin ( – x) = – sin x

= –0.3

f tan ⎝⎜⎛ π

2– θ

⎠⎟⎞ =

sin ⎝⎜⎛ π

2– θ

⎠⎟⎞

cos ⎝⎜⎛ π

2– θ

⎠⎟⎞

= cos θsin θ

= 1tan θ

= 710

g cos ⎝⎜⎛ π

2+ x

⎠⎟⎞ = – sin x

= –0.3

h sin ⎝⎜⎛ π

2– α

⎠⎟⎞ = cos α

= 0.6

i sin ⎝⎜⎛ 3π

2+ α

⎠⎟⎞ = – sin

⎝⎜⎛ π

2+ α

⎠⎟⎞

= – cos α = –0.6

j cos ⎝⎜⎛3π

2– x

⎠⎟⎞ = – cos

⎝⎜⎛ π

2– x

⎠⎟⎞

= – sin x = –0.3


Exercise 11B Solutions 1 a y1 = 2 sin θ

y2 = cos θy = y1 + y2

θ = 0 ⇒ y = 0 + 1 = 1

θ = π2⇒ y = 1 + 1 = 2

θ = π ⇒ y = 0 + –1 = –1

θ = 3π2

⇒ y = –2 + 0 = –2

θ = 2π ⇒ y = 0 + 1 = 1

b y1 = 3 cos 2θ

y2 = 2 sin 2θy = y1 + y2

θ = 0 ⇒ y = 3 + 0 = 3

θ = π4⇒ y = 0 + 2 = 2

θ = π2⇒ y = –3 + 0 = –3

θ = 3π4

⇒ y = 0 + –2 = –2

θ = π ⇒ y = 3 + 0 = 3

c y1 = 12

sin 2θ

y2 = – cos θy = y1 + y2

θ = 0 ⇒ y = 0 + –1 = –1

θ = π2⇒ y = 0 + 0 = 0

θ = π ⇒ y = 0 + 1 = 1

θ = 3π2

⇒ y = 0 + 0 = 0

θ = 2π ⇒ y = 0 + –1 = –1

d y1 = 3 sin θ

y2 = cos 2θy = y1 + y2

θ = 0 ⇒ y = 0 + 1 = 1

θ = π2⇒ y = 3 + –1 = 2

θ = π ⇒ y = 0 + 1 = 1

θ = 3π2

⇒ y = –3 + –1 = –4

θ = 2π ⇒ y = 0 + 1 = 1


e y1 = 4 sin θy2 = 2 cos θy = y1 – y2

θ = 0 ⇒ y = 0 – 2 = –2

θ = π2⇒ y = 4 – 0 = 4

θ = π ⇒ y = 0 – –2 = 2

θ = 3π2

⇒ y = –4 – 0 = –4

θ = 2π ⇒ y = 0 – 2 = –2


Exercise 11C Solutions 1 a The graph will repeat itself every

π2 units (period =π2 ).

b The graph will repeat itself every

π3 units (period = π3 ). It will be twice as

steep.

c The graph of y = tan x, translated

π4 units left.

d The graph will be three times as steep and translated 1 unit up.

e The graph will be twice as steep, and

translated π2 units left and 1 unit up.

f The graph will repeat itself every

π2

units. It will be twice as steep and

translated π4 units left and 2 units down.


Exercise 11D Solutions 1 a sin x = 0.5

x = 2nπ + sin–1(0.5) or (2n + 1)π – sin–1(0.5)

= 2nπ + π6

or (2n + 1)π – π6

= 12nπ6

+ π6

or 6(2n + 1)π6

– π6

= (12n + 1)π6

or (12n + 5)π6

b cos 3x = 32

3x = 2nπ ± π6

x = 2nπ3

± π18

= 12nπ ± π18

= (12n ± 1)π18

c tan x = – 33

= – 3

x = nπ – π3

= 3nπ – π3

= (3n – 1)π3

2 a x = 2nπ + π

6

= 12nπ + π6

= (12n + 1)π6

or x = (2n + 1)π – π6

= 12nπ + 6π + π6

= (12n + 7)π6

x = π6

, π – π6

= π6

, 5π6

b cos 3x = 32

3x = 2nπ ± π6

x = 2nπ3

± π18

= 12nπ ± π18

= (12n ± 1)π18

= π6

, 2π – π6

= π6

, 11π6

c tan x = – 3

3= – 3

x = nπ – π3

= 3nπ – π3

= (3n – 1)π3

= π – π3

, 2π – π3

= 2π3

, 5π3

3 cos ⎝⎜⎛2x + π

4 ⎠⎟⎞ = 2

2= 1

2

2x + π4

= 2nπ ± π4

2x = 2nπ – π4

± π4

= (8n – 1 ± 1)π4

x = (8n – 1 ± 1)π8

= 0, 6π8

, 14π8

, – 2π8

, – 8π8

, – 10π8

= – 5π4

, π, – π4

, 0, 3π4

,π, 7π4


Note: If the ± values of x = (8n – 1 ± 1)π8

are taken separately, then the solution becomes

x = (8n – 1 + 1)π8

= nπ

or x = (8n – 2)π8

= (4n – 1)π4

4 tan ⎝⎜⎛ π

6– 3x

⎠⎟⎞ = 1

3π6

– 3x = nπ + π6

–2x = nπ

x = – nπ3

This is equivalent to x = nπ3 where x ∈ Z.

x = – π, – 2π3

, – π3

, 0

5 sin 4πx = 32

4πx

= 2nπ – π3

= 6nπ – π3

= (6n – 1)π3

or 4πx = (2n + 1)π – π3

= 6nπ + 3π + π4

= (6n + 4)π3

x = 6n – 112

or 3n + 26

= – 23, – 7

12, – 1

6, – 1

12, 1

3, 5

12, 5

6, 11

12


Exercise 11E Solutions 1

a cot 3π4

=cos 3π

4

sin 3π4

= – 12÷ 1

2 = –1

b cot 5π4

=cos 5π

4

sin 5π4

= – 12÷ – 1

2 = 1

c sec 5π6

=1

cos 5π6

=1

– 32

= – 23

= – 2 33

d cosec π2

=1

sin π2

= 11

= 1

e sec 4π3

=1

cos 4π3

=1

– 12

= –2

f cosec 13π6

=1

sin 13π6

=1

sin π6

=112

= 2

g cot 7π3

=cos 7π

3

sin 7π3

= 12÷ 3

2

= 13

= 33

h sec 5π3

=1

cos 5π3

=112

= 2

2 a cot 135° = cos 135

sin 135 = – 1

2÷ 1

2 = –1

b sec 150° = 1

cos 150

=1

– 32

= – 23

= – 2 33

c cosec 90° = 1

sin 90 = 1

1= 1

d cot 240° = cos 240

sin 240

= – 12÷ – 3

2

= 13

= 33


e sec 225° = 1cos 225

=1

– 12

= – 2

f sec 330° = 1cos 330

=132

= 23

= 2 33

g cot 315° = cos 315sin 315

= 12÷ – 1

2 = –1

h cosec 300° = 1sin 300

=1

– 32

= – 23

= – 2 33

i cot 420° = cos 420sin 420

= cos 60sin 60

= 12÷ 3

2

= 13

= 33

3 a cosec x = 2

sin x = 12

x = π6

, 5π6

b cot x = 3tan x = 1

3

x = π6

, 7π6

c sec x = – 2cos x = – 1

2

x = 3π4

, 5π4

d cosec x = sec x

sin x = cos xtan x = 1

x = π4

, 5π4

4 a cos θ = 1

sec θ = – 8

17

b cos2 θ + sin2 θ = 1

64289

+ sin2 θ = 1

sin2 θ = 225289

sin θ = 1517

c tan θ = sin θcos θ

= 1517

÷ – 817

= – 158

5 1 + tan2 θ = sec2 θ

sec2 θ = 1 + 49576

= 625576

sec θ = 2524

(since cos θ > 0)

cos θ = 2425

sin θcos θ

= tan θ = – 724

sin θ = – 724

× 2425

= – 725


6 1 + tan2 θ = sec2 θsec2 θ = 1 + 0.16 = 1.16

sec θ = – 116100

= – 2925

= – 295

7 cot θ = 3

41 + tan2 θ = sec2θ

sec2 θ = 1 + 169

= 259

sec θ = – 53

(cos θ < 0)

cos θ = – 35

sin θcos θ

= tan θ = 43

sin θ = 43× – 3

5 = – 4

5

sin θ – 2 cos θcot θ – sin θ

=– 4

5– – 6

534

– – 45

= 25÷ 31

20 = 2

5× 20

31= 8

31

8 cos2 θ + sin2 θ = 149

+ sin2 θ = 1

sin2 θ = 59

sin θ = – 53

⎝⎜⎛3π

2< θ < 2π

⎠⎟⎞

tan θ = – 53

÷ 23

= – 52

cot θ = – 25

tan θ – 3 sin θcos θ – 2 cot θ

=– 5

2– – 5

23

– – 45

= 52

÷ 2 5 + 123 5

= 52

× 3 52 5 + 12

= 154( 5 + 6)

× 6 – 56 – 5

= 15(6 – 5 )4 × (36 – 5)

= 15(6 – 5 )124

9 a (1 – cos2 θ)(1 + cot2 θ) = sin2 θ × cot2t h

= sin2 θ × cos2 θsin2 θ

= cos2 θ, provided sin θ ≠ 0

If sin θ = 0, cot θ would be undefined. b cos2 θ tan2 θ + sin2 θ cot 2θ

= cos2 θ × sin2 θcos2 θ

+ sin2 θ × cos2 θsin2 θ

= sin2 θ + cos2 θ = 1, provided sin θ ≠ 0 and cos θ ≠ 0


c In cases like this, it is a good strategy to start with the more complicated expression.

tan θ + cot φcot θ + tan φ

=

sin θcos θ

+ cos φsin φ

cos θsin θ

+ sin φcos φ

=

sin θ sin φ + cos φ cos θcos θ sin φ

cos φ cos θ + sin θ cos φcos φ sin θ

= sin θ sin φ + cos φ cos θcos θ sin φ

× cos φ sin θcos φ cos θ + sin θ cos φ

tan θ + cot φcot θ + tan φ

= cos φ sin θcos θ sin φ

= sin θcos θ

× cos φsin φ

= sin θcos θ

÷ sin φcos φ

= tan θtan φ

This is provided tan θ + tan ≠ 0 φ and the tangent and cotangent are

defined. d (sin θ + cos θ)2 + (sin θ – cos θ)2

= sin2θ + 2 sin θ cos θ + cos2θ + sin2θ – 2 sin θ cos θ + cos2θ

= 2 sin2θ + 2 cos2 θ = 2

There are no restrictions on θ.

e 1 + cot2 θcot θ cosec θ

= cosec2 θcot θ cosec θ

= cosec θcot θ

= 1sin θ

× sin θcos θ

= 1cos θ

= sec θ

Conditions: sin θ ≠ 0, cos θ ≠ 0

f sec θ + tan θ = 1cos θ

+ sin θcos θ

= 1 + sin θcos θ

× 1 – sin θ1 – sin θ

= (1 + sin θ)(1 – sin θ)cos θ(1 – sin θ)

= 1 – sin2 θcos θ(1 – sin θ)

= cos2 θcos θ(1 – sin θ)

= cos θ1 – sin θ

Conditions: cos θ ≠ 0 (includes sin θ ≠ 1)


Exercise 11F Solutions 1 Different angles may be used. a cos 15° = cos(45 – 30)

= cos 45 cos 30 + sin 45 sin 30

= 12× 3

2+ 1

2× 1

2

= 3 + 12 2

= 6 + 24

b cos 105° = cos (45 + 60)

= cos 45 cos 60 – sin 45 sin 60

= 12× 1

2– 1

2× 3

2

= 1 – 32 2

= 2 – 64

2 Different angles may be used. a sin 165° = sin (180 – 15)

= sin 15°sin 15° = sin (45 – 30)

= sin 45 cos 30 – cos 45 sin 30

= 12× 3

2– 1

2× 1

2

= 3 – 12 2

= 6 – 24

b tan 75° = tan (45 + 30)

= tan 45 + tan 301 – tan 45 tan 30

=1 + 1

3

1 – 13

× 33

= 3 + 13 – 1

× 3 + 13 + 1

= 3 + 2 3 + 13 – 1

= 2 + 3

3 Different angles may be used.

a cos 5π12

= cos ⎝⎜⎛ π

4+ π

6 ⎠⎟⎞

= cos π4

cos π6

– sin π4

sin π6

= 12× 3

2– 1

2× 1

2

= 3 – 12 2

= 6 – 24

b sin 11π12

= sin ⎝⎜⎛ π – π

12⎠⎟⎞

= sin π12

sin π12

= sin ⎝⎜⎛ π

3– π

4 ⎠⎟⎞

= sin π3

cos π4

– cos π3

sin π4

= 32

× 12

– 12× 1

2

= 3 – 12 2

= 6 – 24

c tan – π12

= tan⎝⎜⎛ π

4– π

3 ⎠⎟⎞

=tan π

4– tan π

3

1 + tan π4

tan π3

= 1 – 31 + 3

× 1 – 31 – 3

= 1 – 2 3 + 31 – 3

= –2 + 3


4 cos2 u = 1 – sin2 u = 1 – 144

169= 25

169cos u = ± 5

13cos2 v = 1 – sin2 v

= 1 – 925

= 1625

cos v = ± 45

sin (u + v) = sin u cos v + cos u sin v = ± 3

5× 5

13 ± 4

5× 12

13 = ±15 ± 48

65

There are four possible answers: 63

65, 33

65, – 33

65, – 63

65

5

a sin ⎝⎜⎛ θ + π

6 ⎠⎟⎞ = sin θ cos π

6+ cos θ sin π

6

= 32

sin θ + 12

cos θ

b cos ⎝⎜⎛ π – π

4 ⎠⎟⎞ = cos φ cos π

4+ sin φ sin π

4 = 1

2 cos φ + 1

2 sin φ

= 12

(cos φ + sin φ)

c tan ⎝⎜⎛ θ + π

6 ⎠⎟⎞ =

tan θ + tan π3

1 – tan θ tan π3

= tan θ + 31 – 3 tan θ

d sin ⎝⎜⎛ θ – π


4– cos θ sin π

4 = 1

2 sin θ – 1

2 cos θ

= 12

(sin θ – cos θ)

6 a sin (v + (u – v)) = sin v b cos ((u + v) – v) = cos u 7 cos2 θ = 1 – sin2 θ

= 1 – 925

= 1625

cos θ = – 45

sin2 φ = 1 – cos2 φ = 1 – 25

169= 144

169sin φ = 12

13

a cos 2φ = cos2 φ – sin2 φ = 25

169– 144

169 = – 119

169

b sin 2θ = 2 sin θ cos θ

= 2 × – 35× – 4

5 = 24

25

c tan θ = sin θcos θ

= –3–4

= 34

tan 2θ = 2 tan θ1 – tan2 θ

=

32

1 – 916

= 32× 16

7 = 24

7

d sec 2φ = 1

cos 2φ = – 169

119


e sin (θ + φ) = sin θ cos φ + cos θ sin φ = – 3

5× – 5

13+ – 4

5× 12

13 = 14 – 48

65 = – 33

65

f cos (θ – φ) = cos θ cos φ + sin θ sin φ

= – 45× – 5

13+ – 3

5× 12

13 = 20 – 36

65 = – 16

65

g cosec (θ + φ) = 1

sin (θ + φ) = – 65

33

h cot 2θ = 1

tan 2θ = 7

24

8 a tan (u + v) = tan u + tan v

1 – tan u tan v

=⎝⎜⎛4

3+ 5

12⎠⎟⎞ ÷

⎝⎜⎛1 – 4

3× 5

12⎠⎟⎞

= 2112

÷ 49

= 2112

× 94

= 6316

b tan 2u = 2 tan u

1 – tan2 u

=

83

1 – 169

= 83× 9

–7 = – 24

7

c sec2 u = 1 + tan2 u = 1 + 16

9= 25

9cos2 u = 9

25cos u = 3

5 (since u is acute )

sec2 v = 1 + tan2 v = 1 + 25

144= 169

144cos2 v = 144

169cos v = 12

13 (since v is acute )

cos (u – v) = cos u cos v + sin u sin v = 3

5× 12

13+ 4

5× 5

13 = 56

65

d sin 2u = 2 sin u cos u

= 2 × 45× 3

5 = 24

25

9 cos2 α = 1 – sin2 α

= 1 – 925

= 1625

cos α = – 45

cos2 β = 1 – sin2 β = 1 – 576

625= 29

625cos β = – 7

25

a cos 2α = cos2 α – sin2 α = 16

25– 9

25 = 7

25

b sin (α – β) = sin α cos β – cos α sin β

= 35× – 7

25– – 4

5× 24

25 = 75

125= 3

5


c tan α = sin αcos α

= – 34

tan β = sin βcos β

= – 247

tan (α + β) = tan α + tan β1 – tan α tan β

=– 3

4+ – 24

7

1 – – 34× 24

7 = – 117

28× – 7

11 = 117

44

d sin 2β = 2 sin β cos β

= 2 × 725

× – 2425

= – 336625

10 a sin 2θ = 2 sin θ cos θ

= 2 × – 32

× 12

= – 32

b cos 2θ = cos2 θ – sin2 θ

= 14

– 34

= – 12

11 a (sin θ – cos θ)2 = sin2 θ – 2 sin θ cos θ + cos2 θ

= 1 – sin 2θ b sin4 θ – cos4 θ = (sin2 θ – cos2 θ)(sin2 θ + cos2 θ)

= cos 2θ × 1 = cos 2θ

12

a 2 sin ⎝⎜⎛ θ – π

4 ⎠⎟⎞ = 2

⎝⎜⎛ sin θ cos π

4– cos θ sin π

4 ⎠⎟⎞

= 2⎝⎜⎛ 1

2 sin θ – 1

2 cos θ

⎠⎟⎞

= sin θ – cos θ

b cos ⎝⎜⎛ θ – π

3 ⎠⎟⎞ = cos θ cos π

3+ sin θ sin π

3

= 12 cos θ + 3

2 sin θ

cos ⎝⎜⎛ θ + π

3 ⎠⎟⎞ = 1

2 cos θ = 3

2 sin θ

Add the last two equations:

cos ⎝⎜⎛ θ – π

3 ⎠⎟⎞ + cos

⎝⎜⎛ θ + π

3 ⎠⎟⎞ = cos θ

c tan ⎝⎜⎛ θ + π

4 ⎠⎟⎞ tan

⎝⎜⎛ θ – π

4 ⎠⎟⎞

=tan θ + tan π

4


×tan θ – tan π

4

1 + tan θ tan π4

= tan θ + 11 – tan θ

× tan θ – 11 + tan θ

= –1

d cos ⎝⎜⎛ θ + π

6 ⎠⎟⎞ = cos θ cos π

6– sin θ sin π

6

= 32

cos θ – 12

sin θ

cos ⎝⎜⎛ θ + π


3+ cos θ sin π

3

= 32

cos θ + 12

sin θ

Add the two equations:

cos ⎝⎜⎛ θ + π

6 ⎠⎟⎞ + sin

⎝⎜⎛ θ + π

3 ⎠⎟⎞ = 3 cos θ


h cos 2θ + 2 sin2 θ = cos2 θ – sinθ + 2 sin2 θ = cos2 θ + sin2 θ = 1e tan

⎝⎜⎛ θ + π

4 ⎠⎟⎞ =

tan θ + tan π4


= tan θ + 11 – tan θ

= 1 + tan θ1 – tan θ

i sin 4θ = sin 2 × 2θ

= 2 sin 2θ cos 2θ = 2 × 2 sin θ cos θ (cos2 θ – sin2 θ) = 4 sin θ cos3 θ – 4 cos θ sin3 θ

f sin (u + v)cos u cos v

= sin u cos v + cos u sin vcos u cos v

= sin u cos vcos u cos v

+ cos u sin vcos u cos v

= tan u + tan v

j 1 – sin 2θsin θ – cos θ

= 1 – sin 2θsin θ – cos θ

× sin θ – cos θsin θ – cos θ

= (1 – sin 2θ)(sin θ – cos θ)sin2 θ – 2 sin θ cos θ + cos2 θ

= (1 – sin 2θ)(sin θ – cos θ)1 – 2 sinθ cos θ

= (1 – sin 2θ)(sin θ – cos θ)1 – sin 2θ

= sin θ – cos θ

g sin (u + v)

sin (u – v)= sin u cos v + cos u sin v

sin u cos v – cos u sin v

Divide numerator and denominator by cos u cos v. sin (u + v)

sin (u – v)= tan u + tan v

tan u – tan v


Exercise 11G Solutions 1

a Maximum = 42 + 32 = 5 Minimum = –5 b Maximum = 3 + 1 = 2 Minimum = –2 c Maximum = 1 + 1 = 2 Minimum = – 2 d Maximum = 1 + 1 = 2 Minimum = – 2 e Maximum = 9 + 3 = 12 = 2 3 Minimum = –2 3 f Maximum = 1 + 3 = 2 Minimum = –2 g Maximum = 1 + 3 + 2 = 4 Minimum = – 1 + 3 + 2 = 0

h Maximum = 5 + 32 + 22 = 5 + 13

Minimum = 5 – 32 + 22 = 5 – 13 2 a r = 1 + 1 = 2

cos α = 12

; sin α = – 12

α = – π4

2 sin ⎝⎜⎛ x – π

4 ⎠⎟⎞ = 1

sin ⎝⎜⎛ x – π

4 ⎠⎟⎞ = 1

2

x – π4

= π4

, 3π4

x = π2

, π

b r = 3 + 1 = 2

cos α = 32

; sin α = 12

α = π6

2 sin ⎝⎜⎛ x + π

6 ⎠⎟⎞ = 1

sin ⎝⎜⎛ x + π

6 ⎠⎟⎞ = 1

2

x + π6

= π6

, 5π5

, 7π6

x = 0, 2π3

, 2π

c r = 3 + 1 = 2

cos α = 12

; sin α = – 32

α = – π3

2 sin ⎝⎜⎛ x – π

3 ⎠⎟⎞ = –1

sin ⎝⎜⎛ c – π

3 ⎠⎟⎞ = – 1

2

x – π3

= – π6

, 7π6

x = π6

, 3π2

d r = 9 + 3 = 12 = 2 3

cos α = 32 3

= 32

sin α = 32 3

= 12

α = π6

2 3 cos ⎝⎜⎛ x + π

6 ⎠⎟⎞ = 3

cos ⎝⎜⎛ x + π

6 ⎠⎟⎞ = 3

2 3= 3

2

x + π6

= π6

, 11π6

, 13π6

x = 0, 5π3

, 2π


e r = 42 + 32 = 25 = 5cos α = 4

5; sin α = 3

5α ≈ 36.87°

5 sin (θ + 36.87) ≈ 5sin (θ + 36.87) ≈ 1

θ + 36.87 ≈ 90°θ ≈ 53.13°

f r = 8 + 4 = 12 = 2 3

cos α = 2 22 3

= 23

sin α = – 22 3

= – 13

α ≈ – 35.26°2 3 sin (θ – 35.26) ≈ 3

sin (θ – 35.26) ≈ 32 3

= 32

θ – 35.26 ≈ 60°, 120°θ ≈ 95.26°, 155.26°

3 r = 3 + 1 = 2

cos α = 32

; sin α = 12

α = π6

2 cos ⎝⎜⎛2x + π

6 ⎠⎟⎞

4 r = 1 + 1 = 2

cos α = 12

; sin α = 12

α = π4

2 sin ⎝⎜⎛ 3x – π

4 ⎠⎟⎞

5 a r = 1 + 1 = 2

cos α = 12

; sin α = – 12

α = – π4

f(x) = 2 sin ⎝⎜⎛ x – π

4 ⎠⎟⎞

The graph will have amplitude 2 , period

2π, and be translated π4 units right.

b r = 3 + 1 = 2

cos α = 32

' sin α = 12

α = π6

f(x) = 2 sin ⎝⎜⎛ x + π

6 ⎠⎟⎞

The graph will have amplitude 2, period

2π, and be translated π6 units left.


c r = 1 + 1 = 2cos α = 1

2; sin α = 1

2

α = π4

f(x) = 2 sin ⎝⎜⎛ x + π

4 ⎠⎟⎞

The graph will have amplitude 2 , period 2π, and be translated π4 units left.

d r = 1 + 3 = 2

cos α = 12; sin α = – 3

2

α = – π3

f(x) = 2 sin ⎝⎜⎛ x – π

3 ⎠⎟⎞

The graph will have amplitude 2, period

2π, and be translated π3 units right.


Solutions to Multiple-choice Questions 1 cosec x – sin x = 1

sin x– sin x

= 1 – sin2xsin x

= cos2xsin x

= cos x × cos xsin x

= cos x cot x A 2 cos x = – 1

3cos2 x + sin2 x = 1

⎝⎜⎛ – 1

3⎠⎟⎞2

+ sin2 x = 1

sin2 x = 1 – 19

= 89

sin x = 89

= – 2 23

, 2 23 A

3 sec θ = b

atan2 θ + 1 = sec2 θ

tan2 θ = b2

a2 = 1

= b2 – a2

a2

tan θ = b2 – a2

a B

4 ∠ABC = u; ∠XBC = v

tan u = x + 42

; tan v = x2

tan θ = tan (u – v) = tan u – tan v

1 + tan u tan v

=

x + 42

– x2

1 + x + 42

× x2

= 42÷ 4 + x(x + 4)

4 = 2 × 4

x2 + 4x + 4

= 8(x + 2)2 A

5 sin2A = 1 – cos2A

= 1 – t2

sin A = 1 – t2

cos2B = 1 – sin2B = 1 – t2

cos B = 1 – t2

sin (B + A) = sin B cos A + cos B sin A

= t × t + ⎝⎜⎛

– 1 – t2 ⎠⎟⎞× 1 – t2

= t2 – (1 – t2)

= 2 C t2 – 1 6 sin 2A

cos 2A – 1= 2 sin A cos A

cos2A – sin2A – 1 = 2 sin A cos A

– sin2A – (1 – cos2A) = 2 sin A cos A

– sin2A – sin2A = 2 sin A cos A

–2 sin2A = cos A

sin A

= –cot A E


9 sin 2A = 2 sin A cos Am = 2 sin A × n

sin A = m2n

tan A = sin Acos A

= m2n

× 1n

7 Check the symmetry properties:

sin ⎝⎜⎛ π

2– x

⎠⎟⎞ = sin

⎝⎜⎛ π

2+ x

⎠⎟⎞

= – sin ⎝⎜⎛3π

2+ x

⎠⎟⎞

= cos x = cos ( – x)

cos ( – x) = cos (2π – x)

= m2n2 A

∴ sin ⎝⎜⎛ π

2– x

⎠⎟⎞ ≠ sin x C

10 r = 1 + 1 = 2cos α = 1

2; sin α = – 1

2

8 (1 + cot x)2 + (1 – cot x)2

= 1 + 2 cot x + cot2 x + 1 – 2 cot x + cot2 x

= 2 + 2 cot2 x = 2(1 + cot2 x)

A positive angle must be chosen,

∴ α = 7π4

2 sin ⎝⎜⎛ x + 7π

4 ⎠⎟⎞ D = 2 E cosec2 x


Solutions to Short-answer Questions 1

a sec θ + cosec θ cot θ = 1cos θ

+ 1sin θ

× cos θsin θ

= 1cos θ⎝⎜

⎛1 + cos θsin θ

× cos θsin θ⎠⎟

⎞

= 1cos θ

(1 + cot2 θ)

= sec θ cosec 2 θ

b tan2 θ + cos2 θsec θ + sin θ

= tan2 θ + 1 – sin2 θsec θ + sin θ

= sec2 θ – sin2 θsec θ + sin θ

= (sec θ – sin θ)(sec θ + sin θ)sec θ + sin θ

= sec θ – sin θ

2 a Maximum = 5, minimum = 1 b Maximum = 4, minimum = –2 c Maximum = 4, minimum = –4 d Maximum = 2, minimum = 0 e Maximum = 1 (when cos θ = –1),

minimum = 13

3 a sin2 θ = 1

4sin θ = ± 1

2

θ = π6

, 5π6

, 7π6

, 11π6

b sin 2θ = 1

2

2θ = π6

, 5π6

, 13π6

, 17π6

θ = π12

, 5π12

, 13π12

, 17π12

c cos 3θ = 32

3θ = π6

, 11π6

, 13π6

, 23π6

, 25π6

, 35π6

θ = π18

, 11π18

, 13π18

, 23π18

, 25π18

, 35π18

d sin2 2θ = 1

sin 2θ = ± 1

2θ = π2

, 3π2

, 5π2

, 7π2

θ = π4

, 3π4

, 5π4

, 7π4

e tan2 θ = 1

3tan θ = ± 1

3

θ = π6

, 5π6

, 7π6

, 11π6

f tan 2θ = –1

2θ = 3π4

, 7π4

, 11π4

, 15π4

θ = 3π8

, 7π8

, 11π8

, 15π8

g sin 3θ = –1

3θ = 3π2

, 7π2

, 11π2

θ = π2

, 7π6

, 11π6

h cos 2θ = 1

2

2θ = π4

, 7π4

, 9π4

, 15π4

θ = π8

, 7π8

, 9π8

, 15π8


4 tan θ = 2 sin θsin θcos θ

= 2 sin θ

sin θcos θ

– 2 sin θ = 0

sin θ ⎝⎜⎛ 1

cos θ– 2

⎠⎟⎞ = 0

sin θ = 0 or cos θ = 12

θ = 60°, 300°, 0°, 180°, 360°

5 cos2 A = 1 – sin2 A

= 1 – 25169

= 144169

cos A = 1213

cos2 B = 1 – sin2 B = 1 – 64

289= 225

289cos B = 15

17

a cos (A + B) = cos A cos B – sin A sin B = 12

13× 15

17– 5

13× 8

17 = 140

221

b sin (A + B) = sin A cos B – cos A sin B

= 513

× 1517

– 1213

× 817

= – 21221

c tan A = sin A

cos A= 5

12tan B = sin B

cos B= 8

15tan (A + B) = tan a + tan B

1 – tan A tan B

=⎝⎜⎛ 5

12+ 8

15⎠⎟⎞ ÷

⎝⎜⎛1 – 5

12× 8

15⎠⎟⎞

= 5760

÷ 79

= 1920

× 97

= 171140

6 a Expression = cos (80° – 20°)

= cos 60° = 12

b Expression = t an (15° + 30°)

= tan 45° = 1 7 a Expression = sin (A + B)

= sin π2

= 1

b Expression = cos (A + B)

= cos π2

= 0

8 a Maximum = 5, minimum = 1 b Maximum = 9, minimum = –1 9 a sin2 A cos2 B – cos2 A sin2 B

= sin2 A(1 – sin2 B) – (1 – sin2 A)sin2 B = sin2 A – sin2 A sin2 B – sin2 B + sin2 A sin2 B = sin2 A – sin2 B

b Left side = sin2 θ + (1 + cos θ)2

sin θ(1 + cos θ)

= sin2 θ + 1 + 2 cos θ + cos2 θsin θ(1 + cos θ)

= 2 + 2 cos θsin θ(1 + cos θ)

= 2(1 + cos θ)sin θ(1 + cos θ)

= 2sin θ


c Left side = sin θ (1 – 2sin2 θ)cos θ (2 cos2 θ – 1)

= sin θ (1 – sin2 θ – sin2 θ)cos θ (cos2 θ + cos2 θ – 1)

= sin θ (cos2 θ – sin2 θ)cos θ (cos2 θ – (1 – cos2 θ))

= sin θ (cos2 θ – sin2 θ)cos θ (cos2 θ – sin2 θ)

= sin θcos θ

= tan θ

10 cos2 A = 1 – sin2 A

= 1 – 59

= 49

cos A = – 23

a sin 2A = 2 sin A cos A

= 2 × 53

× – 23

= – 4 59

b cos 2A = cos2 A – sin2 A

= 49

– 59

= – 19

c sin 4A = 2 sin 2A cos 2A

= 2 × – 4 59

× – 19

= 8 581

11

a Left side =

1 – sin2 θcos2 θ

1 + sin2 θcos2 θ

=

1 – sin2 θcos2 θ

1 + sin2 θcos2 θ

× cos2 θcos2 θ

= cos2 θ – sin2 θcos2 θ + sin2 θ

= cos 2θ1

= cos 2θ

b Left side = sin2 A + (1 + cos A)2

sin A (1 + cos A)

= sin2 A + 1 + 2 cos A + cos2 Asin A (1 + cos A)

= 2 + 2 cos Asin A (1 + cos A)

= 2 (1 + cos A)sin A (1 + cos A)

= 2sin A

12 a tan 15° = tan (60 – 45)

= tan 60 – tan 451 + tan 60 tan 45

= 3 – 11 + 3

= 3 – 11 + 3

× 3 – 13 – 1

= 3 – 2 3 + 13 – 1

= 2 – 3

b sin (x + y) = sin x cos y + cos x sin y sin (x – y) = sin x cos y – cos x sin y Add the two equations: sin (x + y) + sin (x – y) = 2 sin x cos y


13 a x = 0

y = 2 3 – 0 = 2 3(0, 2 3 )

b Solve 2 3 cos x – 2 sin x = 0. Express in the form r cos (x + α) r = 12 + 4 = 4

cos α = 2 34

= 32

sin α = 24

= 12

α = π6

4 cos ⎝⎜⎛ x + π

6 ⎠⎟⎞ = 0

x + π6

= π2

, 3π2

x = π3

, 4π3

x-intercepts: ⎝⎜⎛ π

3, 0

⎠⎟⎞ and

⎝⎜⎛ 4π

3, 0

⎠⎟⎞

c Maximum = 4

⎝⎜⎛ 11π

6, 4

⎠⎟⎞

d Minimum = –4

⎝⎜⎛ 5π

6, – 4

⎠⎟⎞

The graph will be y = 4 cos x, translated π6 units left.

14 a Express in the form r sin (x + α) = 1. r = 1 + 1 = 2

cos α = 12

; sin α = 12

α = π4

2 sin ⎝⎜⎛ x + π

4 ⎠⎟⎞ = 1

sin ⎝⎜⎛ x + π

4 ⎠⎟⎞ = 1

2

x + π4

= π4

, 3π4

, 9π4

x = 0, π2

, 2π

b 2 sin x2

cos x2

= – 12

sin ⎝⎜⎛2 × x

2⎠⎟⎞ = – 1

2sin x = – 1

2

x = 7π6

, 11π6

c 3 × 2 tan x1 – tan2 x

= 2 tan x

2 tan x ⎝⎜⎛ 3

1 – tan2 x– 1

⎠⎟⎞ = 0

2 tan x ⎝⎜⎛ 3 – (1 – tan2 x)

1 – tan2 x ⎠⎟⎞ = 0

tan x = 0 (since 2 + tan2 x ≠ 0)x = 0, π, 2π

d sin2 x – cos2 x = 1cos 2x = –1

2x = π, 3π

x = π2

, 3π2

e sin (3x – x) = 32

sin 2x = 32

2x = π3

, 2π3

, 7π3

, 8π3

x = π6

, π3

, 7π6

, 4π3


f cos ⎝⎜⎛2x – π

3 ⎠⎟⎞ = – 3

2

2x – π3

= 5π6

, 7π6

, 17π6

, 19π6

2x = 7π6

, 9π6

, 19π6

, 21π6

x = 7π12

, 3π4

, 19π12

, 7π4

15 a y = 2 cos2 x

= cos2 x + (1 – sin2 x) = cos2 x – sin2 x + 1 = cos2x + 1

The graph of y = cos 2x (amplitude 1, period π) raised 1 unit.

b The graph is y = 1 – 2 sin x

2.

It is y = sin x2

(period 4π) reflected in

the x-axis and raised 1 unit.

c The normal tangent graph, but with

period π2 .

16 tan (θ + A) = 4tan θ + tan A

1 – tan θ tan A= 4

tan θ + 21 – 2 tan θ

= 4

tan θ + 2 = 4(1 – 2 tan θ) = 4 – 8 tan θ

9 tan θ = 2tan θ = 2

9

17 a r = 4 + 81 = 85

cos α = 285

; sin α = 985

85 cos (θ + α), where

α = cos–1

⎝⎜⎛ 2

85 ⎠⎟⎞

b i 85 ii cos (θ + α) = 1

θ + α = 0θ = – α

cos θ = cos ( – α) = cos α = 2

85

iii Solve 85 cos (θ + α) = 1. The first positive solution will be when θ + α = 2π. θ = – α + 2π

= 2π – cos–1 ⎝⎜⎛ 2

85 ⎠⎟⎞


Chapter 12 – Trigonometric ratios and applications Exercise 12A Solutions 1 a x

5= cos 35°

x = 5 × 0.8191 = 4.10 cm

b x

10= sin 45°

x = 10 × 0.0871 = 0.87 cm

c x

8= tan 20.16°

x = 8 × 0.3671 = 2.94 cm

d x

7= tan 30°15'

x = 7 × 0.9661 = 4.08 cm

e tan x° = 10

15 = 0.666x = 33.69°

f 10

x= tan 40°

10 = x × 0.8390x = 10

0.8390 = 11.92 cm

2

20

x= sin 60°

20 = x × 32

x = 403

= 40 33

cm

3

cos x = 6

15= 0.4

x = 66.42°

The third angle = 180 – 2 × 66.42= 47.16°

4 h

20= tan 49°

x = 20 × 1.1503 ≈ 23 m

5 a tan x = 1

6 = 0.8333. . .x = 9.59°

Keep this number in your calculator for the next question.


b BC2 = 62 – 12 = 35BC = 35 m

= 5.92 m

6 a cos θ = 10

20= 0.5

θ = 60°

b PQ

20= sin 60°

x = 20 × 0.866 = 17.32 m

7 3

L= sin 26°

3 = L × 0.4383L = 3

0.4383 = 6.84 m

8 sin θ = 13

60= 0.21666. . .

θ = 12.51°

9 h

200= sin 66°

x = 200 × 0.9135 = 182.7 m

10 400

d= sin 16°

400 = d × 0.2756d = 400

0.2756 = 1451 m

11 BC = BA = AD = CD a AC 2 = BC 2 + BA2 = 2BC 2

100 = 2BC 2

BC 2 = 50BC = 50 = 5 2 cm

b cos ∠CBD = 10

10 2= 1

2∠CBD = 45°∠ABC = 2 × 45 = 90°

12 Find the vertical height, h cm.

h

90= cos 15°

h = 90 × 0.9659h = 86.93 cmx = 90 – 86.93 = 3.07 cm

13 15L2

= sin 52.5°

15 = L2× 0.7933

L = 300.7933

= 37.8 cm

14 w

50= tan 32°

w = 50 × 0.6248 = 31.24 cm

15 h2 + 1.72 = 4.72

h2 = 4.72 – 1.72

= 19.2h = 4.38 m

16 50

d= sin 60°

50 = d × 0.866d = 50

0.866 = 57.74 m


Exercise 12B Solutions 1 a x

sin 50° = 10sin 70°

x = 10 × sin 50°sin 70°

= 8.15 cm

b y

sin 37° = 6sin 65°

x = 6 × sin 37°sin 65°

= 3.98 cm

c x

sin 100° = 5.6sin 28°

x = 5.6 × sin 100°sin 28°

= 11.75 cm

d X = 180° – 38° – 90°

= 52°x

sin 52° = 12sin 90°

x = 12 × sin 52°sin 90°

= 9.46 cm

2

a sin θ7

= sin 72°8

sin θ = 7 × sin 72°8

= 0.8321θ = 56.32°

In this case θ cannot be obtuse.

b sin θ8.3

= sin 42°9.4

sin θ = 8.3 × sin 42°9.4

= 0.5908θ = 36.22°


c sin θ8

= sin 108°10

sin θ = 8 × sin 108°10

= 0.7608θ = 49.54°


d sin B9

= sin 38°8

sin B = 9 × sin 38°8

= 0.6929B = 43.84° or 180 – 43.84 = 131.16°θ = 180 – 43.84 – 38 = 98.16°

or 180 – 136.16 – 38 = 5.84°

3 a A = 180 – 59 – 73

= 48°b

sin 59° = 12sin 48°

b = 12 × sin 59°sin 48°

= 13.84 cmc

sin 73° = 12sin 48°

c = 12 × sin 73°sin 48°

= 15.44 cm

b C = 180 – 75.3 – 48.25

= 56.45°a

sin 75.3° = 5.6sin 48.25°

a = 5.6 × sin 75.3°sin 48.25°

= 7.26 cmc

sin 56.45° = 5.6sin 48.25°

c = 5.6 × sin 56.45°sin 48.25°

= 6.26 cm


c B = 180 – 123.2 – 37 = 19.8°

bsin 19.8° = 11.5

sin 123.2°

b = 11.5 × sin 19.8°sin 123.2°

= 4.66 cmc

sin 37° = 11.5sin 123.2°

c = 11.5 × sin 37°sin 123.2°

= 8.27 cm

d C = 180 – 23 – 40

= 117°b

sin 40° = 15sin 23°

b = 15 × sin 40°sin 23°

= 24.68 cmc

sin 117° = 15sin 23°

c = 15 × sin 117°sin 23°

= 34.21 cm

e C = 180 – 10 – 140

= 30°a

sin 10° = 20sin 140°

a = 20 × sin 10°sin 140°

= 5.40 cmc

sin 30° = 20sin 140°

c = 20 × sin 30°sin 140°

= 15.56 cm

4

a sin B17.6

= sin 48.25°15.3

sin B = 17.6 × sin 48.25°15.3

= 0.8582B = 59.12° or 180 – 59.12 = 120.88°A = 180 – 48.25 – 59.12 = 72.63°

or 180 – 48.25 – 120.88 = 10.87°15.3

sin 48.25° = asin 72.63° or a

sin 10.87°

a = 15.3 × sin 72.63°sin 48.25° or 15.3 × sin 10.87°

sin 48.25°

= 19.57 cm or 3.87 cm

b sin C4.56

= sin 129°7.89

sin C = 4.56 × sin 129°7.89

= 0.4991C = 26.69°A = 180 – 129 – 26.69 = 24.31°

asin 24.31° = 7.89

sin 129°

a = 7.89 × sin 24.31°sin 129°

= 4.18 cm

c sin B14.8

= sin 28.35°8.5

sin B = 14.8 × sin 28.35°8.5

= 0.8268B = 55.77° or 180 – 55.77 = 124.23°C = 180 – 55.77 – 28.35 = 95.88°

or 180 – 124.23 – 28.35 = 27.42°8.5

sin 28.35° = csin 95.88° or c

sin 27.42°

c = 8.5 × sin 95.88°sin 28.35° or 8.5 × sin 27.42°

sin 28.35° = 178.1 cm or 8.24 cm


7 5

C = 180 – 69 – 74 = 37°

bsin 69° = 1070

sin 37°

b = 1070 × sin 69°sin 37°

= 1659.86 m

A = 180 – 68 – 70 = 42°

bsin 68° = 400

sin 42°

b = 400 × sin 68°sin 42°

= 554.26 m

8 6 ∠APB = 46.2 – 27.6 = 18.6° (exterior angle property)

asin 27.6° = 34

sin 18.6°

PB = a = 34 × sin 27.6°sin 18.6°

= 49.385 mh

PB= sin 46.2°

h = 49.385 × 0.7217 = 35.64 m

a X = 180 – 120 – 20 = 40°

AXsin 20° = 50

sin 40°

= 50 × sin 20°sin 40°

= 26.60 m

b Y = 180 – 109 – 32

= 39°AY

sin 109° = 50sin 39°

AY = 50 × sin 109°sin 39°

= 75.12 m


Exercise 12C Solutions 1 BC 2 = a2

= b2 + c2 – 2bc cos A = 152 + 102 – 2 × 15 × 10 × cos 15° = 325 – 300 × cos 15° = 35.222

BC = 5.93 cm

2 ∠ABC = ∠B

cos B = a2 + c2 – b2

2ac

= 52 + 82 – 102

2 × 5 × 8 = –0.1375

∴∠ABC ≈ 97.90°∠ACB = ∠C

cos C = a2 + b2 – c2

2ac

= 52 + 102 – 82

2 × 5 × 10 = 0.61

∴∠ACB ≈ 52.41°

3 a a2 = b2 + c2 – 2bc cos a

= 162 + 302 – 2 × 16 × 30 × cos 60° = 1156 – 960 × cos 60° = 676

a = 26 b b2 = a2 + c2 – 2ac cos B

= 142 + 122 – 2 × 14 × 12 × cos 53° = 340 – 336 × cos 53° = 137.7901

a ≈ 11.74 c ∠ABC = ∠B

cos B = a2 + c2 – b2

2ac

= 272 + 462 – 352

2 × 27 × 46 = 0.6521

∴∠ABC ≈ 49.29°

d b2 = a2 + c2 – 2ac cos B = 172 + 632 – 2 × 17 × 63 × cos 120° = 4258 – 2142 cos 120° = 5329

b = 73 e c2 = a2 + b2 – 2ab cos C

= 312 + 422 – 2 × 31 × 42 × cos 140° = 2642 – 2604 × cos 140° = 4719.77c ≈ 68.70

f ∠BCA = ∠C

cos C = a2 + b2 – c2

2ab

= 102 + 122 – 92

2 × 10 × 12 = 0.6791

∴∠BCA ≈ 47.22°

g c2 = a2 + b2 – 2ab cos C

= 112 + 92 – 2 × 11 × 9 × cos 43.2° = 202 – 198 × cos 43.2° = 57.6642c ≈ 7.59

h ∠CBA = ∠B

cos B = a2 + c2 – b2

2ac

= 82 + 152 – 102

2 × 8 × 15 = 0.7875

∴∠ABC ≈ 38.05°

4 c2 = a2 + b2 – 2ab cos C

= 42 + 62 – 2 × 4 × 6 × cos 20° = 52 – 48 × cos 20° = 6.8947c ≈ 2.626 km


B = 180 – 88 – 43.0045 = 48.9954°

bsin 48.9954° = 7.3263

sin 88°

b = 7.3263 × sin 48.9956°sin 88°

≈ 5.53 cm

5 AB2 = a2 + b2 – 2ab cos O = 42 + 62 – 2 × 4 × 6 × cos 30° = 52 – 48 × cos 30° = 10.4307

AB ≈ 3.23 km

6

a ∠B = 180 – 48 = 132°

AC 2 = a2 + c2 – 2ac cos B = 52 + 42 – 2 × 5 × 4 × cos 132° = 41 – 40 × cos 132° = 67.7652

AC ≈ 8.23 cm

9

a cos ∠AO'B = 62 + 62 – 82

2 × 6 × 6 = 0.111

∠AO'B ≈ 83.62°

b cos ∠AOB = 7.52 + 7.52 – 82

2 × 7.5 × 7.5 = 0.43111

∠AOB ≈ 64.46°

b BD2 = b2 + d2 – 2bd cos A = 52 + 42 – 2 × 5 × 4 × cos 48° = 41 – 40 × cos 48° = 14.2347

BD ≈ 3.77 cm

10

a Treat AB as c. 7 Label the points suitably: A and B are the

hooks, and C is the angle. 70° c2 = a2 + b2 – 2ab cos OAB2 = 702 + 902 – 2 × 70 × 90 × cos 65°

= 13 000 – 12 600 × cos 65° = 7675.0099

AB ≈ 87.61 m

c2 = a2 + b2 – 2ab cos CBD2 = 422 + 542 – 2 × 42 × 54 × cos 70°

= 4680 – 4536 × cos 70° = 3128.5966

BD ≈ 55.93 cm

b cos ∠B = a2 + c2 – b2

2ac

= 702 + 87.60712 – 902

2 × 70 × 87.6071 = 0.3648

∠AOB ≈ 68.6010°

8 a Use ΔABD. BD2 = b2 + d2 – 2bd cos A

= 62 + 42 – 2 × 6 × 4 × cos 92° = 52 – 48 × cos 92° = 53.6751

BD ≈ 7.326 cm

Now use ΔOCB. Let CB = a, OB = b, OC = c. CB = AB

2= 43.80

b ∠D = ∠BDCsin D

5= sin 88°

7.3263

sin D = 5 × sin 88°7.3263

= 0.6820D = 43.0045°

c2 = a2 + b2 – 2ab cos OOC2 = 43.80352 + 702 – 2 × 43.8035 × 70 × 0.3648

= 4581.24OC ≈ 67.7 m


Exercise 12D Solutions 1 a Area = 1

2 ab sin C

= 12× 6 × 4 × sin 70°

= 11.28 cm2

b Area = 1

2 yz sin X

= 12× 5.1 × 6.2 × sin 72.8°

= 15.10 cm2

c Area = 1

2 nl sin M

= 12× 3.5 × 8.2 × sin 130°

= 10.99 cm2

d ∠C = 180 – 25 – 25 = 130°

Area = 12 ab sin C

= 12× 5 × 5 × sin 130°

= 9.58 cm2

2 a Use the cosine rule to find ∠B. (Any angle will do.)

cos ∠B = a2 + c2 – b2

2ac

= 3.22 + 4.12 – 5.92

2 × 3.1 × 4.1 = –0.2957

∠B = 107.201°

Area = 12

ac sin B

= 12× 3.2 × 4.1 × sin 107.201

≈ 6.267 cm2

b Use the sine rule to find ∠ C.

sin C7

= sin 100°9

sin C = 7 × sin 100°9

= 0.7659C = 49.992°

∠A = 180 – 100 – 49.992 = 30.007°

Area = 12 bc sin A

= 12× 9 × 7 × sin 30.007°

≈ 15.754 cm2

c E = 180 – 65 – 66

= 60°e

sin 60° = 6.3sin 55°

e = 6.3 × sin 60°sin 55°

= 6.6604 cmArea = 1

2 ef sin D

= 12× 6.6604 × 6.3 × sin 65°

≈ 19.015 cm2

d Use the cosine rule to find ∠D.

cos ∠D = e2 + f2 – d2

2ef

= 5.12 + 5.72 – 5.92

2 × 5.1 × 5.7 = –0.4074

∠D = 65.95°

Area = 12 ef sin D

= 12× 5.1 × 5.7 × sin 65.95

≈ 13.274 cm2


e sin I12

= sin 24°5

sin I = 12 × sin 24°5

= 0.9671I = 77.466° or 180 – 74.466 = 102.533°

I is obtuse, so must be 102.533°G = 180 – 24 – 108.533 = 53.466°

Area = 12

hi sin G

= 12× 5 × 12 × sin 53.466

≈ 24.105 cm2

f I = 180 – 10 – 19

= 151°i

sin 151° = 4sin 19°

i = 4 × sin 151°sin 19°

= 5.9564Area = 1

2 ih sin G

= 12× 5.9564 × 4 × sin 10°

≈ 2.069 cm2


Exercise 12E Solutions 1 l = 105

360× 2πr

= 105360

× 2 × π × 25

≈ 45.81 cm

2 a θ = 50

30= 5

3 radians

= 53× 180

π degrees

= 95.4929° = 95°30'

b

sin θ

2= 25

30= 0.8333

θ2

= 56.4426°

θ = 112.885° = 112°53'

3 a Set your calculator to radian mode.

sin θ2

= 37

= 0.4285

θ2

= 0.4429

θ = 0.8858l = rθ = 7 × 0.8858 = 6.20 cm

b This represents the minor segment area.

A = 12

r2 (θ – sin θ)

= 12× 72 × (0.8858 – sin 0.8858)

= 2.73 cm2

4 A represents the interior of a circle of radius 4, centre the origin.

cos θ

2= 2

4= 1

2θ2

= π3

θ = 2π3

A ∩ B is a segment where r = 4, θ = 2π3

A = 12 r2(θ – sin θ)

= 12× 42 ×

⎝⎜⎛ 2π

3– sin 2π

3 ⎠⎟⎞

= 9.83 cm2

5 cos θ = cos 2 ⎝⎜⎛ θ

2⎠⎟⎞

= cos2 θ2

– sin2 θ2

cos2 θ2

= 1 – sin2 θ2

∴ cos θ = 1 – sin2 θ2

– sin2 θ2

= 1 – 2sin2 θ2

2r2(1 – cos θ) = 2r2

⎝⎜⎛ 1 –

⎝⎜⎛ 1 – 2 sin2 θ

2 ⎠⎟⎞⎠⎟⎞

= 2r2

⎝⎜⎛ 1 – 1 + 2sin2 θ

2 ⎠⎟⎞

= 2r2

⎝⎜⎛ 2sin2 θ

2 ⎠⎟⎞

= 4r2sin2 θ2

= 2r sin θ2


6

Altitude CD = 5 tan 60°

= 5 3 cm

OD = 5 tan 30°

= 53

= 5 33

cm

Radius = 5 3 – 5 33

= 15 3 – 5 33

= 10 33

cm

∠AOD = 60°∴∠AOB = 120°

Area = 3 × segment area = 3

2× r2 × (θ – sin θ)

= 32× 300

9×⎝⎜⎛ 2π

3– sin 2π

3 ⎠⎟⎞

= 50 ⎝⎜⎛ 2π

3– sin 2π

3 ⎠⎟⎞

= 61.42 cm2 7 a C = 2πr

= 2 × π × 20 = 40π ≈ 125.66 cm

b

cos θ = 20

20 + 60= 0.25

θ = 1.31812θ = 2.6362

Proportion visible = 2.63622π

= 0.41956≈ 41.96%

8 a Use fractions of an hour (minutes). l = 25

60× 2πr

= 2560

× 2 × π × 4

= 10π3

≈ 10.47 m

b Angle = 2560

× 2π = 5π6

Area = 12

r2θ

= 12× 42 × 5π

6

= 20π3

≈ 20.94 m2

9

The required area is the sum of two segments. Let the left area be A1 and the right area A2.

tan θ = 43

θ = 0.92722θ = 1.8545A1 = 1

2× 32 × (1.8545 – sin 1.8545)

= 4.0256tan φ = 3

4φ = 0.6435

2φ = 1.2870A2 = 1

2× 42 × (1.2870 – sin 1.2870)

= 2.6160

Total area = 4.0256 + 2.6160 = 6.64 cm2


10 A = 12

r2θ = 63

r2θ = 126θ = 126

r2

P = r + r + rθ = 322r + r × 126

r2 = 32

2r + 126r

= 32

2r2 + 126 = 32r2r2 – 32r + 126 = 0

r2 – 16r + 63 = 0(r – 7)(r – 9) = 0

r = 7 or 9 cmθ = 126

r2

When r = 7, θ = 12672 =

⎝⎜⎛18

7 ⎠⎟⎞c

When r = 9, θ = 12692 =

⎝⎜⎛14

9 ⎠⎟⎞c

12 a

The balls can be enclosed as in the

diagram above. 2θ = 360 – 90 – 60 – 90

= 120°θ = 60°x5

= tan 60° = 3

x = 5 3

Perimeter = 6 × 5 3 + 3 × 10≈ 81.96 cm

b Height of large triangle = (2x + 10) × sin 60°

= (10 3 + 10) × 32

= 15 + 5 3 cm

11 The following diagram can be deduced from the data:

Area of large triangle = 12 (10 3 + 10)(15 + 5 3 )

≈ 173.2050 cm2

Area of three discs = 10 cm triangle – half a circle Height of 10 cm triangle = 10 × sin 60°

= 5 3 cm

x2 = 602 – 102 = 3500x = 10 35

cos θ = 1060

= 16

θ = 1.40332θ = 2.8066

2π – 2θ = 3.4764

Area = 1

2× 10 × 5 3 – 1

2× π × 52

= 50 3 – 12.5π ≈ 4.03 cm2

Length of belt on left wheel: l = rθ

= 15 × 2.8066 = 42.1004 Length of belt on right wheel:

l = rθ

= 25 × 3.4764 = 86.9122 Total = 12 × 10 25 + 42.1004 + 86.9112

≈ 247.33 cm


Exercise 12F Solutions 1

130

d= tan 18°

d = 130tan 18°

= 400.10 m

2

h

40= tan 41°

h = 40 × 0.869 = 34.77 m

3

500

d= tan 41°

d = 500tan 41°

= 575.18 m

4

40

d= tan 20°

d = 40tan 20°

= 109.90 m

5

50

AB= tan 20°

AB = 50tan 20°

= 137.373 m50AC

= tan 18°

AC = 50tan 18°

= 153.884 md = AC – AB = 153.884 – 137.373 ≈ 16.51 m

6

tan θ = 15

10= 1.5

θ ≈ 56°

The bearing is 056°. 7 a

tan θ = 22

15= 1.466

θ = 55.713° The bearing is 90 – θ ≈ 034°. b 180 ° + 34° = 214°


First, use the cosine rule to find LK. 8

LK2 = m2

= k2 + l2 – 2kl cos M = 122 + 92 – 2 × 12 × 9 × cos 128° = 357.9829

LK = 18.920

It is easier to use the sine rule to find ∠MLK.

sin L9

= sin 128°18.920

sin L = sin 128° × 918.920

= 0.3748∠MLK = ∠L

≈ 22.01°

Use the cosine rule, where ∠C = 180 – 40 – 35 = 105°

AB2 = c2

= a2 + b2 – 2ab cos C = 25002 + 20002

– 2 × 2500 × 2000 × cos 105° = 12 838 190.4510

AB = 3583.04 m

12 a ∠BAN = 360 – 346 = 14°

∠BAC = 14 + 35 = 49°

b Use the cosine rule: 9 207 – 180 = 027° BC 2 = a2

= b2 + c2 = 2bc cos A = 3402 + 1602 – 2 × 340 × 160 × cos 49° = 69 820.7776

BC = 264.24 km

10

13

∠SAB = 90 – 55 = 35°∠SBA = 302 – 270 = 32°∠ASB = 180 – 35 – 32 = 113°

11 Use the cosine rule:

∠PSQ = 115°

PQ2 = s2

= p2 + q2 – 2pq cos A = 52 + 7.52 – 2 × 5 × 7.5 × cos 115° = 112.9464

PQ = 10.63 km ∠ LMK = 360 – 90 – 142 = 128°


Exercise 12G Solutions 1 a FH2 = 122 + 52

= 169FH = 13 cm

b BH2 = 132 + 82

= 233BH = 233 ≈ 15.26 cm

c tan ∠BHF = 8

13 = 0.615

∠BHF = 31.61°

d ∠BGH = 90° and BH = 233 cos ∠BGH = 12

233 = 0.786

∠BGH = 38.17°

2 a AB = 2EF EF = 4 cm b tan ∠VEF = VE

EF = 12

4= 3

∠VEF = 71.57°

c VE2 = 42 + 122

= 160VE = 160

= 4 10 ≈ 12.67 cm

d All sloping sides are equal in length.

Choose VA. VA2 = VE2 + EA2

= 160 + 42 = 176VA = 176

= 4 11 ≈ 13.27 cm

e ∠VAD = ∠VAEtan ∠VAE = VE

EA

= 4 104

= 10 ≈ 3.162∠VAE = 72.45°

f Area of a triangular face = 1

2× AD × VE

= 12× 8 × 4 10

= 16 10 cm2

Area of base = 8 × 8 = 64 cm2

Surface area = 4 × 16 10 + 64≈ 266.39 cm2

3 First, sketch the square base, and find

the height h of the tree. Mark M as the mid-point of TC and O as the centre of the square.

OM = TM = 50 m OT2 = 502 + 502 = 5000

OT = 5000 mh

5000= tan 20°

h = 5000 × tan 20° = 25.7365

At A and C, tan θ = 25.7365

100= 0.2573

θ = 14.43°

At B, TB = 2 × OT = 2 5000 m tan θ = 25.7365

5000= 0.1819

θ = 10.31°


4 a ∠ABC = 180 – 90 – 45

= 45° ABC is isosceles, and CB = AC = 85 m. b XB

BC= sin 32°

XB85

= sin 32°

XB = 85 × sin 32° = 45.04 m

5

50

x= tan 26°

x = 50tan 26°

= 102.515 my2 = x2 + 1202

= 24 909.364y = 24 909.364 = 157.827 m

tan φ = 50y

= 0.316

φ = 17.58°

6 From the top of the cliff:

For the first buoy:

160d

= tan 3°

d = 160tan 3°

= 3052.981 m

For the second buoy:

160

d= tan 5°

d = 160tan 5°

= 1828.808 m

From the cliff:

∠ C = 337 – 308 = 29° Use the cosine rule.

c2 = 3052.9812 + 1828.8082

– 2 × 3052.981 × 1828.808 × cos 29° = 2 898 675.1436c = 1702.55 m

7 a AC2 = 122 + 52 = 169

AC = 13 cmtan ∠ACE = 6

13 = 0.4615

∠ACE = 24.78°

b Triangle HDF is identical (congruent) to

triangle AEC. ∴∠HFD = ∠ACE

∠HDF = 90 – 24.28 = 65.22°

c CH2 = 122 + 62 = 180

CH = 180tan ∠ECH = EH

CH = 5

180= 0.3726

∠ECH = 20.44°


8 Looking from above, the following diagram applies.

Because the angle of elevation is 45°, AT will equal the height of the tower, h m. Use the cosine rule. BT2 = h2 + 1002 – 2 × h × 100 × cos 60°

= h2 + 1002 – 200h × 12

= h2 – 100h + 1002

From point B:

h

d= tan 26°

d = htan 26°

= 2.050h∴ 2.0502 h2 = h2 – 100h + 1002

4.2037h2 = h2 – 100h + 10 0003.2037h2 + 100h = 10 000

Using the quadratic formula: h ≈ 42.40 m 9 Find the horizontal distance of A from

the balloon.

750

d= tan 40°

d = 750tan 40°

= 893.815 m

The distance of B from the balloon may be calculated in the same way: 750

d= tan 20°

d = 750tan 20°

= 2060.608 m

Draw the view from above and use the cosine rule.

x2 = 893.8152 + 2060.6082

– 2 × 893.815 × 2060.608 × cos 70° = 3 785 143.5836x = 1945.54 m

10 a Find the length of an altitude:

a2 = 2.52 + 62 = 42.45

a ≈ 6.5 cm The sloping edges are also the

hypotenuse of a right-angled triangle. s2 = 2.52 + 6.52 = 48.5

s ≈ 6.96 cm b Area = 1

2× 5 × 6.5

= 16.25 cm2


11 c Let the angle of elevation be θ. a Distance = 300 × 1

60= 5 km OC2 = 3535.5332 + 2535.5332

= 18 928 932OC = 4350.739

tan θ = 5004350.739

= 0.1149θ = 6.56° = 6°33'

b Looking from above:

AE = 5000 × sin 45°

= 50002

≈ 3535.433

CE = 5000 × sin 45°

= 50002

≈ 3535.433

CD = CE – DE = 3535.533 – 1000 = 2535.533

tan ∠COD = 2535.5333535.533

= 0.7171∠COD = 35.65°

Bearing = 1 80 + 35.65= 215.65°


Solutions to Multiple-choice Questions 1 Use the sine rule. sin Y

y= sin X

xsin Y18

= sin 62°21

sin Y = 18 × sin 62°21

= 0.7568

Y = 49.2° D 2 Use the cosine rule. c2 = a2 + b2 – 2ab cos C

= 302 + 212 – 2 × 30 × 21 × 5153

= 128.547

c ≈ C 11 3 Use the cosine rule.

cos C = a2 + b2 – c2

2ab

= 5.22 + 6.82 – 7.32

2 × 5.2 × 6.8 = 0.2826

C ≈ C 74° 4 Area = 1

2 bc sin A

= 12× 5 × 3 × sin 30°

= 3 B .75 cm2

5 The other angles in the (isosceles)

triangle are both 180 – 130

2= 25°.

Use the sine rule. 10

sin 130° = rsin 25°

r = 10 × sin 25°sin 130°

≈ A 5.52 cm

6 First find the angle at the centre using the cosine rule.

cos C = 62 + 62 – 52

2 × 6 × 6 = 0.6527

C = 49.248° = 0.8595c Segment area = 1

2 r2(θ – sin θ)

= 12× 62 × (0.8595 – sin 0.8595)

≈ A 1.8 cm2

7

500

d= tan 20°

d = 500tan 20°

≈ 1374 D m 8 tan θ = 80

1300 = 0.0615

θ = B 3.521° ≈ 4° 9

tan θ = 7

5= 1.4

θ = 54°

Bearing = 270° + 54° = 324° C 10 215° – 180° = 035° A


Solutions to Short-answer Questions 1 a a2 = b2 + c2 – 2bc cos A

62 = x2 + 102 – 2x × 10 × 32

x2 – 10 3 x + 64 = 0

x = 10 3 ± 300 – 4 × 1 × 642

= 10 3 ± 442

= 10 3 ± 2 112

= 5 3 ± 11

b sin y10

= sin 30°6

sin y = 10 × sin 30°6

= 1012

= 56

y = sin–1

⎝⎜⎛5

6⎠⎟⎞

or π – sin–1

⎝⎜⎛5

6⎠⎟⎞

Since both answers to a are positive, this must be an ambiguous case.

2 a Triangle is isosceles, so ∠ B = 30° and ∠C = 150°. Area = 1

2× 40 × 40 × sin 150°

= 400 cm2

= 12 bh

∴ 20h = 400h = 20 cm

b CM

40= sin 30°

CM = 40 × sin 30° = 20 cm

3

QR2 = 122 + 202 – 2 × 12 × 20 × cos 60°

= 144 + 400 – 240 = 304

QR = 304 = 16 × 19 = 4 19 km

4

a Use the cosine rule. AC2 = 52 + 52 – 2 × 5 × 5 × cos 120°

= 25 + 25 + 25 = 75

AC = 75 = 5 3 cm

b Area = 1

2× 5 × 5 × sin 120°

= 25 34

cm2

c In isosceles triangle ABC, ∠ACB = ∠BAC

= 12 (180 – 120) = 30°

∠ACD = 90 – 30 = 60°

Area of ADC = 12× 7 × AC × sin 60°

= 12× 7 × 5 3 × 3

2 = 105

4 cm2


d Total area = 25 34

+ 1054

= 25 3 + 1054

= 5(5 3 + 21)4

cm2

5 x = 180 – 37 = 143° 6

cos S = 102 + 72 – 82

2 × 10 × 7 = 85

140= 17

28

7 sin B

b= sin A

asin B

6= sin 60°

5sin ∠ABC = sin B

= 6 × sin 60°5

= 6 35 × 2

= 3 35

8 A = 1

2 r2θ

33 = 12× 62 × θ

= 18θθ = 33

18= 11

6 (radians)

9 a i ∠ TAB = 90 – 60 = 30° ii ∠ ATB = 180 – 30 – (90 + 45)

= 15° b AT

sin 135° = 300sin 15°

AT = sin 135° × 300 × 46 – 2

= 12× 1200

6 – 2 = 1200

12 – 2 = 1200

2 3 – 2= 600

3 – 1

= 6003 – 1

× 3 + 13 + 1

= 600( 3 + 1)3 – 1

= 300( 3 + 1) m

10

Use the cosine rule. AC2 = b2

= a2 + c2 – 2ac cos B = 112 + 152 – 2 × 11 × 15 cos 60° = 121 + 225 – 165 = 181

AC = 181 km

11

Draw a line AD in an easterly direction

from A (parallel to BC.)


∠DAC = 30°

∠ACB = ∠DAC = 30°∠ABC = 180° – 30 – 30

= 120°∴ BC = 2.4 km

Use the cosine rule to find AC. AC2 = b2

= a2 + c2 – 2ac cos B = 2.42 + 2.42 – 2 × 2.4 × 2.4 × cos 120° = 5.76 + 5.76 + 5.76 = 17.28

AC = 17.28 = 5.76 × 3

= 2.4 3 or 12 35

km

12 a Draw a perpendicular from O to bisect AB. sin ∠AOD = 12

13

∠AOD = sin–1 1213

∠AOB = 2 sin–1 1213

arc AB = rθ

= 13 × 2 sin–1 1213

= 26 sin–1 1213

b Reflex∠AOB = 2π – 2 sin–1 1213

area = 12× 132 ×

⎝⎜⎛ 2π – 2 sin–1 12

13⎠⎟⎞

= 169 ⎝⎜⎛ π – sin–1 12

13⎠⎟⎞ cm2

Note: the perpendicular distance from O to AB can be calculated to be 5 cm using Pythagoras’ theorem, and so sin–1 12

13= cos–1 5

13= tan–1 12

5.

Either of these three angles may be used interchangeably.

13 l = rθ30 = 12θ

θ = 3012

=⎝⎜⎛5

2⎠⎟⎞c

A = 12× 122 × 5

2 = 180 cm2

14 The reflex angle = 2 π – 2

≈ 2 × 3.14 – 2≈ 4.28 radians

Arc length ≈ 5 × 4.28 = 21.4 cm 15 First calculate the distance of each boat

from the cliff. The first boat will form a right-angled isosceles triangle and is 11 m from the cliff.

For the second boat, 11

d= tan 30° = 1

3d = 11 3 m

Use the cosine rule.

x2 = 112 + (11 3 )2 – 2 × 11 × 11 3 × cos 30° = 121 + 363 – 363 = 121x = 11 m


Chapter 13 – Revision of chapters 8–12 Solutions to Multiple-choice Questions 1 cos θ = 2.1

2.6 = 0.8076

θ ≈ A 36° 2 Maximum = 2 Minimum = –4 Amplitude = 2 – –4

2

= 2 B 3 c

sin C= a

sin Ac

sin 38° = 58sin 130°

c = 58 sin 38°sin 130° B

4 Convert both to mm. The scale is 45 : 17 100 Divide both parts of the ratio by 45. Scale = 1 : 380 C 5 ( x, y) → (y, x) D (5, – 2) → (–2, 5)

6 cos A = 1 – sin A2

= 1 – 25169

= 144169

= 1213

cos B = 1 – sin B2

= 1 – 64289

= 225289

= 1517

sin (A – B) = sin A cos B – cos A sin B = 5

13× 15

17– 12

13× 8

17 = 75 – 96

221

= – 21221

B

7 The sine and cosine rules allow us to find the unknown quantities:

Area = 12

bc sin A

a2 = b2 + c2 – 2bc cos A

sin B = sin A × ba

All three statements are correct. D 8 AG

DG= 15

5= 3

AG = 3DG = 18∴ AE = 18 – 9 = 9

BGFG

= AGEG

FG + 6FG

= 189

= 2

FG + 6 = 2FG∴ FG = 6

xFG

= 69

= 23

x = 23 FG

= 2

3× 6 = 4 B

9 ( x, y) → ( – y, – x) C (2, – 6) → (6, – 2) 10 The graph is y = cos (a) raised 1 unit, i.e. y = cos (a) + 1 E

11 cos A = 1 – sin A2

= 1 – 25169

= 144169

= 1213

cos B = 1 – sin B2

= 1 – 64289

= 225289

= 1517


tan A = sin Acos A

= 513

÷ 1213

= 513

× 1312

= 512

tan B = sin Bcos B

= 817

÷ 1517

= 817

× 1715

= 815

tan (A + B) = tan A + tan B1 – tan A tan B

=

512

+ 815

1 – 512

× 815

Multiply numerator and denominator by 12 × 15.

tan (A + B) = 5 × 15 + 8 × 12

12 × 15 – 5 × 8 = 75 + 96

180 – 40

= 171140

E 12 The image after translation is (5 + 2, –4 + –3) = (7, –7) The y-ordinate, –7, is 8 units below y = 1. After reflection, it will be 8 units above

y = 1, i.e. 9. The x-ordinate will be unchanged. The final image will be (7, 9). B 13 Write both in cm. The ratio is 8 : 320 = 1 : 40 E

14 sin ⎝⎜⎛ x – π

6 ⎠⎟⎞ = 3

2

x – π6

= π3

or π – π3

x = π3

+ π6

or 2π3

+ π6

= π2

or 5π6 D

15 Area = 12 bc sin A

= 12× 6 × 7 × sin 48° A

16 sin θ = 1 – c2

cot θ = cos θsin θ

= c

1 – c2 D

17 When a = 0, y = –1. This must mean that the equation is a

sine function (since sin 0 = 0), where the result is then reduced by 1 (0 – 1 = –1). Only C fits that description. C

18 Start with the point in the first quadrant,

left of x = m. The y-ordinate will be unchanged. The x-ordinate initially will be (m – a)

left of x = m. After reflection, it will be (m – a) to the

right of x = m, i.e. m + (m – a). The point will be (2m – a, b). A 19 θ = l

r = 3

4 radians

This is 34× 180

π= 135

π

E ≈ 42°58° 20 y = a sin bθ Amplitude = a

= 1 for both graphs

Period = 2πb

= 2π1

= 2π for first equation

=2π12

= 4π for second equation

y = sin ⎝⎜⎛1

2 θ⎠⎟⎞ has the same amplitude

but double the period. A


21 Refer to Q.18 regarding reflection in the line x = 4.

(x, y) →⎝⎜⎛1

2 x, y

⎠⎟⎞

→⎝⎜⎛8 – 1

2 x, y

⎠⎟⎞

y' = y ⇒ y = y'x' = 8 – 1

2 x ⇒ x = 16 – 2x'

x + y = 4 ⇒ (16 – 2x') + y' = 4y' = 4 – (16 – 2x') = 2x' – 12

Image: {(x, y): y = 2x – 12} E 22 cos A cos B – sin A sin B = cos (A + B)

= cos π2

= 0 D 23 cos2 A = 1 – sin2 A

= 1 – 59

= 49

cos A = – 23

sin 2A = 2 sin A cos A

= 2 × 53

× – 23

= – 4 59 E

24

Use similar triangles.

x2

= 1.65.6

= 1656

= 27

x = 47

B ≈ 0.57 m

25 y = 0 when x = π12

and 2π3

Since sin 0 = 0, this is a sine function

where y = sin a ⎝⎜⎛ x – π

12⎠⎟⎞ .

The graph is positive between 0 and π, so the function will be positive.

Only y = sin 2 ⎝⎜⎛ x – π

12⎠⎟⎞ fits these

criteria. A 26 DE

AB= 10

4= 2.5

DE = 2.5 × ABArea = 2.52 × 24

= C 150 cm2

27 Period = 180°3

= 60°

The tangent graph does not have an amplitude.

The graph is reflected in the x-axis.

y = 2 tan (3x)°

Only i and iv are correct. A 28 (x, y) → (x + 3, y + 2)

→ (x + 3, – y – 2)x' = x + 3 ⇒ x = x' – 3y' = – y – 2 ⇒ y = – y' – 2y = x2

– y' – 2 = (x' + 3)2

y' + 2 = – (x' + 3)2

Image: {( B x, y): – (x + 3)2 = y + 2} 29 60

360× πr2 = 1

6× π × 52

≈ C 13.09 cm2

30 KO : KN = 1 : 3

area KOParea MLK

= 19

area KOParea (MLK + MNK)

= 19× 1

2= 1

18

area KOParea KLMN

= 118

D


31 OE = 40 mmtan θ = VO

OE = 100

40= 2.5

E θ ≈ 68°

32 sin θ = 1 – c2

sin 2θ = 2 sin θ cos θ

= 2c 1 – c2 E

33 (x, y) →⎝⎜⎛1

3 x, y

⎠⎟⎞

→⎝⎜⎛1

3 x, 2y

⎠⎟⎞

x' = 13

x ⇒ x = 3x'

y' = 2y ⇒ y = 12 y'

y = 2x

12

y' = 23x'

y' = 2 × 23x'

Image: y = 2 × 23x C 34 Write in the form R cos (x + α).

R = 42 + 32 = 5cos α = 4

5α = 36.869°

5 cos (x + 36.869) = 1cos (x + 36.869) = 1

5= 0.2

x + 36.869 = 78.463 or 360 – 78.463

= 281.536

x ≈ 41.59° or 244.67° C

35 A + BA

= 53

33 = 12527

A + 49A

= 12527

27A + 49 × 27 = 125A98A = 49 × 27

A = 49 × 2798

= 272

= 13.5 cm3 C

36 110° = 110 × πc

180

= 11πc

18≈ 1.9198c

A = 12 r2(θ – sin θ)

= 12× 452 × (1.9198 – sin 1.9198)

≈ B 992 cm2

37 8 sin θ cos3 θ – 8 sin3 θ cos θ

= 8 sin θ cos θ (cos2 θ – sin2 θ) = 4 sin 2 θ cos θ

= 2 sin 4θ C

38 The graph has asymptotes at – π2

and 3π2

, so the graph is not shifted left

or right (no addition or subtraction after the tan function). (0, –3) is an x-intercept, so the equation must produce y = –3 when x = 0. Only C fills these criteria. C

39 2

3 πR3 : 1

3 π2r2 = 27 : 4

2R3 : r3 = 27 : 4R3 : r3 = 27

2 : 4

= 27 : 8R : r = 3 27 : 3 8

R : r = 3 : 2 E 40 TS means S followed by T. If the x-ordinate is on the right of x = 2,

its distance from x = 2 after reflection will be x – 2 units to the left of x = 2.

The new x-ordinate will be 2 – (x – 2) = 4 – x. The y-ordinate will be unchanged. (x, y) → (4 – x, y)

→ (4 – x + 2, y + 3) D TS(x, y) → (6 – x, y + 3)


41 The net transformation is

⎣⎢⎡ 0

–2–11 ⎦⎥⎤⎣⎢⎡ –1

001⎦⎥⎤ =

⎣⎢⎡ 0

2–11 ⎦⎥⎤

⎣⎢⎡ 0

2–11 ⎦⎥⎤⎣⎢⎡ 0

0⎦⎥⎤ =

⎣⎢⎡ 0

0⎦⎥⎤

⎣⎢⎡ 0

2–11 ⎦⎥⎤⎣⎢⎡ 1

1⎦⎥⎤ =

⎣⎢⎡ –1

3 ⎦⎥⎤

⎣⎢⎡ 0

2–11 ⎦⎥⎤⎣⎢⎡ 0

1⎦⎥⎤ =

⎣⎢⎡ –1

1 ⎦⎥⎤

⎣⎢⎡ 0

2–11 ⎦⎥⎤⎣⎢⎡ 1

0⎦⎥⎤ =

⎣⎢⎡ 0

2⎦⎥⎤

These points correspond to E. E


Chapter 14 – Circle theorems Exercise 14A Solutions 1 a 50° = 1

2 x

x = 100°

y = 12

x

= 50°

b y = 360 – 108 = 252°

x = 12 × 290 = 1451

x = 12× 108 = 54°

c Acute ∠O = 2 × 35 = 70° z = 360 – 70 = 290°

y = 12× 290 = 145°

d O = 180°

x = 360 – 180 = 180°y = 90° (Theorem 3)

e 3x + x = 180°

4x = 180°x = 45°z = 2 × 3x = 2 × 3 × 45 = 270°y = 360 – 270 = 90°

f x + y + 25 + 125 = 360°

x + y = 210°

A line from x to the 125° angle will create two isosceles triangles – one with angles 25°, 25° and x1, and angles 100°, y and x2.

The base angles in an isosceles triangle are complementary, so y = 100°.

x = 210 – 100 = 110°

2 The opposite angles of a cyclic quadrilateral are supplementary.

a x + 112 = 180°x = 68°

y + 59 = 180°y = 121°

b x + 68 = 180°

x = 112°y + 93 = 180°

y = 87°

c x + 130 = 180°

x = 50°y + 70 = 180°

y = 110°

3 Let the equal angles be x°. 2x + 40 = 180°

2x = 140°x = 70°

The angles in the minor segments will be the opposite angles of cyclic quadrilaterals.

180 – 70 = 110°180 – 70 = 110°180 – 40 = 140°


4

28°

20°

70°

A

B

C

D

E

In cyclic quadrilateral ABDE, ∠DEA = 110° On arc DC, ∠ DBC = 28° ∴∠ABC = 70 + 28 = 98° Join EB. Equal chords will subtend

equal angles at the circumference.

∴∠ABE = ∠EBD = 35°∠EAD = 35° (also on equal arcs )

On arc BC, ∠BAC = ∠BDC = 20° ∴∠EAB = 35 + 28 + 20 = 83° In cyclic quadrilateral ABDE, ∠ EDB = 180 – 83 = 97° ∴∠EDC = 97 + 20 = 117° In cyclic quadrilateral ABCD, ∠BCD = 180 – (28 + 20) = 132° 5

(subtended by the same

arc) ∠BAC = ∠BDC

(subtended by equal arcs) ∠DAC = ∠BDA ∴∠BAC + ∠DAC = ∠BDC + ∠BDA ∠BAD = ∠ADC (opposite angles in

a cyclic quadrilateral) ∠ADC + ∠ABC = 180°

∴∠BAD + ∠ABC = 180° BC and AD are thus parallel, as co-interior

angles are supplementary

6

A

B C

D

E

∠ADE + ∠ADC = 180°

∠ABC = ∠ADC (opposite angles in a parallelogram)

∴∠ADE + ∠ABC = 180° (opposite

angles in a cyclic quadrilateral) ∠AED + ∠ABC = 180°

∴∠ADE = ∠AEDAE = AD

7

120°O

A

B

C

D

∠ADC = 120

2= 60°

If B and D are on opposite sides of AOC, then ∠ADC = 240

2= 120°.

(Reflex angle ADC will be used.)

= 360° – 120°

8 The opposite angles in a parallelogram

are equal. In a cyclic parallelogram, the opposite

angles will add to 180 .° ∴ the opposite angles equal 90 .° ∴all angles are 90 i.e. the

parallelogram is a rectangle (subtended by the same arc).

,°


9

A

B

C

D

S

PR

Q× ×

°°

In triangle BCS, ∠BSC = 180° – ∠SBC – ∠BCS

= 180° – 12∠ABC – 1

2∠BCD

Likewise, ∠AQD = 180° – 1

2∠BAC – 1

2∠CDA

∴∠BSC + ∠ACD = 180 – 12∠ABC – 1

2∠BCD + 180° – 1

2∠BAD – 1

2∠CDA

= 360° – 12

(∠ABC + ∠BCD + ∠BAD + ∠CDA)

∠ABC + ∠BCD + ∠BAD + ∠CDA = 360° (angle sum of quadrilateral)∠BSC + ∠AQD = 360° – 180°

= 180° ∴both pairs of opposite angles in PQRS will add to 180 .°

∴PQRS is a cyclic quadrilateral.


Exercise 14B Solutions 1 a x = 73° (alternate segments) y = 81° (alternate segments) b ∠T = 90°

∴ x = 90 – 33 = 57° q = 57° (alternate segment theorem) c y = 74° (alternate segments)

z = 180 – 742

= 53°x = 53° (alternate segments)

d x = 180 – 80 – 40 = 60° Use the alternate segment theorem to

find the other angles. y = 180 – 60 – 60 = 60°

w = 180 – 40 – 40 = 100°z = 180 – 80 – 80 = 20°

e w = z = x = 54° (alternate segment, alternate

angles and isosceles triangle PTS) y = 180 – 54 – 54 = 72° 2 a ∠BCX = 40° b ∠CBD = 40° c ∠ABC = 2 × 40° = 80° 3 In ΔCAT, ∠ACB = 180 – 30 – 110 = 40° The alternate segment theorem shows

∠BAT = 40°. ∴∠CAB = 110 – 40 = 70° In Δ CAB, ∠ABC = 180 – 40 – 70 = 70°

4

A

C

BD

E116°

Triangle ABC is isosceles; ∠ABC = ∠ACB

= 180 – 1162

= 32°

Using the alternate angle theorem, ∠BDC = ∠ACB = 32°.

∠BEC + ∠BDC = 180° (opposite angles in cyclic quadrilateral

BCED) ∴BEC = 180 – 32 = 148° 5

C B A

DE

There are multiple ways of proving this result. ∠ADB = ∠DCB (alternate segment theorem)

∠DCB = ∠DCE (subtended by equal arcs)∴∠ADB = ∠DCE

∠DBA + ∠DBC = 180° ∠DEC + ∠DBC = 180°

∴ ∠DBA = ∠DEC ∴ triangles ABD and CDE are similar,

since two pairs of opposite angles are equal.


6

T

A B

C

∠TCB = ∠CBA (alternate angles)

∠TCB = ∠CAB (alternate segment)∴∠CBA = ∠CAB

ABC is an isosceles triangle with CA = CB. 7

T

P

Q

AB

Triangles PAT and BAQ are similar, since two pairs of opposite angles are equal.

∠TAP = ∠AQP (alternate segment)∠AQP = ∠BAQ (alternate angles)

∴∠TAP = ∠BAQ∠APT + ∠APQ = 180° (adjacent angles)∠AQB + ∠APQ = 180° (opposite angles)

∴ ∠APT = ∠ABQ

8

T

A

Q

B

O NP

Let T be the point where the perpendicular

from P meets the tangent at A Let O be the centre of the circle. Join PA and PB. Consider triangles OAN and OBN: ∠ANO = ∠BNO = 90°

OA = OB (radii) ON is common to both triangles. ∴∠AON ≡ ∠BON (RHS)

AN = BN

Now consider triangles PAN and PBN: AN = BN

∠PNA = ∠PNB = 90°

PN is common to both triangles.

∴ PAN ≡ PBN (SAS)∠PAN = ∠PBN

Now consider triangles PAT and PAN: ∠PBN = ∠PAT (alternate segment theorem)

∴∠PAT = ∠PAN∠PTA = ∠PNA = 90°

PA is common to both triangles. ∴ ∠PAT ≡ ∠PAN (AAS)

PT = PN


Exercise 14C Solutions 1

A D

C B

P

a AP⋅PB = CP⋅PD

5 × 4 = 2PDPD = 10 cm

b AP⋅PB = CP⋅PD

4PB = 3 × 8PB = 6 cm

2

O

A

B

CP

3 cm

8 cm5 cm

Let the centre of the circle be O and the

length of the radius r cm. Extend OP to meet the circumference of the

circle at C and D. CP = r – 3 and PD = r + 3 CP⋅PD = AP⋅PB

(r – 3)(r + 3) = 8 × 5r2 – 9 = 40

r2 = 49r = 7 cm

3

B

D

O

A

C

P

7 cm5 cm

PD = 7 cm – 5 cm = 2 cm Let PB = x cm PA = 4x cm

AP⋅PB = CP⋅PD4x × x = 12 × 2

x2 = 6x = 6

∴ AB = 4 6 + 5 = 5 6 cm

4

A

BR

P

Q

Use theorem 9: PQ2 = PA⋅PB

PR2 = PA⋅PB∴PQ2 = PR2

PQ = PR


5

O

AB

C

SP

QR

Let the centre of the circles be O. Let the radii of the larger and smaller

circles be R and r respectively. Let QP produced meet the larger circle at S. By symmetry, SP = RQ. Extend OQ to meet the larger circle at A

and C, and the smaller circle at B. Since SP = RQ, SP + PQ = RQ + PQ

∴ SQ = PR

Using the large circle, SQ ⋅RQ = AQ⋅CQ

∴PR⋅RQ = (R + r)(R – r), which is constant

6

A

CB PD

E

Let P be a point on BC such that AP is perpendicular to BC.

Because ABC is isosceles, AP will bisect AB. Let AP = x and PC = PB = y. DP = y – BD CD = 2y – BD Using Pythagoras’ theorem twice, we

get AB in triangle ABP and in triangle ADP.

2 = x2 + y2

AD2 = x2 + (y – BD)2

= x2 + y2 – 2y × BD + BD2

= AB2 – BD(2y – BD) = AB2 – BD⋅CD

BD⋅CD = DE⋅AD∴ AD2 = AB2 – DE⋅AD

AB2 = AD2 + DE⋅AD = AD(AD + DE) = AD⋅AE


Solutions to Multiple-choice Questions 1 In isosceles triangle ABD, ∠ABD = ∠ADB

= 180 – 702

= 55°

is subtended by the same arc, so B

∠ACD∠ACD = 55°.

2 In quadrilateral OAPB, ∠OAP = ∠OBP = 90°

∠APB = 360 – 150 – 90 – 90 = 30°

The angle subtended at the circumference on minor arc AB is

1502

= 75°.

This angle is opposite Q in a cyclic quadrilateral.

A ∴∠AQB = 180 – 75 = 105° 3 There are multiple ways to solve this

problem. ∠OAB = 68°

∠BAT = 90 – 68 = 22°∠ABT = 180 – 20 – 68 = 92°

∴∠ E ATB = 180 – 22 – 92 = 66° 4 ∠BAC = 60° Reflex ∠BOC = 360 – 120

= 240° In quadrilateral ABOC, ∠ABO = 360 – 240 – 42 – 60 = 18° A 5 ∠DAB = 180 – 65 = 115° Corresponding angles on parallel lines ∴∠ C CBE = 115°

6

∠BAD = 50° (alternate segment theorem)

∠ABD = 50° (alternate segment theorem) In triangle ABD, ∠ADB = 180 – 50 – 50 = 80° A 7

A D

C B

P

AP⋅PB = CP⋅PD

12 × 6 = 2PD

PD = 36 cm C 8 NB = 13 – 5 = 8 cm NQ = PN AN⋅NB = PN⋅NQ

= PN2

18 × 8 = PN2

PN = 144 = 12cmPB2 = 122 + 82 = 208PB = 208

= 16 × 13

= 4 13 cm B 9 In triangle BAX, ∠BAX = 180 – 40 – 105

= 35° Angles are subtended by the same arc A ∴∠XSC = ∠BAX = 35°


10 (angle subtended by a diameter) ∠CDA = 90°

In triangle ACD, ∠CAD = 180 – 90 – 25

= 65°∠CBD = ∠ACD = 65°∠ BCD = 180 – 75 = 105°

In triangle BCD, ∠BDC = 180 – 105 – 65 = 10° A


Solutions to Short-answer Questions 1 a Theorem 1: y = 140

2= 70°

Theorem 4: x + y = 180°x = 180 – 70 = 110°

b Name the quadrilateral ABCD, in which

y is at A and x is at B. Let P be the point of intersection of AC

and BD. In triangle XCD, ∠ CDX = 180 – 50 – 75

= 55° ∠BCD = 90°

(angle subtended by a diameter)

In triangle BCD, x = ∠BDC

= 180 – 90 – 55 = 35°

y = x = 35° (angles subtended by the same arc)

c Angles in the same segment are equal: x = 47° y = 53° z = 360 – (180 – 53 – 47) – (180 – 53 – 47)

2 = 360 – 160

2 = 100°

d First note that y = x. Consider the concave quadrilateral

containing the 30° angle. Its angles are 30°, 180° – 70° = 110, x +

70° and x + 70°, using supplementary angles, vertically opposite angles and exterior angles of a triangle.

x + 70 + x + 70 + 110 + 30 = 2602x + 280 = 260

x = 40°y = 40°z = 180 – (x + 70) = 70°

2

a Using angles on arc AP,

∠POA = 2∠CBA Using alternate angles, ∠ POA = ∠CAB ∴∠CAB = 2∠CBA b Using angles on arc AP,

∠POA = 2∠CBA

Using alternate angles, ∠ OPC = ∠CBA Using the exterior angle of triangle

OCP, ∠PCA = ∠POC + ∠OPC

= ∠POA + ∠OPC∴∠PCA = 2∠CBA + ∠CBA

= 3∠CBA 3

O

A B C

E

∠OBC = ∠OAB + ∠AOB (exterior angle of triangle AOB) ∠OBC = ∠OAB + ∠AOB (exterior angle of triangle AEB) ∠BAE = 1

2∠OAB

∠BEA = 12∠AOB (angles on arc AB)

∴∠EBC = 12 (∠OAB + ∠AOB)

= 12∠OBC

i.e. EB bisects ∠OBC.


4

A

B

PD

Q

∠PBD = ∠BAP (alternate segment)∠BAP = ∠BDQ (angles on arc BQ)

∴ PBD = ∠BDQ These are alternate angles on BP and QD. BP is parallel to QD. ∴ 5 a The base angle of the isosceles triangle

is 57° (alternate segment theorem) ∴ x = 180 – 57 – 57

= 66° b Make a construction as shown below.

64°x

y

y = 64° (alternate segment theorem)

x = 180 – 64 = 116° (cyclic quadrilateral)

c Make a construction as shown below.

48° y x

Marked angle = 180 – 48

2 = 66°x = 66°(alternate segment)y = 180 – 66 = 114° (cyclic quadrilateral)

6

Q

M

P

N

Consider triangles MNQ and NPM. ∠MQN = ∠NMP (alternate segment)

∠MNQ = ∠NPM (alternate segment) ∴ the triangles are similar and

MNNP

= QMMN

.

Cross multiplying gives MN2 = NP⋅QM 7 AE⋅EB = CE⋅ED

15 × 5 = 25EDED = DE

= 3 cm


Chapter 15 – Vectors Exercise 15A Solutions 1 a is the vector "1 across to the right

and 5 up." ⎣⎢⎡ 1

5⎦⎥⎤

b is the vector "2 down."

⎣⎢⎡ 0

–2⎦⎥⎤

c is the vector "1 across to the left

and 2 down." ⎣⎢⎡ –1

–2⎦⎥⎤

d is the vector "1 across to the left

and 3 up." ⎣⎢⎡ –4

3 ⎦⎥⎤

2 u =

⎣⎢⎡ 6 – 1

6 – 5⎦⎥⎤ =

⎣⎢⎡ 5

1⎦⎥⎤

a = 5, b = 1 3 v =

⎣⎢⎡ 2 – –1

–10 – 5⎦⎥⎤ =

⎣⎢⎡ –3

15⎦⎥⎤

a = 3, b = –15

4 a OA =

⎣⎢⎡ 1 – 0

–2 – 0⎦⎥⎤ =

⎣⎢⎡ 1

–2⎦⎥⎤

b AB =

⎣⎢⎡ 3 – 1

0 – –2⎦⎥⎤ =

⎣⎢⎡ 2

2⎦⎥⎤

c BC =

⎣⎢⎡ 2 – 3

–3 – 0⎦⎥⎤ =

⎣⎢⎡ –1

–3⎦⎥⎤

d CO = – OC =

⎣⎢⎡ –2

3 ⎦⎥⎤

e CB = – BC =

⎣⎢⎡ 1

3⎦⎥⎤

5 a

b

c


d

e

6 a

b

c

d

e

f

7 From the graphs above it can be seen

that a and c are parallel. 8 a & b

c i AB =

⎣⎢⎡ 1 – –1

4 – 0 ⎦⎥⎤ =

⎣⎢⎡ 2

4⎦⎥⎤

DC =⎣⎢⎡ 4 – 2

3 – –1⎦⎥⎤ =

⎣⎢⎡ 2

4⎦⎥⎤

∴AB = DC

ii BC =⎣⎢⎡ 4 – –1

3 – 4 ⎦⎥⎤ =

⎣⎢⎡ 3

–1⎦⎥⎤

AD =⎣⎢⎡ 2 – –1

–1 – 0⎦⎥⎤ =

⎣⎢⎡ 3

–1⎦⎥⎤

∴BC = AD

d ABCD is a parallelogram.


9 a i a + b =

⎣⎢⎡ 1

2⎦⎥⎤ +

⎣⎢⎡ 1

–3⎦⎥⎤

=⎣⎢⎡ 1 + 1

2 + –3⎦⎥⎤ =

⎣⎢⎡ 2

–1⎦⎥⎤

ii 2 c – a = 2 ×⎣⎢⎡ –2

1 ⎦⎥⎤ –

⎣⎢⎡ 1

2⎦⎥⎤

=⎣⎢⎡ –4 – 1

2 – 2 ⎦⎥⎤ =

⎣⎢⎡ –5

0 ⎦⎥⎤

iii a + b – c =⎣⎢⎡ 2

–1⎦⎥⎤ –

⎣⎢⎡ –2

1 ⎦⎥⎤

=⎣⎢⎡ 2 – –2

–1 – 1⎦⎥⎤ =

⎣⎢⎡ 4

–2⎦⎥⎤

b a + b =⎣⎢⎡ 2

–1⎦⎥⎤ = – c

∴ a + b is parallel to c.

10 m⎣⎢⎡ 3

–3⎦⎥⎤ + n

⎣⎢⎡ 2

4⎦⎥⎤ =

⎣⎢⎡ 3m

–3m⎦⎥⎤ +

⎣⎢⎡ 2n

4n⎦⎥⎤

=⎣⎢⎡ 3m + 2n

–3m + 4n⎦⎥⎤ =

⎣⎢⎡ –19

61 ⎦⎥⎤

3m + 2n = –196m + 4n = –38

–3m + 4n = 61 – :

9m = –99m = –11

–33 + 2n = –192n = –19 + 33

= 14n = 7

11 a i MD = MA + AD

= 12

BA + b

= – 12

AB + b

= b – 12

a

ii MN = MA + AD + DN = 1

2 BA + b + 1

2 DN

= – 12

AB + b + 12 DC

= – 12

a + b + 12 a

= b

b MN = AB (both are equal to b) 12 a CB = CA + AB

= – a + bMN = MA + AN

= – 12

b + 12

a

= – 12

( – a + b)

b MN is half the length of CB , is parallel to

CB and in the opposite direction to CB. 13 a CD = AF = a b ED = AB = b c The regular hexagon can be divided into

equilateral triangles, showing that BE = 2AF = 2a. d Likewise, FC = 2AB = 2b e FA = – AF = – a f FB = FA + AB

= – a + b = b – a

g FE = FA + AB + BE

= – a + b + 2a = a + b

14

a DC = AB = a b DA = – BC = – b c AC = AB + BC = a + b


d CA = – AC = – a – b b QS = QR + RS = v + w

e BD = BA + AD = – a + b = b – a

c PS = PQ + QR + RS

= u + v + w

15 17

a OB = OA + AB = u + v

AM = MB = 1

2 AB = 1

2 v

OM = OA + AM = u + 1

2 v

a BA = BO + OA = a – b b AB = – BA = b – a

PB = 13 AB = 1

3 (b – a)

c AP = 2

3 AB = 2

3 (b – a)

OP = OA + AP = a + 2

3 (b – a)

= 13

a + 23

b

= 13

(a + 2b)

b CM = CB + BM

= u + 12 BA

= u – 12

v

c CP = 2

3 CM

= 23

⎝⎜⎛u – 1

2 v⎠⎟⎞

= 23

u – 13

v

d PQ = – 1

3 OP

= – 13× 1

3 (a + 2b)

= – 19 (a + 2b)

d OP = OC + CP

= v +⎝⎜⎛2

3 u – 1

3 v⎠⎟⎞

= 23

u + 23 v

= 23

(u + v) = 23

AB

e BP = – PB = 1

3 (a – b)

BQ = BP + OA = 1

3 (a – b) + a

= 13

(4a – b)

Since OP is parallel to OB and they share a common point O, OP must be on the line OP. Hence P is on OB.

16

e Using the result from part d, OP : PB = 2 : 3.

a PR = PQ + QR = u + v


Exercise 15B Solutions 1 AB = (3i – 5j) – (i + 2j)

= 3i – 5i – i – 2j = 2i – 7j

2

a OP = OA + AP

= 5i + 6j

b AB = AO + OB

= –5i + 6j

c BA = – AB

= 5i – 6j

3

a 5i = 52 = 5

b –2j = (–2)2 = 2

c 3i + 4j = 32 + 42

= 9 + 16 = 5

d –5i + 12j = (–5)2 + 122

= 25 + 144 = 13

4 a u – v = (7i + 8j) – (2i – 4j)

= 7i + 8j – 2i + 4j = 5i + 12j

|u – v| = |5i + 12j| = 25 + 144 = 13

b xu + yv = x(7i + 8j) + y(2i – 4j) = 7xi + 8xj + 2yi – 4yj = 44j

7x + 2y = 014x + 4y = 0

8x – 4y = 44 + :

22x = 44x = 2

7 × 2 + 2y = 02y = –14

y = –7

5

AB = AO + OB

= – OA + OB = –10i + (4i + 5j) = –6i + 6j

AM = 12

AB

= –3i + 52 j

OM = OA + AM

= 10i +⎝⎜⎛–3i + 5

2 j⎠⎟⎞

= 7i + 52

j

6

a i OM = 1

5 OP

= 25

i

ii MQ = MO + OQ = – OM + OQ = – 2

5 i + j


iii MN = 16

MQ

= 16

⎝⎜⎛ – 2

5 i + j

⎠⎟⎞

= – 115

i + 16 j

iv ON = OM + MN

= 25 i +

⎝⎜⎛ – 1

15i+ 1

6 j⎠⎟⎞

= 13 i + 1

6 j

v OA = OP + PA = 2i + j

b i ON = 1

3 i + 1

6 j

= 16

(2i + j)

= 16

OA

Since ON is parallel to OA and they share a common point O, ON must be on the line OA. Hence N is on OA.

ii 1 : 5 7 OA =

⎣⎢⎡ 1

3⎦⎥⎤ = i + 3j

OB =⎣⎢⎡ 5

–1⎦⎥⎤ = 5i – j

AB = – OA + OB = – i – 3j + 5i – j = 4i – 4j

⎪⎪⎪AB⎪

⎪⎪ = 42 + (–4)2

= 16 + 16 = 32 = 4 2 units

8 a 2i + 3j = 2li + 2kj

2j = 2l = 1

2k = 3k = 2

3

b x – 1 = 5

x = 6y = x – 4 = 2

c x + y = 6x – y = 0

+ :2x = 6x = 3

3 + y = 6y = 3

d k = 3 + 2l

k = –2 – l3 + 2l = –2 – l

3l = –5l = – 5

3k = –2 – – 5

3 = –2 + 5

3 = – 1

3

9 a AB =

⎣⎢⎡ 5 – 2

1 – 3⎦⎥⎤

=⎣⎢⎡ 3

–2⎦⎥⎤

= 3i – 2j

b ⎪⎪⎪AB⎪

⎪⎪ = 32 + (–2)2

= 9 + 4 = 13

10 a AB = i + 4j – 3i

= –2i + 4j

b AC = –3i + j – 3i

= –6i + j

c BC = AC – AB

= –6i + j – (–2i + 4j) = –4i – 3j

⎪⎪⎪BC⎪

⎪⎪ = (–4)2 + (–3)2

= 16 + 9 = 5


11 a Let D = (a, b). AB = –5i + 3j

CD = (a + 1)i + bja + 1 = –5

a = –6b = 3

D is (–6, 3). b Let F = (c, d). BC = – i – 4j

AF = (c – 5)i + (d – 1)jc – 5 = –1

c = 4d – 1 = –4

d = –3

F is (4, –3). c Let G = (e, f).

AB = –5i + 3j2GC = 2(–1 – e)i + 2( – f)j

2(–1 – e) = –5e = 3

2–2f = 3

f = – 32

G is ⎝⎜⎛3

2, – 3

2⎠⎟⎞ .

12 OA = – AO

= – i – 4j

A is (–1, –4).

B is (–2, 2).

BC = OC – OBOC = BC + OB

= 2i + 8j + (–2i + 2j) = 10j

C is (0, 10)

13

a i 2i – j ii –5i + 4j iii i + 7j iv 6i + 3j v AD = BC

= 6i + 3j

b AD = OD – OA

OD = AD + OA = 6i + 3j + 2i – j = 8i + 2j

D is (8, 2). 14 a OP = 12i + 5j

PQ = OQ – OP = 18i + 13j – 12i – 5j = 6i + 8j

b ⎪

⎪⎪RQ⎪

⎪⎪ = ⎪

⎪⎪OP⎪

⎪⎪

= 122 + 52

= 13

⎪⎪⎪OR⎪

⎪⎪ = ⎪

⎪⎪PQ⎪

⎪⎪

= 62 + 82

= 10

15 a i ⎪

⎪⎪AB⎪

⎪⎪ = |2i – 5j|

= 22 + 52 = 29

ii ⎪⎪⎪BC⎪

⎪⎪ = |10i + 4j|

= 102 + 42

= 116 = 2 29

iii ⎪⎪⎪CA⎪

⎪⎪ = |12i – j|

= 122 + 12 = 145

b AB2 + BC2 = 29 + 116

= 145 = AC2

∴ ABC is a right-angled triangle.


16 17 a i OA = –3i + 2j a i AB = – i – 3j

ii ii OB = 7j BC = 4i + 2j iii CA iii BA = –3i – 5j = –3i + j

iv BM = 12

BA

= 12

(–3i – 5j)

b i ⎪⎪⎪AB⎪

⎪⎪ = 12 + 32 = 10

ii ⎪⎪⎪BC⎪

⎪⎪ = 42 + 22

b OM = OB + BMOD = 7j + – 3

2 i – 5

2 j

= – 32

i + 92 j

M =⎝⎜⎛ – 3

2, – 5

2⎠⎟⎞

= 20 = 2 5

iii ⎪⎪⎪CA⎪

⎪⎪ = 32 + 12

=

10 c AB = CA

= 10AB2 + CA2 = 10 + 10

= 20 = BC2

is an isosceles right-angled triangle. ∴ ABC


Exercise 15C Solutions 1 a a – b = (i + j + 2k) – (2i – j + 3k)

= – i + 2j – k

b 3b – 2a + c = 3(2i – j + 3k)

– 2(i + j + 2k) + ( – i + k) = 6i – 3j + 9k – 2i – 2j

– 4k – i + k = 3i – 5j + 6k

c |b| = 22 + (–1)2 + 32

= 4 + 1 + 9 = 14

d |b + c| = |(2i – j + 3k) + ( – i + k)|

= |i – j + 4k|

= 12 + (–1)2 + 42

= 18 = 3 2

e 3(a – b) + 2c = 3((i + j + 2k) – (2i – j + 3k))

+ 2( – i + k) = 3( – i + 2j – k) – 2i + 2k = –3i + 6j – 3k – 2i + 2k = –5i + 6j – k

2

a i |a| = 32 + 12 + 12

= 11a = 1

11 (3i + j – k)

= 311

i + 111

j – 111

k

ii –2a = – 611

i – 211

j + 211

k

b 5a = 15

11 i + 5

11 j – 5

11 k

3 |a| = 12 + 12 + 52

= 27 = 3 3

|b| = 22 + 12 + 32

= 14c = |a|

|b| a

= 143 3

(i – j + 5k)

= 429

(i – j + 5k)

4 a PQ = i – 3j

b ⎪⎪⎪PQ⎪

⎪⎪ = 12 + 32 + 02

= 10

c OM = OP + PM

= OP + 12 PQ

= i + 2j – k + 12 i – 3

2 j

= 32

i + 12

j – k

5 a OB = OA + OC

= 2j + 2k

b OE = OA + OD

= i + 2j

c OG = OC + OD

= i + 2k

d OF = OA + OC + OD

= i + 2j + 2k

e ED = – OA

= –2j

f EG = – OA + OC

= –2j + 2k


g CE = – OC + OA + OD

= i + 2j – 2k

h BD = – OC – OA + OD

= i – 2j – 2k

6 a OE = OA + AE

= i + 3jOM = 1

3 OE

= 13

i + j

BF = OD = i

BN = 12

BF

= 12

i

ON = OC + CB + BN = 1

2 i + 3j + 2k

MN = ON – OM

= 12

i + 3j + 2k –⎝⎜⎛1

3 i + j

⎠⎟⎞

= 16

i + 2j + 2k

b ⎪⎪⎪MN⎪

⎪⎪ =

⎝⎜⎛1

6⎠⎟⎞2

+ 22 + 22

= 1 + 144 + 14436

= 28936


Exercise 15D Solutions 1 a i OR = 4

5 OP

= 45

p

ii RP = 15 OP

= 15 p

iii PQ = q – p iv PS = 1

5 PQ

= 15 (q – p)

v RS = RP + PS = 1

5 p + 1

5 (q – p)

= 15

q

b They are parallel (and OQ = 5RS). c A trapezium (one pair of parallel lines). d The area of triangle POQ is 25 times the

area of PRS = 125 cm2. ∴ area of ORSQ = 125 – 5 = 120 cm2

2 a AP = 2

3 AB and CQ = 6

7 CB.

OP = OA + AP

= OA + 23

AB

= a + 23 (b – a)

= 13

a + 23

b

OQ = OC + CQ = OC + 6

7 CB

= ka + 67

(b – ka)

= k7

a + 67

b

b OPQ is a straight line if OP = nOQ.

13

a + 23

b = n⎝⎜⎛ k

7 a + 6

7 b⎠⎟⎞

= nk7

a + 6n7

b

23

= 6n7

n = 1418

= 79

13

a + 23

b = 79

⎝⎜⎛ k

7 a + 6

7 b⎠⎟⎞

= k9

a + 23

b

k9

= 13

k = 3

3 a i OD = 1

3 OB

= 13

(6i – 1.5j)

= 2i – 0.5jAB = 3i – 6jAE = 1

4 (3i – 5j)

= 0.75i – 1.25jOE = OA + AE

= 3i + 3.5j + 0.75i – 1.25j = 3.75i + 2.25j = 15

4 i + 9

4 j

ii ED = 2i – 0.5j –⎝⎜⎛15

4 i + 9

4 j⎠⎟⎞

= – 64

i – 114

j

⎪⎪⎪ED⎪

⎪⎪ =

⎝⎜⎛7

4⎠⎟⎞2

+⎝⎜⎛11

4 ⎠⎟⎞2

= 49 + 12116

= 17016

= 1704


b i OX = 15p4

i + 9p4

j

ii AD = 2i – 0.5j – (3i + 3.5j) = – i – 4j

XD = – qi – 4qjOD = OX + ODOX = OD – XD

= 2i = 0.5j – ( – qi – 4qj) = (q + 2)i + (4q – 0.5)j

c (q + 2)i + (4q – 0.5)j = 15p4

i + 9p4

j

q + 2 = 15p4

4q + 8 = 15p4q – 0.5 = 9p

4 – : 8.5 = 51p

4

p = 8.5 × 451

= 23

q + 2 = 15p4

= 104

= 52

q = 12

4 a PQ = q – p

= PM + MQ

MQ = βα

PM

∴ PQ = PM + βα

PM

= α + βα

PM

PM = αα + β

PQ

OM = OP + PM

= p + αα + ba

(q – p)

= α + βα + β

p + αα + ba

(q – p)

= α + β – αα + β

p + αα + β

q

= βp + αqα + β

b i

It can be seen from the parallelogram formed by adding a and b that a + will lie on the bisector of angle POQ.

Hence any multiple, will also lie on this bisector.

(λ +a b),

ii If p = ka and q = lb, then

OM = βp + αqα + β

= βka + αlban + β

If M is the bisector of , ∠POQ OM = λa + λb

∴ αl = βk Divide both sides by βl:

αβ

= kl

5

a s = OS

= OR + RS = OR + OT = r + t

b ST = OT – OS

= t – sv = OV = OS + SV = OS + 1

2 ST

= s – 12

(t – s)

= 12

(s + t)


c Similarly:

u = OU = OS + SU = OS + 1

2 SR

= s – 12

(r – s)

= 12 (s + r)

∴ u + v = 12

(s + r) + 12 (s + t)

= 12

(2s + r + t)

2u + 2v = 2s + r + t We may also express u as

u = OR + RU = OR + 1

2 RS

= OR + 12 OT

= r + 12 t

∴ u + v = 1 + 12

t + 12

(s + t)

= 12

(s + 2r + 2t)

2u + 2v = s + 2r + 2t

Add the two expressions for 2u + 2v:

4u + 4v = 3s + 3r + 3t

= 3(s + r + t)


Solutions to Multiple-choice Questions 1 v =

⎣⎢⎡ 3 – 1

5 – 1⎦⎥⎤ =

⎣⎢⎡ 2

4⎦⎥⎤

a = 2, b = 4 C 2 CB = CA + AB

= – AC + AB

= u – v C 3 a + b =

⎣⎢⎡ 1 + 2

–2 + 3⎦⎥⎤

E =⎣⎢⎡ 3

1⎦⎥⎤

4 2a – 3b = 2⎣⎢⎡ 3

–2⎦⎥⎤ – 3

⎣⎢⎡–1

3 ⎦⎥⎤

=⎣⎢⎡ 6 – –3

–4 – 9⎦⎥⎤

= A ⎣⎢⎡ 9

–13⎦⎥⎤

5

SQ = SR + RQ

= PQ + – QR

= p – q B

6 |3i – 5j| = 32 + (–5)2

= 9 + 25

= 34 B 7 AB = – OA + OB

= (i – 2j) – (2i + 3j)

= –i – 5j A 8 ⎪

⎪⎪AB⎪

⎪⎪ = | – i – 5j|

= (–1)2 + (–5)2

= 1 + 25

= 26 C

9 |a| = 22 + 32

= 13

a = 113

(2i + 3j) D

10 |a| = 32 + 12 + 32

= 19

a = 119

(–3i + j + 3k) C


Solutions to Short-answer Questions 1 a a is parallel to b if a = kb, where k is a constant. 7i + 6j = k(2i + xj)

2k = 7k = 7

2kx = 67x2

= 6

x = 127

b |a| = 72 + 62

= 85

|b| = 22 + x2

= |a| = 85∴ x2 + 4 = 85

x2 = 81x = ± 9

2

A = (2, –1) OB = OA + AB

= 5i + 3j

B = (5, 3) AC = AB + BC

= AB + AD = i + 9j

OC = OA + AC = 2i – j + i + 9j = 3i + 8j

C = (3, 8)OD = OA + AD

= 4jD = (0, 4)

3 a + pb + qc = (2 + 2p – q)i + (–3 – 4p – 4q)j + (1 + 5p + 2q)k

To be parallel to the x-axis,

a + pb + qc = ki1 + 5p + 2q = 0

2 + 10p + 4q = 0–3 – 4p – 4q = 0

+ :–1 + 6p = 0

p = 16

1 + 56

+ 2q = 0

2q = – 116

q = – 1112

4 a PQ = (3i – 7j + 12k) – (2i – 2j + 4k)

= i – 5j + 8k

⎪⎪⎪PQ⎪

⎪⎪ = 12 + 52 + 82

= 90 = 3 10

b 13 10

(i – 5j + 8k)

5 AB = 4i + 8j + 16kAC = xi + 12j + 24k

For A, B and C to be collinear, we need AC = kAB . xi + 12j + 24k = k(4i + 8j + 16k)

8k = 12k = 1.5x = 4k = 6

6

a OA = 42 + 32

= 5

Unit vector = 15 (4i + 3j)


b OC = 165

OA

= 165× 1

5 (4i + 3j)

= 1625

(4i + 3j)

7 a i SQ = b + a = a + b ii TQ = 1

3 SQ

= 13

(a + b)

iii RQ = –2a + b + a = b – a iv PT = PQ + QT

= PQ – TQ = a – 1

3 (a + b)

= 13

(2a – b)

v TR = TQ + QR = TQ – RQ = 1

3 (a + b) – (b – a)

= 13

(4a – 2b)

= 23

(2a – b)

b 2PT = TR P, T and R are collinear. 8 a = b a i – sj = 2j

s = –2

ii 5 i = tit = 5

iii 2 k = uku = 2

b a = 52 + 22 + 22

= 25 + 4 + 4 = 33

9

Use the cosine rule. |p + q|2 = 72 + 122 – 2 × 7 × 12 × cos 60°

= 109|p + q| = 109

10 a a + 2b = (5i + 2j + k) + 2 × (3i – 2j + k)

= 11i – 2j + 3k

b |a| = 52 + 22 + 12

= 30

c a = 1

30 (5i + 2j + k)

d a – b = (5i + 2j + k) – (3i – 2j + k)

= 2i + 4j

11 a OC = OA – OB

= (3i + 4j) – (4i – 6j) = – i + 10j

C = (–1, 10)

b i + 24j = h(3i + 4j) + k(4i – 6j)

3h + 4k = 14h – 6k = 24

Multiply the first equation by 3 and the second equation by 2.

9h + 12k = 38h – 12k = 48

+ :17h = 51

h = 39 + 4k = 1

k = –2


13 12 mp + nq = 3mi + 7mj + 2ni – 5nj = 8i + 9j

3m + 2n = 87m – 5n = 9

a


15m + 10n = 4014m – 10n = 18

+ :29m = 58

m = 26 + 2n = 8

n = 1

b = OB = OA + AB = OA + OC = a + c

b AB = b – a

BC = c – bAB :BC = 3 : 2

ABBC

= 32

2AB = 3BC2(b – a) = 3(c – b)2b – 2a = 3c – 3b

5b = 2a + 3cb = 2

5 a + 3

5 c


Chapter 16 – Polar coordinates and complex numbers Exercise 16A Solutions 1 a

b

c

d

e

f

g

h

2 a

b

c

d


3 a The point is in the first quadrant.

r = 42 + 42

= 32 = 4 2cos θ = 4

4 2= 1

2

θ = π4

= 45°

∴(4, 4) =⎣⎢⎡4 2 ,π

4 ⎦⎥⎤

= [4 2 , 45°]

b The point is in the fourth quadrant.

r = 12 + ( – 3 )2

= 4 = 2cos θ = 1

2

θ = – π3

= –60°

∴ (1, – 3 ) =⎣⎢⎡ 2,π

3 ⎦⎥⎤

= [2, – 60°]

c The point is in the fourth quadrant.

r = (2 3 )2 + (–2)2

= 12 + 4 = 4

cos θ = 2 34

= 32

θ = – π6

= –30°

∴ (2 3 , – 2) =⎣⎢⎡4, – π

6 ⎦⎥⎤

= [4, – 30°]

d The point is in the first quadrant.

r = (–5)2 + 122

= 169 = 13cos θ = – 5

12θ ≈ 113°

∴ (–5, 12) =⎣⎢⎡ 13, cos–1

⎝⎜⎛ – 5

13⎠⎟⎞⎦⎥⎤

≈ [13, 113°]

e The point is in the fourth quadrant.

r = 62 + (–5)2

= 36 + 25 = 61sin θ = – 5

61 θ ≈ – 40°

∴ (6, – 5) =⎣⎢⎡ 61 , sin–1

⎝⎜⎛ – 5

61 ⎠⎟⎞⎦⎥⎤

≈ [ 61 , – 40°]

f The point is in the third quadrant.

r = ( 3 )2 + 12

= 3 + 1 = 4

cos θ = 32

θ = π6

= 30°

∴ ( 3 , 1) =⎣⎢⎡ 2,π

6 ⎦⎥⎤

= [2, 30°]

g The point is in the second quadrant.

r = 52 + 122

= 169 = 13cos θ = – 5

13

θ = – π + cos–1 513

≈ – 113°

∴ (–5, – 12) =⎣⎢⎡ 2, – π + cos–1 5

13⎦⎥⎤

≈ [2, – 113°]

h The point is in the first quadrant.

r = 42 + 32

= 25 = 5cos θ = 4

5θ ≈ 37°

∴ (4, 3) =⎣⎢⎡ 5, cos–1 4

5⎦⎥⎤

≈ [5, 37°]


4 a x = –2 cos 30°

= –2 × 32

= – 3

y = –2 sin 30°

= –2 × 12

= –1

∴ [–2, 30°] = ( – 3 , – 1)

b x = –4 cos π2

= –4 × 0 = 0

y = –4 sin π2

= –4 × 1 = 4

∴ ⎣⎢⎡ –4,π

2 ⎦⎥⎤ = (0, 4)

c x = –1 cos 5π4

= –1 × – 12

= 12

y = –1 sin 5π4

= –1 × 12

= – 12

∴ ⎣⎢⎡–1, 5π

4 ⎦⎥⎤ =

⎝⎜⎛ 1

2, – 1

2 ⎠⎟⎞

d x = 4 cos – 2π

= 4 × 1 = 4y = 4 sin – 2π = 4 × 0 = 0

∴ [4, – 2π] = (4, 0)

e x = 2 cos – 7π6

= 2 × – 32

= – 3

y = 2 sin – 7π6

= 2 × 12

= 1

∴ ⎣⎢⎡2, – 7π

6 ⎦⎥⎤ = ( – 3 , 1)

f x = 5 cos 240°

= 5 × – 12

= – 52

y = 5 sin 240°

= 5 × – 32

= – 5 32

∴ [5, 240°] =⎝⎜⎛ – 5

2, – 5 3

2 ⎠⎟⎞

g x = 2 cos 180°

= 2 × –1 = –2y = 2 sin 180° = 2 × 0 = 0

∴ [2, 180°] = (–2, 0) h x = 1 cos – 120°

= 1 × – 12

= – 12

y = 1 sin – 120°

= 1 × – 32

= – 32

∴ [1, – 120°] =⎝⎜⎛ – 1

2, – 3

2 ⎠⎟⎞

5 Use your calculator to plot these graphs. a

b

c


d

e

f

g

h

i

j

k

l

6 a r2 = x2 + y2

= 16r = 4

b x + y = 6

r cos θ + r sin θ = 6r (cos θ + sin θ) = 6

r = 6cos θ + sin θ

c x2 = yr2 cos2θ = r sin θr cos2θ = sin θ

r = sin θcos2θ


e r = a1 + cos θ

r + r cos θ = a

x2 + y2 + x = a

x2 + y2 = a – xx2 + y2 = a2 – 2ax + x2

y2 = a2 – 2ax = a(a – 2x)

d x2

4+ y2 = 1

r2cos2θ4

+ r2sin2θ = 1

r2cos2θ + 4r2sin2θ = 4r2(cos2θ + sin2θ + 3sin2θ) = 4

r2(1 + 3 sin2θ) = 4r2 = 4

1 + 3 sin2θ

f r = a1 + sin θ

r + r sin θ = a

x2 + y2 + y = a

x2 + y2 = a – yx2 + y2 = a2 – 2ay + y2

x2 = a2 – 2ay2ay = a2 – x2

y = 12a

(a2 – x2)

7 a r = 2

x2 + y2 = 2x2 + y2 = 4

b r = a(1 + cos θ)

r2 = a(r + r cos θ)

x2 + y2 = a x2 + y2 + ax

x2 + y2 – ax = a x2 + y2

c r = a cos θ

r2 = ar cos θx2 + y2 = ax

d r = 2a(1 + sin 2θ)

= 2a(1 + 2 sin θ cos θ)r3 = 2a(r2 + 2(r sin θ)(r cos θ))

(x2 + y2)32 = 2a(x2 + y2 + 2xy) = 2a(x + y)2


Exercise 16B Solutions 1 a Re(z) = a = 2 Im(z) = b = 3 b Re(z) = a = 4 Im(z) = b = 5 c Re(z) = a = 1

2

Im(z) = b = – 32

d Re(z) = a = –4 Im(z) = b = 0 e Re(z) = a = 0 Im(z) = b = 3 f Re(z) = a = 2 Im(z) = b = –2 2 2 a 2a – 3bi = 4 + 6i

2a = 4a = 2

–3bi = 6ib = –2

b a + b = 5

b = 5 – a–2ab = –12

ab = 6a(5 – a) = 65a – a2 = 6

a2 + 5a + 6 = 0(a – 2)(a – 3) = 0

When a = 2b = 5 – 2 = 3

When a = 3b = 5 – 3 = 2

c 2a + bi = 10

= 10 + 0i2a = 10a = 5b = 0

d 3a = 2a = 2

3a – b = 123

– b = 1

b = 23

– 1 = – 13

3 a (2 – 3i) + (4 – 5i) = 2 + 4 – 3i – 5i

= 6 – 8i b (4 + i) + (2 – 2i) = 4 + 2 + i – 2i

= 6 – i c (–3 – i) – (3 + i) = –3 – 3 – i – i

= –6 – 2i d (2 – 2 i ) + (5 – 8 i ) = 2 + 5 – 2 i – 8 i

= 7 – 2 i – 2 2 i = 7 – 3 2 i

e (1 – i) – (2i + 3) = 1 – 3 – i – 2i

= –2 – 3i f (2 + i) – (–2 – i) = 2 + 2 + i + i

= 4 + 2i g 4(2 – 3i) – (2 – 8i) = 8 – 2 – 12i + 8i

= 6 – 4i h – (5 – 4i) + (1 + 2i) = –5 + 1 + 4i + 2i

= –4 + 6i i 5(i + 4) + 3(2i – 7) = 20 – 21 + 5i + 6i

= –1 + 11i j 1

2 (4 – 3i) – 3

2 (2 – i) = 2 – 3 – 3

2 i + 3

2 i

= –1


4 5 a i(2 – i) = 2i – i2

= 2i – –1 = 1 + 2i

a –16 = 16 × –1 = 4i

b 2 –9 = 2 9 × –1

= 6i

b i2(3 – 4i) = –1(3 – 4i) = –3 + 4i

c –2 = 2 × –1 = 2 i

c 2 i (i – 2 ) = 2 i 2 – 2i

= – 2 – 2i

d i3 = i2 × i

= – i d – 3 ( –3 + 2 ) = – 3 ( 3 i + 2 ) = –3i – 6 = – 6 – 3i

e i14 = i4 × 3 + 2

= –1 f i20 = i4 × 5

= 1 g –2i × i3 = –2i4

= –2

h 4i4 × 3i2 = 4 × 3 × i4 × i2

= 12i6

= –12 i 8 i 5 × –2 = 8 i 4 × i × 2 i

= 16 × 1 × –1 = –4


Exercise 16C Solutions 1 i2 = –1a (4 + i)2 = 16 + 8i + i2

= 15 + 8i b (2 – 2i)2 = 4 – 8i + 4i2

= –8i c (3 + 2i)(2 + 4i) = 6 + 12i + 4i + 8i2

= –2 + 16i d (–1 – i)2 = 1 + 2i + i2

= 2i e ( 2 – 3 i )( 2 + 3 i ) = 2 – 3i2

= 2 + 3 = 5

f (5 – 2i)(–2 + 3i) = –10 + 15i + 4i – 6i2

= –4 + 19i 2 z = a – bi a z = 2 – 5i

z = 2 + 5i

b z = –1 + 3i

z = –1 – 3i

c z = 5 – 2i

z = 5 + 2i

d z = 0 – 5i

z = 0 + 5i = 5i

3 a z1 = 2 + i b z2 = –3 – 2i c z1 z2 = (2 – i)(–3 + 2i)

= –6 + 4i + 3i – 2i2

= –4 + 7i

d z1 z2 = –4 – 7i

e z1 z2 = (2 + i)(–3 – 2i) = –6 – 4i – 3i – 2i2

= –4 – 7i

f z1 + z2 = (2 – i) + (–3 + 2i)

= –1 + i

g z1 + z2 = –1 – i h z1 + z2 = (2 + i) + (–3 – 2i)

= –1 – i

4 z = 2 – 4ia z = 2 + 4i b zz = (2 – 4i)(2 + 4i)

= 4 – 16i2

= 20

c z + z = (2 – 4i) + (2 + 4i)

= 4

d z(z + z) = 4z

= 8 – 16i

e z – z = (2 – 4i) – (2 + 4i)

= –8i

f i(z – z) = i × –8i

= –8i2 = 16

g z–1 = 12 – 4i

= 12 – 4i

× 2 + 4i2 + 4i

= 2 + 4i4 – 16i2

= 2 + 4i20

= 110

(1 + 2i)

h z

i= z

i× i

i = i(2 – 4i)

–1 = –1 × (2i – 4i2) = –4 – 2i


5 (a + bi)(2 + 5i) = 2a + 5ai + 2bi – 5b = 3 – i

2a – 5b = 35a + 2b = –1


4a – 10b = 625a + 10b = –5

+ :29a = 1

a = 129

229

– 5b = 3

5b = 229

– 3

= – 8529

b = – 1729

6 a 2 – i

4 + 1= 2 – i

4 + 1× 4 – i

4 – i

= 8 – 2i – 4i + i2

16 – i2

= 7 – 6i17

= 717

– 617

i

b 3 + 2i

2 – 3i= 3 + 2i

2 – 3i× 2 + 3i

2 + 3i

= 6 + 9i + 4i + 6i2

4 – 9i2

= 13i13

= i

c 4 + 3i

1 + i= 4 + 3i

1 + i× 1 – i

1 – i

= 4 – 4i + 3i – 3i2

1 – i2

= 7 – i2

= 72

– 12

i

d 2 – 2i4i

= 2 – 2i4i

× ii

= 2i – 2i2

–4 = 2 + 2i

–4 = –1 – i

2 = – 1

2– 1

2 i

e 1

2 – 3i= 1

2 – 3i× 2 + 3i

2 + 3i = 2 + 3i

4 – 9i2

= 2 + 3i13

= 213

+ 313

i

f i

2 + 6i= i

2 + 6i× 2 – 6i

2 – 6i = 2i + 6

4 – 36i2

= 2i + 640

= 320

+ 120

i

7 (3 – i)(a + bi) = 3a + 3bi – ai + b

= 6 – 7i3a + b = 6

– a + 3b = –7–3a + 9b = –21

+ :10b = –15

b = – 32

3a – 32

= 6

3a = 6 + 32

= 152

a = 52


d z = 5 + 2i2(4 – 7i)

= 5 + 2i2(4 – 7i)

× 4 + 7i4 + 7i

= 20 + 35i + 8i + 14i2

2(16 – 49i2) = 6 + 43i

130 = 1

130 (6 + 43i)

8 a z = 4 + 2i

2 – i = 4 + 2i

2 – i× 2 + i

2 + i

= 8 + 4i + 4i + 2i2

4 – i2

= 6 + 8i5

= 25

(3 + 4i)

e z = 4

1 + i = 4

1 + i× 1 – i

1 – i = 4 – 4i

1 – i2

= 4 – 4i2

= 2 – 2i

b z = –2 – i

1 + 3i = –2 – i

1 + 3i× 1 – 3i

1 – 3i

= –2 + 6i – i + 3i2

1 – 9i2

= –5 + 5i10

= – 12

(1 – i)

c z = 1 + i5 + 3i

= 1 + i5 + 3i

× 5 – 3i5 – 3i

= 5 – 3i + 5i – 3i2

25 – 9i2

= 8 + 2i34

= 117

(4 + i)


Exercise 16D Solutions 1 A: 3 + i B: 2i C: –3 – 4i D: 2 – 2i E: –3i F: –1 – i 2

3 a z1 + z2 = (6 – 5i) + (–3 + 4i)

= 3 – i

b z1 – z2 = (6 – 5i) – (–3 + 4i)

= 9 – 9i

4 a z = 1 + 3i b z = 1 – 3i c z2 = 1 + 6i + 9i2

= –8 + 6i

d – z = – (1 + 3i) = –1 – 3i

e 1z

= 11 + 3i

= 11 + 3i

× 1 – 3i1 – 3i

= 1 – 3i1 + 9i2

= 110

– 310

i

5 a z = 2 – 5i b zi = i(2 – 5i)

= 2i – 5i2

= –5 + 2i

c zi2 = – z = –2 + 5i d zi3 = – iz

= – (–5 + 2i) = 5 – 2i

e zi4 = z = 2 – 5i


Exercise 16E Solutions 1 a z2 + 4 = 0

z2 – 4i2 = 0(z – 2i)(z + 2i) = 0

z = ± 2i b 2x2 + 18 = 0

z2 + 9 = 0z2 – 9i2 = 0

(z – 3i)(z + 3i) = 0z = ± 3i

c 3z2 + 15 = 0

z2 + 5 = 0z2 – 5i2 = 0

(z – 5 i )(z + 5 i ) = 0z = ± 5 i

d (z – 2)2 = –16

z – 2 = ± 4iz = 2 ± 4i

e (z + 1)2 = –49

z + 1 = ± 7iz = –1 ± 7i

f Complete the square. z2 – 2z + 1 + 2 = 0

(z – 1)2 – 2i2 = 0(z – 1 – 2 i )(z – 1 + 2 i ) = 0

z = 1 ± 2 i

g Use the quadratic formula.

z = –3 ± 9 – 122

= –3 ± –32

= 12

(–3 ± 3 i )

h Use the quadratic formula.

z = –5 ± 25 – 324

= –5 ± –74

= 14

(–5 ± 7 i )

i Use the quadratic formula. 3z2 – z + 2 = 0

z = 1 ± 1 – 246

= 1 ± –236

= 16

(1 ± 23 i )

j Complete the square. z2 – 2z + 5 = 0

z2 – 2z + 1 + 4 = 0(z – 1)2 – 4i2 = 0

(z – 1 – 2i)(z – 1 + 2i) = 0z = 1 ± 2i

k Use the quadratic formula. 2z2 – 6z + 10 = 0

z2 – 3z + 5 = 0

z = 3 ± 9 – 202

= 3 ± –112

= 12 (3 ± 11 i )

l Complete the square. z2 – 6z + 14 = 0

z2 – 6z + 9 + 5 = 0(z – 3)2 – 5i2 = 0

(z – 3 – 5 i )(z – 3 + 5 i ) = 0z = 3 ± 5 i


Exercise 16F Solutions 1 a The point is in the first quadrant.

r = 12 + ( 3 )2

= 1 + 3 = 2cos θ = 1

2

θ = π3

∴ 1 + 3 i = 2 cis π3


r = 12 + 12

= 2cos θ = 1

2

θ = – π4

∴ 1 – i = 2 cis – π4

c The point is in the second quadrant.

r = (2 3 )2 + 22

= 16 = 4

cos θ = –2 34

= – 32

θ = π – π6

= 5π6

∴ – 2 3 + 2i = 4 cis 5π6

d The point is in the third quadrant.

r = 42 + 42

= 32 = 4 2cos θ = – 4

4 2= – 1

2

θ = – π + π4

= – 3π4

∴ – 4 – 4i = 4 2 cis – 3π4

e The point is in the fourth quadrant.

r = 122 + 122 × 3 = 4 × 144 = 24

cos θ = – 1224

= – 12

θ = – π3

∴ 12 – 12 3 i = 24 cis – π3

f The point is in the second quadrant.

r =⎝⎜⎛1

2⎠⎟⎞2

+⎝⎜⎛1

2⎠⎟⎞2

= 12

= 12

cos θ = – 12÷ 1

2 = – 1

2× 2 = – 1

2

θ = π – π4

= 3π4

∴ – 12

+ 12

i = 12

cis 3π4

2

a 3 cis π2

= 3 cos π2

+ 3i sin π2

= 3i

b 2 cis π3

= 2 cos π3

+ 2 i sin π3

= 22

+ 62

i

= 22

(1 + 3 i )

c 2 cis π6

= 2 cos π6

+ 2i sin π6

= 3 + i


d 5 cis 3π4

= 5 cos 3π4

+ 5i sin 3π4

= – 52

+ 52

i

= – 52

(1 – i)

e 12 cis 5π6

= 12 cos 5π6

+ 12i sin 5π6

= –6 3 + 6i = –6( 3 – i)

f 3 2 cis – π4

= 3 2 cos – π4

+ 3 2 i sin – π4

= 3 – 3i = 3(1 – i)

g 5 cis 4π3

= 5 cos 4π3

+ 5i sin 4π3

= – 52

– 5 32

i

= – 52

(1 + 3 i )

h 5 cis – 2π3

= 5 cos – 2π3

+ 5i sin – 2π3

= – 52

– 5 32

i

= – 52 (1 + 3 i )

3 z1 z2 = r 1 r2 cis(θ1 + θ2)

a ⎝⎜⎛ 2 cis π

6 ⎠⎟⎞⋅⎝⎜⎛ 3 cis π

12⎠⎟⎞ = 6 cis

⎝⎜⎛ π

6+ π

12⎠⎟⎞

= 6 cis π4

= 6 cos π4

+ 6i sin π4

= 62

+ 62

i

= 3 2 (1 + i)

b ⎝⎜⎛ 4 cis π

12⎠⎟⎞ ⋅⎝⎜⎛ 3 cis π

4 ⎠⎟⎞ = 12 cis

⎝⎜⎛ π

12+ π

4 ⎠⎟⎞

= 12 cis π3

= 12 cos π3

+ 12i sin π3

= 6 + 6 3 i = 6(1 + 3 i )

c ⎝⎜⎛ cis π

4 ⎠⎟⎞ ⋅⎝⎜⎛ 5 cis 5π

12 ⎠⎟⎞ = 5 cis

⎝⎜⎛ π

4+ 5π

12 ⎠⎟⎞

= 5 cis 2π3

= 5 cos 2π3

+ 5i sin 2π3

= – 52

+ 5 32

i

= – 52

(1 – 3 i )

d ⎝⎜⎛ 12 cis – π

3 ⎠⎟⎞⋅⎝⎜⎛ 3 cis 2π

3 ⎠⎟⎞ = 36 cis

⎝⎜⎛ – π

3+ 2π

3 ⎠⎟⎞

= 36 cis π3

= 36 cos π3

+ 36i sin π3

= 18 + 18 3 i = 18(1 + 3 i )

e ⎝⎜⎛ 12 cis 5π

6 ⎠⎟⎞ ⋅⎝⎜⎛ 3 cis π

2 ⎠⎟⎞ = 36 cis

⎝⎜⎛5π

6+ π

2 ⎠⎟⎞

= 36 cis 4π3

= 36 cos 4π3

+ 36i sin 4π3

= –18 – 18 3 i = –18(1 + 3 i )

f ( 2 cis π)⋅⎝⎜⎛ 3 cis – 3π

4 ⎠⎟⎞ = 6 cis

⎝⎜⎛ π – 3π

4 ⎠⎟⎞

= 6 cis π4

= 6 cos π4

+ 6 i sin π4

= 3 + 3 i = 3 (1 + i)


g 10 cis π

4

5 cis π12

= 105

cis ⎝⎜⎛ π

4– π

12⎠⎟⎞

= 2 cis π6

= 2 cos π6

+ 2i sin π6

= 3 + i

h 12 cis – π

3

3 cis 2π3

= 123

cis ⎝⎜⎛ – π

3– 2π

3 ⎠⎟⎞

= 4 cis – π = 4 cos – π + 4i sin – π = –4 + 0 = –4

i 12 8 cis 3π

4

3 2 cis π12

= 12 83 2

cis⎝⎜⎛3π

4– π

12⎠⎟⎞

= 8 cis 2π3

= 8 cos 2π3

+ 8i sin 2π3

= –4 + 4 3 i = –4(1 – 3 i )

j 20 cis – π

6

8 cis 5π6

= 208

cis ⎝⎜⎛ – π

6– 5π

6 ⎠⎟⎞

= 52 cis – π

= 52 cos – π + 5

2 i sin – π

= – 52

+ 0 = – 52


Solutions to Multiple-choice Questions 1 30 – 180 = –150 D ∴ [–3, 30°] = [3, – 150°] 2 180 – 40 = 140 Coordinates are [2, 140 C °]. 3 The point is in the third quadrant.

r = (–1)2 + ( – 3 )2

= 1 + 3 = 2cos θ = – 1

2

θ = – π + π3

= – 2π3

⎣⎢⎡ 2, – 2π

3 ⎦⎥⎤

This is equivalent to⎣⎢⎡ –2,π

3 ⎦⎥⎤ . B

4 x = 3 cos π6

= 3 32

y = 3 sin π6

= 32

⎝⎜⎛ 3 3

2, 32⎠⎟⎞ D

5 ⎣⎢⎡ 3,π

2 ⎦⎥⎤ = (0, 3)

The Cartesian equation is x2 + (y – 3)2 = 9

x2 + y2 – 6y + 9 = 9x2 + y2 = 6y

r2 = 6r sin θ

r = 6 sin θ E 6 Since x = r cos θ, it is the graph of x = 2. D 7 x2 + y2 = 16

r2 = 16

r = 4 E 8 1

2 – u= 1

1 – i = 1

1 – i× 1 + i

1 + i = 1 + i

2

= 12

+ 12 i C

9 i = cis π2

, so the point will be

rotated by π2. D

10 |z| = 5

⎪⎪⎪ 1

z ⎪⎪⎪ = 1

|z|

= 15

C



b

c

d

2 a x = 3 cos π

= –3y = 3 sin π = 0

[3,π] = (3, 0)

b x = 2 cos π3

= 1

y = 2 sin π3

= 3

⎣⎢⎡2,π

3 ⎦⎥⎤ = (1, 3 )

c x = –2 cos 210°

= 3y = –2 sin 210° = 1

[–2, 210°] = ( 3 , 1)

d x = –3 cos 11π6

= – 3 32

y = –3 sin 11π6

= 32

⎣⎢⎡–3, 11π

6 ⎦⎥⎤ =

⎝⎜⎛ – 3 3

2, 32⎠⎟⎞

3 a {[r,θ] : r = 3}

b ⎩⎨⎧ [r,θ] : θ = π

3 ⎭⎬⎫

c {[r,θ] : r = –4}

d ⎩⎨⎧ [r,θ] : θ = – 5π

4 ⎭⎬⎫


r = 32 + 32

= 18 = 3 2cos θ = 3

3 2= 1

2

θ = π4

∴ (3, 3) =⎣⎢⎡3 2 ,π

4 ⎦⎥⎤



r = 34

+ 14

= 1 = 1

cos θ = 32

θ = – π6

∴⎝⎜⎛ 3

2, – 1

2⎠⎟⎞ =

⎣⎢⎡1, – π

6 ⎦⎥⎤

c The point is in the third quadrant.

r = 254

+ 754

= 25 = 5cos θ = – 5

2 × 5= – 1

2

θ = – π + π3

= – 2π3

∴ ⎝⎜⎛ – 5

2, – 5 3

2 ⎠⎟⎞ =

⎣⎢⎡ 5, – 2π

3 ⎦⎥⎤

d The point is in the fourth quadrant. r = 32 + 32

= 64 = 8

cos θ = 4 28

= 22

θ = – π4

∴ (4 2 , – 4 2 ) =⎣⎢⎡8, – π

4 ⎦⎥⎤

5 a r2 = x2 + y2

= 16r = 4

b r2 = x2 + y2

= 9r = 3

c y2 = 8x(r sin θ)2 = 8r cos θ

r2sin2θ = 8r cos θr sin2θ = 8 cos θ

r = 8 cos θsin2θ

= 8 cot θ cosec θ

d x2 = 4y

(r cos θ)2 = 4r sin θr2cos2θ = 4r sin θr cos2θ = 4 sin θ

r = 4 sin θcos2θ

= 4 tan θ sec θ

e x2 + 4y2 = 64

(r cos θ)2 + 4(r sin θ)2 = 64r2cos2θ + r2sin2θ + 3r2sin2θ = 64

r2 + 3r2sin2θ = 64r2(1 + 3 sin2θ) = 64

r2 = 641 + 3sin2θ

r = 8

1 + 3 sin2θ

f 2x – y + 2 = 0

2r cos θ – r sin θ + 2 = 0r(2 cos θ – sin θ) = –2

r = – 22 cos θ – sin θ

6 a r = 5

x2 + y2 = 5x2 + y2 = 25


b r = 3 sin θr2 = 3r sin θ

x2 + y2 = 3y

This is correct, but it is possible to go on and complete the square.

x2 + y2 – 3y = 0

x2 + y2 – 3y +⎝⎜⎛3

2⎠⎟⎞2

= 94

x2 +⎝⎜⎛ y – 3

2⎠⎟⎞2

= 94

c r2cos 2θ = 9

r2(cos2θ – sin2θ) = 9r2cos2θ – r2sin2θ = 0

x2 – y2 = 9 d r(1 – 2 cos θ) = 8

r – 2r cos θ = 8

x2 + y2 – 2x = 8

x2 + y2 = 2x + 8x2 + y2 = (2x + 8)2

= 4x2 + 32x + 643x2 + 32x – y2 + 64 = 0

3⎝⎜⎛ x2 + 32

5 x +

⎝⎜⎛16

3 ⎠⎟⎞2

⎠⎟⎞ – y2 = –64 + 3 × 256

9

3⎝⎜⎛ x + 16

3 ⎠⎟⎞2

– y2 = –64 + 2563

x⎝⎜⎛ x + 16

3 ⎠⎟⎞2

– y2 = 643

e r(2 – cos θ) = 7

2r – r cos θ = 7

2 x2 + y2 – x = 7

2 x2 + y2 = x + 74x2 + 4y2 = (x + 7)2

= x2 + 14x + 493x2 – 14x + 4y2 = 49

3⎝⎜⎛ x2 – 14

3 x +

⎝⎜⎛7

3⎠⎟⎞2

⎠⎟⎞ + 4y2 = 49 + 3 × 49

9

3⎝⎜⎛ x – 7

3⎠⎟⎞2

+ 4y2 = 1963

f r(1 – sin θ) = 1r – r sin θ = 1

x2 + y2 – y = 1

x2 + y2 = y + 1x2 + y2 = (y + 1)2

= y2 + 2y + 1x2 = 2y + 12y = x2 – 1y = 1

2 (x2 – 1)

7 a 2z1 + 3z2 = 2m + 2ni + 3p + 3qi

= 2m + 3p + (2n + 3q)i b z2 = p – qi c z1 z2 = (m + ni)(p – qi)

= mp + npi – mqi – nqi2

= mp + nq + (np – mq)i

d z1

z2= m + ni

p + qi = m + ni

p + qi× p – qi

p – qi

= mp + npi = mqi – nqi2

p2 + q2

= mp + nq + (np – mq)ip2 + q2

e z1 + z1 = (m + ni) + (m – ni)

= 2m

f (z1 + z2)(z1 – z2) = (m + p + (n + q)i))m – p + (n – q)i)

= (m + p)(m – p) + (m + p)(n – q)i+ (n + q)(m – p)i + (n + q)(n – q)i2

= m2 – p2 + (mn – mq + pn – pq)i+ (mn – pn + mq – pq)i – (n2 – q2)

= m2 – n2 – p2 + q2 + (2mn – 2pq)i g 1

z1= 1

m + ni = 1

m + ni× m – ni

m – ni = m – ni

m2 + n2


h z2

z1= p + qi

m + ni = p + qi

m + ni× m – ni

m – ni = mp + nq + (mq – np)i

m2 + n2

i 3z1

z2= 3(m + ni)

p + qi = 3(m + ni)

p + qi× p – qi

p – qi

= 3(mp + npi – mqi – nqi2)p2 + q2

= 3(mp + nq + nq – mq)ip2 + q2

8 a z = 1 – 3 i b z2 = (1 – 3 i )2

= 1 – 2 3 i + 3i2

= –2 – 2 3 i

c z2 = z2 × z

= (–2 – 2 3 i )(1 – 3 i ) = –2 + 2 3 i – 2 3 i + 6i2

= –8

d 1

z= 1

1 – 3 i

= 11 – 3 i

× 1 + 3 i1 + 3 i

= 1 + 3 i4

e z = 1 + 3 i f 1

z = 11 + 3 i

= 11 + 3 i

× 1 – 3 i1 – 3 i

= 1 – 3 i4


r = 12 + 12

= 2cos θ = 1

2

θ = π4

∴ 1 + i = 2 cis π4

b The point is in the fourth quadrant. r = 1 + 3

= 2cos θ = 1

2

θ = – π3

∴1 – 3 i = 2 cis – π3

c The point is in the first quadrant. r = 12 + 1

= 13

tan θ = 12 3

× 33

= 36

∴ 2 3 + i = 13 cis ⎝⎜⎛ tan–1 3

6 ⎠⎟⎞

d The point is in the first quadrant.

r = 18 + 18 = 36 = 6

cos θ = 3 26

= 12

θ = π4

∴ 3 2 + 3 2 i = 6 cis π4


e The point is in the third quadrant. r = 18 + 18

= 36 = 6

cos θ = – 3 26

= – 12

θ = – π + π4

= – 3π4

∴ – 3 2 – 3 2 i = 6 cis – 3π4

f The point is in the fourth quadrant. r = 3 + 1

= 2

cos θ = 32

= – π6

∴ 3 – i = 2 cis – π6

10

a x = –2 cos π3

= –1

y = –2 sin π3

= – 3∴ z = –1 – 3 i

b x = 3 cos π4

= 3 22

y = 3 sin π4

= 3 22

∴ z = 3 22

+ 3 22

i

c x = 3 cos 3π4

= – 3 22

y = 3 sin 3π4

= 3 22

∴ z = – 3 22

+ 3 22

i

d x = –3 cos – 3π4

= 3 22

y = –3 sin – 3π4

= 3 22

∴ z = 3 22

+ 3 22

i

e x = 3 cos – 5π6

= – 3 32

y = 3 sin – 5π6

= – 32

∴ z = – 3 32

– 32

i

f x = 2 cos – π4

= 1

y = 2 sin – π4

= –1∴ z = 1 – i


Chapter 17 – Loci Exercise 17A Solutions 1 a The line is equidistant from A and B. y = 0

x = 3 + 62

= 92

b The line is equidistant from A and B. x = 0

y = 8 + 122

= 10

c The line must be a circle of radius 3

with the centre as the origin. r = 3

r2 = 9x2 + y2 = 9

d For an area of 12 square units, a triangle with a base of 4 units must have a height of 6 units. This can be above or below the axis, so its equation will be y = ± 6.

2 PA = PO

(x + 2)2 + (y – 5)2 = (x – 0)2 + (y – 0)2

x2 + 4x + 4 + y2 – 10y + 25 = x2 + y2

4x + 4 – 10y + 25 = 04x – 10y + 29 = 0

10y = 4x + 29

3 PA = PB

(x – 0)2 + (y – 6)2 = (x + 2)2 + (y – 4)2

x2 + y2 – 12y + 36 = x2 + 4x + 4 + y – 8y + 16–12y + 36 = 4x + 4 – 8y + 16

4x + 4y – 16 = 0x + y – 4 = 0

x + y = 4

4 Distance of P from (–2, 0) is

(x + 2)2 + y2 Distance of P from (2, 0) is

(x – 2)2 + y2 (x + 2)2 + y2 + (x – 2)2 + y2 = 26

x2 + 4x + 4 + y2 + x2 – 4x + 4 + y2 = 262x2 + 2y2 = 18

x2 + y2 = 9


5 Distance of P from (2, 5) is

(x – 2)2 + (y – 5)2 Distance of P (x, y) from x = 1 is |x – 1|

∴ (x – 2)2 + (y – 5)2 = 2|x – 1| Square both sides:

(x – 2)2 + (y – 5)2 = 4(x – 1)2

x2 – 4x + 4 + (y – 5)2 = 4x2 – 8x + 43x2 – 4x – (y – 5)2 = 0

3⎝⎜⎛ x2 – 4x

3+⎝⎜⎛2

3⎠⎟⎞2

⎠⎟⎞ – (y – 5)2 = 3 ×

⎝⎜⎛2

3⎠⎟⎞2

2⎝⎜⎛ x – 2

3⎠⎟⎞2

– (y – 5)2 = 43

Multiply both sides by 3:

9⎝⎜⎛ x – 2

3⎠⎟⎞2

– 3(y – 5)2 = 4

(3x – 2)2 – 3(y – 5)2 = 4(3x – 2)2

4– 3(y – 5)2

4= 1

6 Distance of P from (0, 20) is

x2 + (y – 20)2 Distance of P from (–4, 5) is

(x + 4)2 + (y – 5)2

x2 + (y – 20)2 = 2 (x + 4)2 + (y – 5)2

x2 + (y – 20)2 = 4((x + 4)2 + (y – 5)2)x2 + y2 – 40y + 400 = 4x2 + 32x + 64

+ 4y2 – 40y + 1003x2 + 32x + 3y2 = 236

x2 + 323

x + y2 = 2363

x2 + 323

x +⎝⎜⎛16

3 ⎠⎟⎞2

+ y2 = 2363

+ 2569

⎝⎜⎛x + 16

3 ⎠⎟⎞2

+ y2 = 9649

7 (x – 1)2 + (y – 2)2 = (x + 2)2 + (y + 1)2

x2 – 2x + 1 + y2 – 4y + 4 = x2 + 4x + 4 + y2 + 2y + 16x + 6y = 0

y = – x

8 (x – 4)2 + (y + 2)2 = 2 x2 + y2

x2 – 8x + 16 + y2 + 4y + 4 = 4x2 + 4y2

3x2 + 8x + 3y2 – 4y = 20

x2 + 83 x +

⎝⎜⎛4

3⎠⎟⎞2

+ y2 + y2 – 43 y +

⎝⎜⎛2

3⎠⎟⎞2

= 203

+ 169

+ 49

⎝⎜⎛ x + 4

3⎠⎟⎞2

+⎝⎜⎛ y – 2

3⎠⎟⎞2

= 809

9 Distance of P from (4, 0) is (x – 4)2 + y2

Distance of P from (–4, 0) is (x + 4)2 + y2

(x – 4)2 + y2 – ((x + 4)2 – y2) = 16x2 – 8x + 16 + y2 – x2 – 8x – 16 – y2 = 16

–16x = 16x = –1

or (x + 4)2 + y2 – ((x – 4)2 – y2) = 16x2 + 8x + 16 + y2 – x2 + 8x – 16 – y2 = 16

16x = 16x = 1


10 x2 + y2 = x + yx2 – x + 1

4+ y2 – y + 1

4= 1

4+ 1

4

⎝⎜⎛ x – 1

2⎠⎟⎞2

+⎝⎜⎛ y – 1

2⎠⎟⎞2

= 12

11 PA = 2PB

x2 + y2 = 2 (x – 4)2 + y2

x2 + y2 = 4((x – 4)2 + y2) = 4x2 – 32x + 64 + 4y2

3x2 – 32x + 3y2 = –64

x2 – 323

x +⎝⎜⎛16

3 ⎠⎟⎞2

+ y2 = – 643

+ 2569

⎝⎜⎛ x – 16

3 ⎠⎟⎞2

+ y2 = 649

12 (x – 1)2 + (y – 2)2 = (x + 1)2 + y2

x2 – 2x + 1 + y2 – 4y + 4 = x2 + 2x + 1 + y2

–4x – 4y + 4 = 0– x – y + 1 = 0

x + y = 1

13 Slope of AB = 5 – 1

2 – 0= 2

Slope of BP = y – 5x – 2

= 2

y – 5 = 2(x – 2) = 2x – 4y = 2x + 1

14

y = 1 or y = 5

15 PM is the distance of P from the line x = –4. (x – 2)2 + y2 = (x – –4)2

x2 – 4x + 4 + y2 = x2 + 8x + 16y2 = 12x + 12

= 12(x + 1)

16 PM is the distance of P from the line y = 2. x2 + (y + 4)2 = (y – 2)2

x2 + y2 + 8y + 16 = y2 – 4y + 4x2 = –12y – 12

= –12(y + 1)y + 1 = – 1

12 x2

y = – 112

x2 – 1


17 The graph will have a vertex at the origin, and will consist of the set of points equidistant from the point (a, 0) and the line x = –k.

(x – a)2 + y2 = (x + k)2

x2 – 2ax + a2 + y2 = x2 + 2kx + k2

y2 = 2ax + 2kx + k2 – a2

Since y 2 = 3x, 2a + 2k = 3

a + k = 32

k2 – a2 = 0k2 = a2

k = a

(If k = –a, –k = a and the line goes through the point, so k ≠ –a.)

a + k = 32

2a = 32

a = k = 34

is therefore the graph of all points that are equidistant from the point y2 = 3x

⎝⎜⎛3

4, 0⎠⎟⎞ and the line x = – 3

4.


Exercise 17B Solutions 1

a x2

9+ y2

64= 1

This will be an ellipse, origin at the centre, and axes of length 3 × 2 and 8 × 2 along the x- and y-axes respectively.

b x2

25+ y2

100= 1


c y2

9+ x2

64= 1


d 25x2 + 9x2 = 225

25x2

225+ 9y2

225= 1

x2

9+ y2

25= 1


2 a Centre (3, 4). x-intercept(s) when y = 0:

(x – 3)2

9+ (–4)2

64= 1

(x – 3)2

9= 1 – 16

64= 3

4

(x – 3)2 = 3 × 94

x – 3 = ± 3 32

x = 3 ± 3 32

y-intercept(s) when x = 0:

(–3)2

9+ (y – 4)2

64= 1

(y – 4)2

64= 1 – 9

9= 0

(y – 4)2 = 0y = 4

b Centre (–3, –4). x-intercept(s) when y = 0:

(x + 3)2

9+ 42

25= 1

(x + 3)2

9= 1 – 16

25= 9

25

(x + 3)2 = 9 × 925

x + 3 = ± 95

x = –3 + 95

= – 65 or

x = –3 – 95

= – 245



32

9+ (y + 4)2

25= 1

(y + 4)2

25= 1 – 9

9= 0

(y + 4)2 = 0y = –4

c Centre (3, 2). x-intercept(s) when y = 0:

(–3)2

16+ (x – 2)2

4= 1

(x – 2)2

4= 1 – 9

16= 7

16

(x – 2)2 = 7 × 416

= 74

x – 2 = ± 72

x = 2 ± 72


(y – 3)2

16+ (–2)2

4= 1

(y – 3)2

16= 1 – 4

4= 0

(y – 3)2 = 0y = 3

d Divide both sides by 225:

25(x – 5)2

225+ 9y2

225= 1

(x – 5)2

9+ y2

25= 1

2 2

2 2

25( 5) 9 1225 225

( 5) 19 25

x y

x y

−+ =

−+ =

Centre (5, 0). x-intercept(s) when y = 0: (x – 5)2 = 9

x – 5 = ± 3x = 8 or 2

y-intercept(s) when x = 0: 25 × (–5)2 + 9y2 = 225

625 + 9y2 = 2259y2 = –400

There are no y-intercepts.

3 The graph will be an ellipse with axes

3 × 2 and 5 × 2.

Its equation will be x2

32 + y2

52 = 1

x2

9+ y2

25= 1


4 (x – 4)2 + y2 + (x + 4)2 + y2 = 10

(x – 4)2 + y2 = 10 – (x + 4)2 + y2

x2 – 8x + 16 + y2 = 100 – 20 x2 + 8x + 16 + y2 + x2 + 8x + 16 + y2

20 x2 + 8x + 16 + y2 = 100 + 16x

5 x2 + 8x + 16 + y2 = 25 + 4x25(x2 + 8x + 16 + y2) = 625 + 200x + 16x2

25x2 + 200x + 400 + 25y2 = 625 + 200x + 16x2

9x2 + 25y2 = 2259x2

225+ 25y2

225= 1

x2

25+ y2

9= 1

5 The graph will be an ellipse with axes

4 × 2 and 8 × 2.

Its equation will be x2

42 + y2

82 = 1

x2

16+ y2

64= 1

6 (x – 2)2 + y2 = 12 |x – –4|

(x – 2)2 + y2 = 14 (x + 4)2

4(x2 – 4x + 4 + y2) = x2 + 8x + 164x2 – 16x + 16 + 4y2 = x2 + 8x + 16

3x2 – 24x + 4y2 = 03(x2 – 8x + 42) + 4y2 = 3 × 16Divide both sides by 48:

(x – 4)2

16+ y2

12= 1

7 x2 + (y – 8)2 = 12 |y – 4|

x2 + (y – 8)2 = 14 (y – 4)2

4(x2 + y2 – 16y + 64) = y2 – 8y + 164x2 + 4y2 – 64y + 256 = y2 – 8y + 16

4x2 + 3y2 – 56y = –240

4x2 + 3⎝⎜⎛ y2 – 56

3 y +

⎝⎜⎛28

3 ⎠⎟⎞2

⎠⎟⎞ = –240 + 3 × 784

9

4x2 + 3⎝⎜⎛ y – 28

3 ⎠⎟⎞2

= 643

Multiply both sides by 364

:

3x2

16+ 9

64 ⎝⎜⎛ y – 28

3 ⎠⎟⎞2

= 1

3x2

16+ (3y – 28)2

64= 1



a x2

9– y2

64= 1

y2

64= x2

9– 1

y2 = 64x2

9 ⎝⎜⎛ 1 – 9

x2⎠⎟⎞

y → ± 83 x

Axis intercepts are (3, 0) and (–3, 0).

b x2

25– y2

100= 1

y2

100= x2

25– 1

y2 = 100x2

25 ⎝⎜⎛ 1 – 25

x2⎠⎟⎞

y → ± 2x


c y2

9– x2

64= 1

x2

64= y2

9– 1

x2 = 64y2

9 ⎝⎜⎛ 1 – 9

y2⎠⎟⎞

x → ± 83 y

Axis intercepts are (3, 3) and (0, –3).

d 25x2 – 9y2 = 22525x2

225– 9y2

225= 1

x2

9– y2

25= 1

y2

25= x2

9– 1

y2 = 25x2

9 ⎝⎜⎛ 1 – 9

x2⎠⎟⎞

y → ± 53

x


2 a Centre (3, 4). x-intercept(s) when y = 0:

(x – 3)2

9– (–4)2

64= 1

(x – 3)2

9= 1 + 16

64= 5

4

(x – 3)2 = 5 × 94

x – 3 = ± 3 52

x = 3 ± 3 52


(–3)2

9– (y – 4)2

64= 1

– (y – 4)2

64= 1 – 9

9= 0

(y – 4)2 = 0y = 4


Asymptotes: y – k = ± b

a (x – h)

y – 4 = ± 83

(x – 3)

y = 4 ± 83

(x – 3)

= 8x3

– 4 and – 8x3

+ 12

b Centre (–3, –4). x-intercept(s) when y = 0:

(x + 3)2

9– 42

25= 1

(x + 3)2

9= 1 + 16

25= 41

25

(x + 3)2 = 9 × 4125

x + 3 = ± 3 415

x = –3 ± 3 415


32

9– (y + 4)2

25= 1

– (y + 4)2

25= 1 – 9

9= 0

(y + 4)2 = 0y = –4


a (x – h)

y + 4 = ± 53

(x + 3)

y = –4 ± 53

(x + 3)

= 5x3

+ 1 and – 5x3

– 9

c Centre (3, 2). x-intercept(s) when y = 0:

(–3)2

16– (x – 2)2

4= 1

– (x – 2)2

4= 1 – 9

16= 7

16– (x – 2)2 = 7

4

There are no x intercepts. y-intercept(s) when x = 0:

(y – 3)2

16– (–2)2

4= 1

(y – 3)2

16= 1 + 4

4= 2

(y – 3)2 = 2 × 16y = 3 ± 4 2


a (x – h)

y – 3 = ± 42

(x – 2)

y = 3 ± 2(x – 2) = 2x – 1 and – 2x + 7


d Divide both sides by 225.

25(x – 5)2

225– 9y2

225= 1

(x – 5)2

9– y2

25= 1

Centre (5, 0). x-intercept(s) when y = 0: 25(x – 5)2 = 225

(x – 5)2 = 9x – 5 = ±3

x = 8 or 2 y-intercept(s) when x = 0: 25 × (–5)2 – 9y2 = 225

625 – 9y2 = 2259y2 = 400

y = ± 203


a (x – h)

y = ± 53 (x – 5)

= 5x3

– 253

and – 5x3

+ 253

e Divide both sides by 4.

x2

4– y2

4= 1

Centre (0, 0). x-intercept(s) when y = 0: x2

4= 1

x2 = 4x = ± 4


– y2

4= 1

y2 = –4

No y-intercepts.

Asymptotes: y = ± b

a x

= ± 22 x

= ± x

f Divide both sides by 4.

x2

2– y2

4= 1

Centre (0, 0). x-intercept(s) when y = 0:

x2

2= 1

x2 = 2x = ± 2


– y2

4= 1

y2 = –4

No y-intercepts. Asymptotes:

y = ± ba

x

= ± 22

x

= ± 2 x


g Complete the square. h Complete the square. 9x2 – 90x – 25y2 + 150y = 225

9(x2 – 10x + 25) – 25(y2 – 6y + 9) = 225 + 225 – 2259(x – 5)2 – 25(y – 3)2 = 225

(x – 5)2

25– (y – 3)2

9= 1

x2 – 4x – 4y2 – 8y = 16x2 – 4x + 4 – 4(y2 + 2y + 1) = 16 + 4 – 4

(x – 2)2 – 4(y + 1)2 = 16(x – 2)2

16– (y + 1)2

4= 1

Centre (2, –1). Centre (5, 3). x-intercept(s) when y = 0: x-intercept(s) when y = 0:

(x – 2)2

16– 12

4= 1

(x – 2)2

16= 1 + 1

4= 5

4x – 2

4= ± 5

2x – 2 = ± 2 5

x = 2 ± 2 5

(x – 5)2

25– (–3)2

9= 1

(x – 5)2

25= 1 + 9

9= 2

(x – 5)2 = 2 × 25x – 5 = ± 5 2

x = 5 ± 5 2

y-intercept(s) when x = 0: y-intercept(s) when x = 0:

(–5)2

25– (y – 3)2

9= 1

– (y – 3)2

9= 1 = 25

25= 0

(y – 3)2 = 0y = 3

(–2)2

16– (y + 1)2

4= 1

– (y + 1)2

4= 1 – 4

16= 3

4– (y + 1)2 = 3

There are no y-intercepts. Asymptotes: Asymptotes:

y – k = ± ba

(x – h)

y – 3 = ± 35

(x – 5)

y = 3 ± 35 (x – 5)

= 3x5

and – 3x5

+ 6

y – k = ± ba (x – h)

y + 1 = ± 24 (x – 2)

y = –1 ± 12 (x – 2)

= x2

– 2 and – x2


3 (x – 4)2 + y2 – (x + 4)2 + y2 = 6

(x – 4)2 + y2 = 6 + (x + 4)2 + y2

x2 – 8x + 16 + y2 = 36 + 12 x2 + 8x + 16 + y2 + x2 + 8x + 16 + y2

–12 x2 + 8x + 16 + y2 = 36 + 16x

–3 x2 + 8x + 16 + y2 = 9 + 4x9(x2 + 8x + 16 + y2) = 81 + 72x + 16x2

9x2 + 72x + 144 + 9y2 = 81 + 72x + 16x2

–7x2 + 9y2 = –63x2

9– y2

7= 1

4 (x – 2)2 + y2 = 2|x – –4|(x – 2)2 + y2 = 4(x + 4)2

x2 – 4x + 4 + y2 – 4(x2 + 8x + 16) = 4x2 + 32x + 64

3x2 + 36x – y2 = –603(x2 + 12x + 36) – y2 = –60 + 3 × 36

3(x + 6)2 – y2 = 48(x + 6)2

16– y2

48= 1

5 x2 + (y – 8)2 = 4|y – 4|x2 + (y – 8)2 = 16(y – 4)2

x2 + y2 – 16y + 64 = 16(y2 – 8y + 16) = 16y2 – 128y + 256

x2 – 15y2 + 112y = 192

x2 – 15⎝⎜⎛ y2 – 112

15 y +

⎝⎜⎛56

15⎠⎟⎞2

⎠⎟⎞ = 192 – 15 × 3136

225

x2 – 15⎝⎜⎛ y – 56

15⎠⎟⎞2

= – 25615

Multiply both sides by – 15256

:

– 15x2

256+ 225

256 ⎝⎜⎛ y – 56

15⎠⎟⎞2

= 1

(15y – 56)2

256– 15x2

256= 1


6 (x + 3)2 + y2 – (x – 3)2 + y2 = 4

(x + 3)2 + y2 = 4 + (x – 3)2 + y2

x2 + 6x + 9 + y2 = 16 + 8 x2 – 6x + 9 + y2 + x2 – 6x + 9 + y2

–8 x2 – 6x + 9 + y2 = 16 – 12x

–2 x2 – 6x + 9 + y2 = 4 – 3x4(x2 – 6x + 9 + y2) = 16 – 24x + 9x2

4x2 – 24x + 36 + 4y2 = 16 – 24x + 9x2

–5x2 + 4y2 = –20x2

4– y2

5= 1


Solutions to Multiple-choice Questions 1 The vertical axes of the ellipse look

about 5 units long (just over 2).

x2

25+ y2

5= 1

C x2 + 5y2 = 25

2 x2

a2 = x2

25a = 5

x-axis intercepts = (a, 0) and (–a, 0) = (5, 0) and (–5, 0) D

3 ⎝⎜⎛ x

9⎠⎟⎞2

+⎝⎜⎛ y

4⎠⎟⎞2

= 1

x2

92 + y2

42 = 1

This will be an ellipse, origin at the centre, and axes of length 9 × 2 and 9 × 2 along the x and y axes respectively. B

4 The y-intercept(s) occur when x = 0:

(y + 2)2

4= 1

(y + 2)2 = 4y + 2 = ± 2

y = 0 or – 4

2

2

( 2) 14

( 2) 42 20 or 4

y

yyy

+=

+ =+ = ±= −

Intercepts are (0, 0) and (0, –4). D

5 y-axis intercept is 2. ∴ b × 22 = 8

b = 2

The graph goes through ⎝⎜⎛ 1, 1

2 10

⎠⎟⎞ .

∴a × 12 + 2 ×⎝⎜⎛1

2 10

⎠⎟⎞2

= 8

a + 5 = 8

a = 3 D 6 If its centre is on the y-axis, then a = 0. It passes through (4, 4), so x2 + (y – b)2 = 16

42 + (4 – b)2 = 16(4 – b)2 = 0

4 – b = 0 b = 4 A 7 The first two transformations will give

x2

16+ y2

9= 1

The next two transformations will give

(x – 4)2

16+ (y – 3)2

9= 1 B

8 The centre is (2, 0) and the length of

half the axis is 9, so the graph is

(x – 2)2

81– y2

b2 = 1

One asymptote has gradient 42

= 2

ba

= 2

Since a = 9, b9

= 2

b = 18

The equation of the graph is

(x – 2)2

81– y2

324= 1 D


9 PA = PB

(x – 2)2 + (y + 5)2 = (x + 4)2 + (y + 1)2

x2 – 4x + 4 + y2 + 10y + 25 = x2 + 8x + 16 + y2

0 = 12x – 12y – 1212y = 12x – 12

y = x – 1 A 10 PA = 2PB

(x – 2)2 + (y + 5)2 = 2 (x + 4)2 + (y – 1)2

x2 – 4x + 4 + y2 + 10y + 25 = 4(x2 + 8x + 16 + y2 – 2y + 1) = 4x2 + 32x + 64 + 4y2 – 8y + 4

3x2 + 36x + 3y2 – 18y + 39 = 0x2 + 12x + y2 – 6y + 13 = 0

x2 + 12x + 36 + y2 – 6y + 9 = –13 + 36 + 9(x + 6)2 + (y – 3)2 = 32

This is a circle. C


Solutions to Short-answer Questions 1 Complete the square. x2 + 4x + 4 + y2 + 8y + 16 = 0 + 4 + 16

(x + 2)2 + (y + 4)2 = 20

The circle has centre (–2, –4) and radius 20 = 2 5 . 2 Complete the square. x2 + 4x + 4 + 2y2 = 0 + 4

(x + 2)2 + 2y2 = 4(x + 2)2

4+ y2

2= 1

The ellipse has centre (–2, 0). x-intercept(s) when y = 0: x2 + 4x = 0

x(x + 4) = 0x = 0 and – 4

x-intercepts are (0, 0) and (–4, 0) y-intercept(s) when x = 0: 2y2 = 0

y = 0 y-intercept is (0, 0). 3 PA = PB

x2 + (y – 2)2 = (x – 6)2 + y2

x2 + y2 – 4y + 4 = x2 – 12x + 36 + y2

–4y = –12x + 32y = 3x – 8

4 This is clearly a circle with centre (3, 2)

and radius 6. (x – 3)2 + (y – 2)2 = 36 5 Asymptotes: y = ± b

a x

= ± 23 x

= 2x3

, – 2x3

6 PA = 2PB

x2 + (y – 2)2 = 2 (x – 6)2 + y2

x2 + y2 – 4y + 4 = 4(x2 – 12x + 36 + y2) = 4x2 – 48x + 144+ 4y2

3x2 – 48x + 3y2 + 4y + 140= 0

3(x2 – 16x + 64) + 3⎝⎜⎛y2 + 4

3 y +

⎝⎜⎛23⎠⎟⎞2

⎠⎟⎞ = –140+ 3 × 64 + 3 × 4

9

3(x – 8)2 + 3⎝⎜⎛y + 2

3⎠⎟⎞2

= 1603

(x – 8)2 +⎝⎜⎛y + 2

3⎠⎟⎞2

= 1609

7 Centre (2, 0). x-intercept(s) when y = 0:

(x – 2)2

9= 1

(x – 2)2 = 9x – 2 = ± 3

x = 1 or 5


(–2)2

9+ y2

4= 1

y2

4= 1 – 4

9= 5

9

y2 = 4 × 59

y = 209

= ± 2 53


8 Let k be the constant. k = ((x + 4)2 + y2) – ((x – 4)2 + y2)

= (x2 + 8x + 16 + y2) – (x2 – 8x + 16 + y2) = x2 + 8x + 16 + y2 – x2 + 8x – 16 – y2

= 16xx = k

16

If the subtraction is done in reverse order, we would get x = – k

16.

The ± is necessary, as the locus will be the two parallel lines x = k and x = –k. ∴x = ± k

16, where k is the constant

difference.


Chapter 18 – Revision of chapters 14–17 Solutions to Multiple-choice Questions 1 The angle subtended at the top of the

circle by QT = 1502

= 75°.

By the alternate segment theorem, B ∠QTS = 75°. 2 Join LN. Using the alternate segment theorem, ∠MLN = 40°. In triangle LMN, ML = MN ∴∠MNL = ∠MLN = 40°

∠LMN = 180 – 40 – 40 = 100° C 3 Using the alternate segment theorem, ∠ ZYX = ∠ZXT

∠ZXT = ∠ZXY∴∠ZYX = ∠ZXY

Triangle ZXY is isosceles, with YZ = XZ. B 4 ∠QOS = 180 – 70

= 110° Reflex ∠QOS = 360 – 110

= 250°

∠QRS = 2502

= 125° D 5 Let the diameter be CE. If AD = DB = 4 cm, then AD⋅DB = CD⋅DE

4 × 4 = 2(2r – 2)2r – 2 = 8

2r = 10

r = 5 cm C

6 Join AB. Since TA = TB, ∠ TBA = 45°. Since AC is perpendicular to tangent TA,

it must be a diameter. ∴ ∠CBA = 90°

∠TBC = 90 + 45 = 135°

TB is parallel to AC, since co-interior angles BTA and CAT are supplementary.

∴ ∠BCA = 180 – 135 = 45° C 7 ∠RTP = 30° (alternate segments)

∠TRS = 40 + 30 = 70° (exterior angle of RTP)

∠RTS = 180 – 70 – 30 = 80° B 8 AB = AC

∴∠ADC = 60°∠ACD = 180 – 60 – 50

= 70°(angle sum of RTS)∠ABD = 180 – 70

= 110° D

9 |a| = 32 + (–4)2

= 25 = 5a = 3

5 i – 4

5 j

= 15

(3i – 4j) B

10 AB = OB – OA = i + 8j D 11 a – b = (2i + 4j) – (3i – 2j)

= 2i – 3i + 4j + 2j

= – i + 6j B

12 |a| = 22 + (–1)2 + 42

= 4 + 1 + 16

= 21 A


13 Since they have the same magnitude and direction,

AB = OC = cCD = OB = b

∴AD = c – b + c – b

= 2(c – b) B 14 2r – s = 2(2i – j + k) – ( – i + j + 3k)

= 4i + i – 2j – j + 2k – 3k

= 5i – 3j – k D 15 QR = OR – OQ

= – i + 5jPQ = 2QR

= –2i + 10j = OQ – OP

∴ OP = OQ – PQ = 2i – 3j – (–2i + 10j)

= 4i – 13j A 16 u and v are parallel only if u = nv,

where n is a constant. Comparing k components: –5 = 6n

n = – 56

Comparing i and j components: 1 = – 5

6 b

b = – 65

a = – 56× –3

= 52

B

17 sa + yb = x

3si + 4sj + 2ti – tj = i + 5j3s + 2t = 1

4s – t = 58s – 2t = 10

+ :11s = 11

s = 14 – t = 5

t = –1 C

18 CB = 13 OA

= 13 a

AB = – a + c + 13

a

= c – 23 a B

19 AB = OB – OA

= b – aCA = 2AB

= 2b – 2a = OA – OC

∴OC = OA – CA = a – (2b – 2a)

= 3a – 2b B 20 Arg z = Arg v + Arg w

= Argv – Arg w = –0.3π – 0.6π

= –0.9π B

21 x = 2 cos ⎝⎜⎛2π

3 ⎠⎟⎞

= –1

y = 2 sin ⎝⎜⎛2π

3 ⎠⎟⎞

= 3

2 cis ⎝⎜⎛2π

3 ⎠⎟⎞ = –1 + 3 i D

22 The angle is in the third quadrant.

tan θ = 12÷ 3

2 = 1

3

Arg z = θ

= – π + π6

= – 5π6

(because π ≤ Arg z ≤ π ) E 23 a + bi = –2 – 3i b = –3 A


24 uv = 3 × 5 × cis ⎝⎜⎛ π

2+ 2π

3 ⎠⎟⎞

= 15 cis 7π6

= 15 cis ⎝⎜⎛ 2π – 7π

6 ⎠⎟⎞

= 15 cis – 5π6 C

25 |(12 – 5i)| = 122 + 52

= 169

= 13 C 26 z2 = x2 + 2ixy – y2

This still has a non-real (i) component, so is not necessarily a real number. A

27 z = –14 – 7i z = –14 + 7i C 28 3z2 + 9 = 3(z2 + 3)

= 3(z2 – 3i2) = 3(z + 3 i )(z – 3 i ) E 29 (1 + 2i)2 = 1 + 4i + 4i2

= 1 + 4i – 4 = –3 + 4i C 30 x = r cos θ

= 2 cos 4π3

= –1

y = r sin θ

= 2 sin 4π3

= – 3

(–1, – 3 ) D

31 The point is in the fourth quadrant.

r = (4 2 )2 + (–4 2 )2

= 32 + 32 = 8

cos θ = 4 28

= 12

θ = – π4

The point is ⎣⎢⎡ 8, – π

4 ⎦⎥⎤ . E

32 ⎣⎢⎡ 2,π

2 ⎦⎥⎤ = (0, 2)

The equation of the circle is x2 + (y – 2)2 = 4

x2 + y2 – 4y + 4 = 4x2 + y2 – 4y = 0r2 – 4r sin θ = 0

r2 = 4r sin θ

r C = 4 sin θ 33 Since x = r cos θ , this will be the

graph of x = 4, a vertical line through [4, 0]. C

34 x2 + y2 = 9

r2 = x2 + y2

= 9

r = 3 B 35 r2 = x2 + y2

= 1 – (r sin θ)2

x2 + y2 = 1 – y2

D x2 + 2y2 = 1 36 When θ = 0, r = 1

1 + cos 0 = 1

2

Curve will be a parabola with intercept at

⎣⎢⎡ 1

2, 0⎦⎥⎤ . E


Chapter 19 – Kinematics Exercise 19A Solutions 1 a When t = 0, x = 12. 12 cm to the right of O b When t = 5, x = 52 – 7 × 5 + 12

= 2

2 cm to the right of O c v = dx

dt = 2t – 7

When t = 0, v = –7. 7 cm/s to the left d v = 0 when 2 t – 7 = 0

t = 3.5 When t = 3.5, x = 3.52 – 7 × 3.5 + 12

= –0.25

t = 3.5; the particle is 0.25 cm to the left of O.

e Average velocity = change in position

change in time= 2 – 12

5= –2 cm/s

f Average speed = distance travelled

change in time

For the first 3.5 s, the particle has travelled 12.25 cm. From 3.5 s to 5 s, the particle has travelled 2 – –0.25 = 2.25 cm.

Average speed = 12.25 + 2.255

= 14.55

= 2.9 cm/s

2 a v = dx

dt= 2t – 7

v = 0 when 2t – 7 = 0 t = 3.5 s b a = dv

dt = 2 m/s2

c When t = 0, x = 10. When t = 3.5, x = 3.52 – 7 × 3.5 + 10

= –2.25

For the first 3.5 s, the particle has travelled 12.25 m.

When t = 5, x = 52 – 7 × 5 + 10 = 0

From 3.5 s to 5 s, the particle has travelled 2.25 m.

Distance travelled = 12.25 + 2.25 = 14.5 m d v = 2t – 7 = –2 2t = 5 t = 2.5 x = 2 .52 – 7 × 2.5 + 19

= –1.25After 2.5 s, when the particle is 1.25 m left of O.

3 a When t = 0, x = –3. v = dx

dt = 3t2 – 22t + 24

When t = 0, v = 24. 3 cm to the left and 24 cm/s to the right. b v = dx

dt = 3t2 – 22t + 24


c v = 0 when 3t2 – 22t + 24 = 0

(3t – 4)(t – 6) = 0t = 4

3 or 6

After 43

s and after 6 s

d When t = 4

3,

x =⎝⎜⎛4

3⎠⎟⎞3

– 11 ×⎝⎜⎛4

3⎠⎟⎞2

+ 24 ×⎝⎜⎛4

3⎠⎟⎞ – 3

= 6427

– 1769

× 33

+ 32 – 3

= – 46427

+ 29

= 11 2227

When t = 6, x = 63 – 11 × 62 × 6 – 3

= –39

39 cm to the left of O and 11 2227

cm to

the right of O e v < 0 when (3t – 4)(t – 6) = 0

This is a parabola with a minimum value. ∴ v < 0 when 4

3< t < 6

= 6 – 43

= 143

= 4 23

s

f a = dv

dt = 6t – 22 m/s2

g 6t – 22 = 0

t = 226

= 113

v = 3t2 – 22t + 24

= 3 ×⎝⎜⎛11

3 ⎠⎟⎞2

– 22 × 113

+ 24

= 1213

– 2423

+ 24

= 16 23

x =⎝⎜⎛11

3 ⎠⎟⎞3

= 11 ×⎝⎜⎛11

3 ⎠⎟⎞2

+ 24 × 113

– 3

= 133127

– 13319

× 33

+ 88 – 3

= –13 1627

The acceleration is zero after 113

s, when

the velocity is 16 23

cm/s to the left and its

position is 13 1627

cm left of O.

4 a v = 6t2 – 10t + 4 When v = 0:

6t2 – 10t + 4 = 03t2 – 5t + 2 = 0

(3t – 2)(t – 1) = 0t = 2

3 or 1

a = 12t – 10

t = 23:

a = 12 × 23

– 10

= –2t = 1:a = 12 × 1 – 10 = 2

Velocity is zero after 23

s when the

acceleration is 2 cm/s2 to the left, and after 1 s when the acceleration is 2 cm/s2 to the right.

b a = 12t – 10

= 0t = 10

12= 5

6

Find v when a = 56

:

v = 6t2 – 10t + 4

= 6 ×⎝⎜⎛5

6⎠⎟⎞2

– 10 × 55

+ 4

= 256

– 506

+ 4 = – 16


Acceleration is zero after 56

s, at which

time the velocity is 16

cm/s to the left.

6 a They will be at the same position when t2 – 2t – 2 = t + 2

t2 – 3t – 4 = 0(t – 4)(t + 1) = 0

t = 4 or – 1

5 The particle passes through O when x = 0. t3 – 13t2 + 46t – 48 = 0

Trial and error will give x = 0 when t = 2. After 4 s, or 1 s before the start. This means (t – 2) is a factor of Note: In some cases, motion is not

considered before t = 0, and negative values of t may be discarded.

t3 – 13t2 + 46t – 48.

t3 – 13t2 + 46t – 48 = (t – 2)(t2 – 11t + 24)

= 0 b The velocities are 1 cm/s and 2t – 2 m/s. Factorising the quadratic gives

2t – 2 = 12t = 3t = 3

2

(t – 2)(t – 3)(t – 8) = 0t = 2, 3 or 8

After 32

s.


Exercise 19B Solutions 1 a x = 2t2 – 6t + c When t = 0, x = 0. ∴ 0 = 0 – 0 + c

c = 0x = 2t2 – 6t

b t = 3

x = 2 × 32 – 6 × 3 = 0

It will be at the origin, O. c Consider when v = 0: 4t – 6 = 0

t = 32

x = 2 ×⎝⎜⎛3

2⎠⎟⎞2

– 6 × 32

= –4 12

The particle will travel 4 12

cm to the

left of the origin and back, for a total of 9 cm.

d Average velocity = change in position

change in time= 0

3= 0 cm/s

e Average speed = distance travelled

change in time= 9

3= 3cm/s

2 a x = t3 – 4t2 + 5t + c When t = 0, x = 4.

∴ 4 = 0 – 0 + 0 + cc = 4x = t3 – 4t2 + 5t + 4a = dv

dt = 6t – 8

b 3t2 – 8t + 5 = 0(3t – 5)(t – 1) = 0

t = 53

or 1

When t = 53

,

x =⎝⎜⎛5

3⎠⎟⎞3

– 4 ×⎝⎜⎛5

3⎠⎟⎞2

+ 5 × 53

+ 4

= 12527

– 1009

+ 253

+ 4

= 5 2327

When t = 1, x = 13 – 4 × 12 + 5 × 1 + 4

= 6

c When t = 5

3,

a = 6 × 53

– 8

= 2 cm/s2

When t = 1, a = 6 × 1 – 8

= –2 cm/s2

3 v = 10t + c

x = 5t2 + ct + dWhen t = 2:

x = 5 × 22 + 3c + d = 02c + d = –20

When t = 3:x = 5 × 32 + 3c + 2 = 25

3c + d = –20 – : c = 0

d = –20x = 5t2 – 20

When t = 0, x = 20 20 m to the right of O


d When t = 2.5, 4 a = 2t – 3v = t2 – 3t + c x – 5t2 + 25t

= –5 × 2.52 + 25 × 2.5 = 31.25 m

When t = 0, v = 3. 3 = 0 – 0 + c

c = 3v = t2 – 3t + 3

x = t3

3– 3t2

2+ 3t + d

e x = – 5t2 + 25t = 0

–5t(t – 5) = 0t = 5 (t = 0 is the start)

When t = 0, x = 2. 6 Define t = 0 as the moment the lift

passes the 50th floor. 2 = 0 – 0 + 0 + d

d = 2

x = t3

3– 3t2

2+ 3t + 2

a = 19

t – 59

v = 118

t2 – 59

t + c

–8 = 0 – 0 + cc = –8v = 1

18 t2 – 5

9– 8

x = 154

t3 – 518

t2 – 8t + d

50 × 6 = 0 – 0 – 0 + dd = 300

When t = 10,

x = 103

3– 3 × 102

2+ 3 × 10 + 2

= 2000 – 9006

+ 32

= 215 13

v = t2 – 3t – 3 = 102 – 3 × 10 + 3 = 73

v = 0 when

118

t2 – 59

t – 8 = 0

t2 – 10t – 8 × 18 = 0(t – 18)(t + 8) = 0

t = 18x = 1

54 t3 – 5

18 t2 – 8t + 300

= 154

× 183 – 518

182 – 8 × 18 + 300

= 1741746

= 29

5 a a = –10

v = –10t + c When t = 0, v = 25. 25 = 0 + c

c = 25v = –10t + 25

b v = –10t + 25

x = –5t2 + 25t + d

When t = 0, x = 0. It will stop on the 29th floor. (Define the point of projection as x = 0,

the origin.) 0 = 0 + 0 + d

d = 0x = –5t2 + 25t

c Maximum height occurs when v = 0. v = –10t + 25 = 0

t = 2510

= 52

2.5 s after projection


Exercise 19C Solutions 1 s = 30, u = 0, a = 1.5

s = ut + 12

at2

30 = 12× 1.5 × t2

t2 = 40t = 40 = 2 10 s

2 u = 25, v = 0, t = 3

s = 12

(u + v)t

= 12

(25 + 0) × 3

= 37.5 m

3 a For constant acceleration, acceleration = change in velocity

change in time= 27

9= 3 m/s2

b u = 30, v = 50, a = 3

v = u + at50 = 30 + 3t3t = 20t = 20

3= 6 2

3 s

c s = ut + 12

at2

= 12× 3 × 152

= 337.5 m

d 200 km/h = 200 ÷ 3.6 = 500

9 m/s

u = 0, v = 5009

, a = 3

v = u + at500

9= 0 + 3t

3t = 5009

t = 50027

= 18 1427

s

4 a 45 km/h = 45 ÷ 3.6 = 12.5 m/s For constant acceleration, acceleration = change in velocity

change in time = 12.5

5= 2.5 m/s2

b s = ut + 1

2 at2

= 12× 2.5 × 52

= 31.25 m

5 a 90 km/h = 90 ÷ 3.6 = 25 m/s u = 0, v = 25, a = 0.5

v = u + at25 = 0 + 0.5t

0.5t = 25t = 25

0.5= 50 s

b s = ut + 1

2 at2

= 12× 0.5 × 502

= 625 m

6 a 54 km/h = 54 ÷ 3.6 = 15 m/s u = 15, a = –0.25, s = 250

s = ut + 12

at2

250 = 15t + 12× –0.25t2

t = 100 represents the train changing velocity and returning to this point.

Multiply both sides by 8: 2000 = 120t – t2

t2 – 120t + 2000 = 0(t – 20)(t – 100) = 0

∴ t = 20 s


b v = u + at = 15 + –0.25 × 20 = 10 m/s = 10 × 3.6 = 36 km/h

7 a v = u + at

= 20 + –9.8 × 4 = 19.2 m/s

b s = ut + 1

2 at2

= 20 × 4 + 12× –9.8 × 42

= 1.6

8 a v = u + at

= –20 + –9.8 × 4 = –59.2 m/s

b s = ut + 1

2 at2

= –20 × 4 + 12× –9.8 × 42

= –158.4 m

9 a u = 49, s = 0, a = –9.8

s = ut + 12

at2

0 = 49t + 12× –9.8 × t2

0 = 49t – 4.9t2

0 = 4.9t(10 – t)t = 10 s

b u = 49, s = 102.9, a = –9.8

s = ut + 12

at2

102.9 = 49t + 12× –9.8 × t2

102.9 = 49t – 4.9t2

0 = 4.9t2 – 49t + 102.9

Divide by 4.9: t2 – 10t + 21 = 0

(t – 3)(t – 7) = 0 At both 3 s (going up) and 7 s (going

down)

10 a v = u + at

= 4.9 – 9.8t = 4.9(1 – 2t)

b s = ut + 1

2 at2

= 4.9t + 12× –9.8 × t2

= 4.9t – 4.9t2

= 4.9t(1 – t)

This is his displacement from the initial 3 m height.

∴ h = 4.9t(1 – t) + 3 11 a Maximum height occurs when v = 0. u = 19.6, a = –9.8, v = 0

v = u + at0 = 19.6 – 9.8tt = 19.6

9.8= 2 s

b s = ut + 1

2 at2

= 19.6 × 2 + 12× –9.8 × 22

= 19.6 m

c u = 19.6, s = 0, a = –9.8

s = ut + 12

at2

0 = 19.6t + 12× –9.8 × t2

0 = 196t – 4.9t2

0 = 4.9t(4 – t)t = 4 s

d u = 19.6, s = –24.5, a = –9.8

s = ut + 12

at2

–24.5 = 19.6t + 12× –9.8 × t2

–24.5 = 19.6t – 4.9t2

0 = 4.9t2 – 19.6t – 24.5

Divide by 4.9: t2 – 4t – 5 = 0

(t – 5)(t + 1) = 0t = 5 s


12 Let the distance between P and Q be x m. u = 20, v = 40, s = x

v2 = u2 + 2as1600 = 400 + 2ax

2ax = 1200a = 1200

2x = 600

x

At the halfway mark,

u = 20, a = 600x

, s = x2

v2 = u2 + 2as = 400 + 2 × 600

x× x

2 = 1000v = 1000 = 10 10 m/s


Exercise 19D Solutions 1 Draw the velocity–time graph.

Distance travelled = area under graph = 1

2× 10 × 13

= 65 m

2 Draw the velocity–time graph.

a The area can be calculated using the

trapezium formula, or as the sum of two triangles and a rectangle.

A = 12

(a + b)h

= 12× (25 + 50) × 15

= 562.5 m

b A = 1

2 (a + b)h

= 12× (25 + 35) × 15

= 450 m

c Let the halfway point be at time T as

below.

1

2× 10 × 15 + 15(T – 10) = 562.5

275 + 15T – 150 = 281.25

15T = 356.25T = 23.75 s

3

Since the total distance travelled is 1 km

or 1000 m, the combined areas of the two triangles will equal a distance of 500 m.

12× 5 × h + 1

2× 10 × h = 500

5h + 10h = 100015h = 1000

h = 100015

= 66 23

Maximum speed = 66 23

m/s

4 36 km/h = 36 ÷ 3.6 = 10 m/s. 72 km/h = 20 m/s.

Distance = A = 18 × 10 + 1

2× 6 × 10

= 210 m


5 60 km/h = 60 ÷ 3.6 = 50

3 m/s

a Distance = A = 1

2× 60 × 50

3 = 500 m

b Distance = A = 1

2×⎝⎜⎛50

3+ 25

3 ⎠⎟⎞ × 30

= 375 m

c Let the required time be T s.

It is easier to work with the triangle on

the right. This triangle will have area = 500 ÷ 2 = 250 Its base = (60 – T)

The sloping line has gradient= – 503÷ 60

= – 50180

= 518

the triangle’s height ∴ = 518

(60 – T)

12× (60 – T) × 5

18 (60 – T) = 250

526

(60 – T)2 = 250

(60 – T)2 = 250 × 365

= 180060 – T = 1800

≈ 42.43T ≈ 15.57 s

6 Let the common time be T s and the distance x m.

a For the first car, x = 15t For the second car,

x = 12× 10 × 25 + 25(t – 10)

= 125 + 25t – 250 = 25t – 125 = 15t

10t = 125t = 12.5 s

b x = 15t = 15 × 12.5 = 187.5 m

7 Convert the speeds to km/min. 60 km/h = 1 km/min 80 km/h = 4

3 km/min

Treat each train separately. The first train:

The second train:

First train distance = 5 × 1 + 1

2× 2.5 × 1

= 6.25 km

Second train distance = 12× 9 × 4

3= 6 km

Since the trains have together travelled less than 14 km, they will not crash.


8 a The maximum speed will be the height

of the triangle.

1

2× 100 × h = 800

50h = 800h = 16

Maximum speed = 16 m/s = 16 × 3.6 = 57.6 km/h b The slope of the deceleration is twice as

steep as the slope of the acceleration. Since the heights are equal, the

acceleration run will be twice as long as the deceleration run.

T = 23× 100

= 66 23

s

= 1 min 6 23

seconds

c Taking the acceleration section, the gradient = a = 16 ÷ 66 2

3= 48

200= 0.24 m/s2


Solutions to Multiple-choice Questions 1 When t = 0, x = 0 = 0 cm A 2 When t = 0, x = 0. When t = 2, x = –23 + 7 × 22 – 12 × 2

= –4

Average velocity = change in positionchange in time

= – 42

= –2 cm/s E 3 v = 4t – 3t2 + c When t = 0, v = –1 –1 = 0 – 0 + c

c = –1v = 4t – 3t2 – 1

When t = 1, v = 4 × 1 – 3 × 12 – 1 = 0 m/s C 4 u = 0, s = 90, a = 1.8

s = ut + 12

at2

90 = 12× 1.8 × t2

90 = 0.9t2

t2 = 100

t = 10 s C 5 60 km/h = 60 ÷ 3.6 = 50

3 m/s

u = 0, v = 503

, t = 4

v = u + at503

= 4a

a = 5012

= 256

m/s2 E

6 60 km/h = 60 ÷ 3.6 = 50

3 m/s

u = 0, v = 503

, t = 4

s = 12

(u + v)t

= 12× 50

3× 4

= 1003

m C

7 Distance = area under graph = triangle + trapezium + triangle = 1

2× 4 × 10 + 1

2× (10 + 25) × 2 + 1

2× 9 × 25

= 20 + 25 + 112.5

= 167.5 m D 8 u = 0, a = 9.8, s = 40

v2 = u2 + 2as = 0 + 2 × 9.8 × 40 = 784

v = 784 = 28 m/s E 9 u = 20, v = 0, a = –4

v = u + at0 = 20 – 4tt = 5s = 1

2 (u + v)t

= 12× 20 × 5

= 50 m A 10 v = 6t2 – 5t + c When t = 0, v = 1. 1 = 0 – 0 + c

c = 1v = 6t2 – 5t + 1

When t = 1, v = 6 × 12 – 5 × 1 + 1 = 2 m/s D


Solutions to Short-answer Questions 1 a When t = 0, x = –5. 5 cm to the left of O b When t = 3, x = 32 – 4 × 3 – 5

= –8

8 cm to the left of O c v = dx

dt = 2t – 4

When t = 0, v = -4. d v = 0 when 2 t – 4 = 0

t = 2 When t = 2, x = 22 – 4 × 2 – 5

= –9

At 2 s, 9 cm to the left of O e Average velocity = change in position

change in time= –9 – –5

2= –2 cm/s

2 cm/s to the left f Distance travelled = distance from t = 0

to t = 2 (when v = 0), plus distance from t = 2 to t = 3 = 4 + 1

= 5 cm Average speed = distance travelled

change in time= 5

3= 1 2

3 cm/s

Note: Average velocity has a direction and hence a sign, but average speed does not.

2 a v = dx

dt = 3t2 – 4ta = dv

dt= 6t – 4

When t = 0, x = 8, v = 0 and a = –4. 8 cm to the right of O, stationary and

accelerating at 4 m/s2 to the left.

b v = 0 when 3 t2 – 4t = 0 t(3t – 4) = 0

t = 0 or 43

At the start and after 43

s

3 a Solve –2t3 + 3t2 + 12t + 7 = 0 Using factors of 7, t = –1 gives –2 × (–1)3 + 3 × (–1)2 + 12 × –1 + 7 = 0 Dividing by (t + 1),

–2t3 + 3tg2 + 12t + 7 = – (t + 1)(2t2 – 5t – 7)

= – (t + 1)(t + 1)(2t – 7) = 0

t = 3.5, as t = –1 is usually discarded. v = dx

dt = –6t2 + 6t + 12

a = dvdt

= –12t + 6

When t = 3.5 v = –6 × 3.52 + 6 × 3.5 + 12

= 40.5 cm/sa = dv

dt = –12 × 3.5 + 6 = –36 cm/s2

b v = 0

–6t2 + 6t + 12 = 0t2 – t – 2 = 0

(t + 1)(t – 2) = 0t = 2

After 2s (discarding t = –1) c Distance travelled in first 2 seconds = |(–2 × 23 + 3 × 22 + 12 × 2 + 7) – (–0 + 0 + 0 + 7)|

= 20 m Distance travelled from t = 2 to t = 3 is

|(–2 × 32 + 3 × 32 + 12 × 3 + 7)

– (–2 × 23 + 3 × 22 + 12 × 2 + 7)| = |16 – 27| = 11 m

Distance travelled in first 3 s = 20 + 11 = 31 m


4

a i x1⎝⎜⎛1

2⎠⎟⎞ =

⎝⎜⎛1

2⎠⎟⎞3

–⎝⎜⎛1

2⎠⎟⎞2

= 18

– 14

= – 18

18

cm to the left

ii a1(t) = d2xdt2

= 6t – 2

a1⎝⎜⎛1

2⎠⎟⎞ = 6 × 1

2– 2

= 1 cm/s2

iii v2(t) = dxdt

= 2t

v2 = 2 × 12

= 1 cm/s

b i x1(t) = x2(t)

t3 – t2 = t2

t3 – 2t2 = 0t2(t – 2) = 0

t = 0 and 2The particles will have the same position at the start and after 2 s.

ii Let the distance between the particles be y = |t3 – 2t2|. Define y = t3 – 2t2: dy

dt= 3t2 – 4t

= t(3t – 4) = 0 when t = 0 and 4

3

When t = 0, y = 0. When t = 4

3, y = 64

27– 32

9 = –1 5

27

When t = 2, y = 8 – 2 × 4 = 0

The maximum distance the particles are apart in the first 2 s is 32

27= 1 5

27 cm

5 a a = 6t

v = 3t2 + c When t = 0, v = 0. 0 = 0 + c

c = 0∴v = 3t2

When t = 2, v = 3 × 4 = 12 m/s b v = 3t2

x = t3 + d

When t = 0, x = 0. 0 = 0 + d

d = 0x = t3

Since the particle starts at the origin, its displacement s = x = t3.

6 a a = 3 – 2t

v = 3t2 – t2 + c When t = 0, v = 4. 4 = 0 – 0 + c

c = 4v = 3t – t2 + 4 = 0

– (t2 – 3t – 4) = 0– (t – 4)(t + 1) = 0

t = 4 After 4 s b v = 3t – t2 + 4

x = 3t2

2– t3

3+ 4t + d

When t = 0, x = 0. 0 = 0 – 0 + 0 + d

d = 0

x = 3t2

2– t3

3+ 4t

When t = 4, x = 3 × 42

2– 43

3+ 4 × 4

= 18 23

18 23

m to the right


c When t = 4, a = 3 – 2 × 4 = –5 m/s2 d a = 3 – 2t = 0

t = 1.5 s e When t = 1.5,

v = 3t – t2 + 4 = 3 × 1.5 – 1.52 + 4 = 6.25 m/s

7

a s = 2t3

3– 3t4

4+ c

When t = 0, s = 0. 0 = 0 – 0 + c

c = 0

s = 2t3

3– 3t4

4

When t = 1, x = 2 × 13

3– 3 × 14

4 = 2

3– 3

4= 1

12

112

m to the left.

b When t = 1, v = 2 – 3 = –1 m/s c a = dv

dt = 4t – 9t2

When t = 1, a = 4 × 1 – 9 × 12

= –5 m/s2

8 a v = 2t–2

a = dvdt

= –4t–3 = – 4t3

b v = 12

t–2

s = – 12

t–1 + c

When t = 1, s = 0. 0 = – 1

2× 1–1 + c

0 = – 12

+ c

c = 12

s = 12

– 12t

9 a a = dv

dt = 3t2 – 22t + 24

b Solve for v = 0. t3 – 11t + 24t = t(t – 3)(t – 8) Since motion is only defined for t ≥ 0, it cannot be said to change direction at t = 0. ∴ t = 3

a = 3 × 32 – 22 × 3 + 24 = –15 m/s2

c v = t3 – 11t2 + 24t

x = t4

4– 11t3

3+ 12t2 + c

When t = 0, s = 0 0 = 0 – 0 + 0 + c

c = 0

x = t4

4– 11t3

3+ 12t2

When t = 5, x = 54

4– 11 × 53

3+ 12 × 52

= –2 12

When t = 3, x = 34

4– 11 × 33

3+ 12 × 32

= 29 14

When t = 0, x = 0.

Total distance = 29 14

+⎝⎜⎛29 1

4+ 2 1

12⎠⎟⎞

= 60 712

m

2 112

m left of O, 60 712

m


10 u = 20, v = 0, t = 4s = 1

2 (u + v)t

= 12× 20 × 4

= 40 m

11 a u = 0, v = 30, t = 12

v = u + at30 = 12a

a = 3012

= 2.5 m/s2

b u = 30, v = 50, a = 2.5

v = u + at50 = 30 + 2.5t

2.5t = 20t = 8 s

c s = ut + 12

at2

= 0 + 12× 2.5 × 202

= 500 m

d 100 km/h = 100 ÷ 3.6 = 250

9 m/s

u = 0, v = 2509

, a = 2.5

v = u + at250

9= 2.5t

t = 2509 × 2.5

= 11 19 s

12 a 100 km/h = 100 ÷ 3.6 = 50

3 m/s

u = 0, v = 503

, a = 0.4

v = u + at503

= 0.4t

t = 503 × 0.4

= 41 23

s

b s = 12

(u + v)t

= 12× 50

3× 125

3 = 347 2

9 m

13 a u = 35, s = 0, a = –9.8

s = ut + 12

at2

0 = 3.5t – 4.9t2

0.7t(50 – 7t) = 0t = 50

7= 7 1

7 s

b u = 35, s = 60, a = –9.8

s = ut + 12 at2

60 = 35t – 49t2

4.9t2 – 35t + 60 = 049t – 250t + 600 = 0

(7t – 20)(7t – 30) = 0t = 2 6

7 or 4 2

7

After 2 67

s (going up) and 4 27

s (going

down) 14 a Maximum height occurs when v = 0. u = 19.6, a = –9.8, v = 0

v = u + at0 = 19.6 – 9.8tt = 19.6

9.8= 2 s

b s = ut + 1

2 at2

= 19.6 × 2 + 12× –9.8 × 22

= 19.6 m

With respect to ground level, height = 19.6 + 20 = 39.6 m


c u = 19.6, s = 0, a = –9.8s = ut + 1

2 at2

0 = 19.6t + 12× –9.8 × t2

0 = 19.6t – 4.9t2

0 = 4.9t(4 – t)t = 4 s

d u = 19.6, s = –20, a = –9.8

s = ut + 12

at2

–20 = 19.6t + 12× –9.8 × t2

–20 = 19.6t – 4.9t2

4.9t2 – 19. t – 20 = 049t2 – 196t – 200 = 0

Δ = b2 – 4ac = 1962 – 4 × 49 × –200 = 77 616

Δ ≈ 278.596

Since the discriminant is irrational, solve using the quadratic formula:

t = 196 ± 278.59698

≈ 4.84 or – 0.84 ≈ 4.84 s (since t > 0)

15

Distance = area = 1

2× 35 × 25

= 437.5 m

16

a Distance = trapezium area

= 12× (33 + 15) × 12

= 288 m

b Halfway point is 144 m. The car has travelled 1

2× 8 × 12 = 48 m

in the first 8 s. It must travel 144 – 48 = 96 m at 12 m/s.

This will take 96 ÷ 12 = 8 s. Total of 16 s. 17

Since the vehicle travels 1 km = 1000 m,

adding the two triangles together should give an area equal to a distance of 200 m.

The triangles have a combined base of 25. A = 1

2× 25 × V

= 200

V = 200 × 225

= 16 m/s

18 After 3 s, the first car has travelled 12 × 3 = 36 m.

Let the second car’s final velocity be V m/s. The two areas will be equal.

12× 27 × V = 12 × 30

= 360

V = 2 × 36027

= 803

For constant acceleration, acceleration = change in velocity

change in time= 80

3 × 27= 80

81 m/s2


19

a v = 102

4– 3 × 10 + 5

= 0 m/s

b a = dv

dt = 2t

4– 3

= t2

– 3

When t = 0, a = –3 m/s2. c Minimum velocity occurs when a = 0. t

2– 3 = 0

t = 6

When t = 6,

v = 62

4– 3 × 6 + 5

= –4 m/s

d v = t2

4– 3t + 5

x = t3

12– 3t2

2+ 5t + c

When t = 0, x = 0. 0 = 0 – 0 + 0 + c

c = 0

x = t3

12– 3t2

2+ 5t

Check for change of direction of velocity.

v = 0 if t2

4– 3t + 5 = 0

t2 – 12t + 20 = 0(t – 2)(t – 10) = 0

t = 2 or 10

There will be no change of direction of velocity in the first 2 s.

When t = 2,

x = 23

12– 3 × 22

2+ 5 × 2

= 23

– 6 + 10

= 4 23

m

e When t = 3,

x = 33

12– 3 × 32

2+ 5 × 3

= 94

– 272

+ 15

= 3 34

m

Distance travelled in the third second = 4 2

3– 3 3

4= 11

12 m (to the left)

20 a a = 2 – 2t

v = 2t = t2 + c When t = 3, v = 5. 5 = 2 × 3 – 32 + c

5 = –3 + cc = 8v = 2t – t2 + 8

b v = 2t – t2 + 8

x = t2 – t3

3+ 8t + d

When t = 0, x = 0. 0 = 0 – 0 + 0 + d

d = 0

x = t2 – t3

3+ 8t

21 a a = 4 – 4t

v = 4t – 2t2 + c When t = 0, v = 6. 6 = 0 – 0 + c

c = 6v = 4t – 2t2 + 6 = 6 + 4t – 2t2

b Minimum velocity occurs when a = 0. i 4 – 4t = 0

t = 1v = 6 + 4t – 2t2

= 6 + 4 × 1 – 2 × 12

= 8 m/s


ii 6 + 4 t – 2t2 = 64t – 2t2 = 0

2t(2 – t) = 0

23 a a = 3 – 3t

v = 3t = 3t2

2+ c

iii 6 + 4t – 2t2 = 0–2t2 + 4t + 6 = 0

t2 – 2t – 3 = 0(t – 3)(t + 1) = 0

t = 3

When t = 0, v = 2. 2 = 0 – 0 + c

c = 2

v = 3t – 3t2

2+ 2

x = – 2t3

3+ 2t2 + 6t + d

x = 0 when t = 0∴ d = 0

x = – 2t3

3+ 2t2 + 6t

When t = 4, v = 3 × 4 – 3 × 42

2+ 2

= –10 m

b v = 3t – 3t2

2+ 2

x = 3t2

2– t3

2+ 2t + d

When t = 3,

x = – 2 × 33

3+ 2 × 32 + 6 × 3

= 18 m

When t = 0, x = 0. 0 = 0 – 0 + 0 + d

d = 0

x = 3t2

2– t3

2+ 2t

22 a When t = 0, a = 27 m/s2.

When t = 4, x = 3 × 42

2– 43

2+ 24

= 24 – 32 + 8 = 0 m/s

b a = 27 – 4t2

v = 27t – 4t3

3+ c

When t = 0, v = 5. 24 5 = 0 – 0 + c

c = 5

v = 27t – 4t3

3+ 5

a t2 – 10t + 24 = 0

(t – 4)(t – 6) = 0t = 4 and 6

When t = 3, v = 27 × 3 – 4 × 33

3+ 5

= 50 m/s

b v = t2 – 10t + 24

x = t3

5– 5t2 + 24t + c

When t = 0, x = 0.

c v = 27t – 4t3

3+ 5 = 5

27t – 4t3

3= 0

81t – 4t3 = 0t(81 – 4t2) = 0

t(9 – 2t)(9 + 2t) = 0t = 4.5 s

0 = 0 – 0 + 0 + cc = 0

x = t3

3– 5t2 + 24t

When t = 3, x = 33

3– 5 × 32 + 24 × 3

= 36 m

c a = 2t – 10 < 0

2t < 10t < 5

Since t ≥ 0, 0 ≤ t ≤ 5


Chapter 20 – Statics of a particle Exercise 20A Solutions 1 Rearrange into a triangle of forces.

Using trigonometry, T1 = T2

= 5 sin 45°

= 5 22

kg wt

2 Rearrange into a triangle of forces.

∠ACB = ∠ACD + ∠ADC

These angles can be calculated using the cosine rule, but the student should notice that Δ is a 'doubled' 3-4-5 triangle with ∠

ACDCAD = 90°.

∴ACB = ∠ACD + ∠ADC = 180 – 90 = 90°


Using the cosine rule in the triangle in the original diagram, it is clear that:

cos ∠CAB = 152 + 102 – 122

2 × 15 × 10 = 0.6033

∠CAB = 52.89°∠ADC = 90 – ∠CAB

= 37.11°

cos ∠CBA = 152 + 122 – 102

2 × 15 × 12 = 0.7472

∠CBA = 41.65°∠ACD = 90 – ∠CBA

= 48.35°∠CAD = 180 – 37.11 – 48.35

= 94.54°

Use the sine rule to find T1 and T2.

T1

sin ∠ACD= 20

sin ∠CAD

T1 = 20 × sin 48.35°sin 94.54°

≈ 14.99 kg wtT2

sin ∠ADC= 20

sin ∠CAD

T2 = 20 × sin 37.11°sin 94.54°

≈ 12.10 kg wt


Using the cosine rule, F2 = 402 + 302 – 2 × 30 × 40 × cos 45°

= 802.94F ≈ 28.34 kg wt

Using the cosine rule,

cos x = F2 + 402 – 302

2 × F × 40 = 0.663x ≈ 48.5°

W 48.5° S or S 41.5° W


5 The angle between the plane and the horizontal is given by

tan x = 512

= 0.4167x ≈ 22.619°

Rearrange into a triangle of forces.

T = 104 sin x

= 40 kg wt

Note: The hypotenuse is 13, so sin x = 5

13 and cos x = 12

13.

N = 104 cos x = 96 kg wt

6 Note that F will be acting at 50 to the

horizontal and 70 to N, which becomes 110 when the force vectors are joined head to tail.

°°

°

Rearrange into a triangle of forces.

Use the sine rule.

12sin 30 sin110

12 sin 30 6.39 kg wtsin110

F

F

=° °× °

= ≈°

7 In each case, the particle will be in

equilibrium if the forces add to zero. Draw the first two forces, and calculate the third force required for equilibrium.

a

Use the cosine rule to calculate the magnitude of the third force. F2 = 102 + 62 – 2 × 10 × 6 cos 100°

= 156.837F ≈ 12.52 kg wt

This is not the force in the diagram, so these forces will not be in equilibrium.

b

Use the cosine rule to calculate the magnitude of the third force. F2 = 42 + (2 3 )2 – 2 × 4 × 2 3 × cos 30°

= 4F = 2 kg wt

It has the same magnitude as the third force in the diagram.

Use the sine rule to find x.

sin x4

= sin 30°2

sin x = 0.5 × 42

= 1

x = 90°

This vector is at the same angle with the 2 3 vector as in the original diagram. ∴ the vectors will be in equilibrium.


8 Draw the triangle of forces and use the cosine rule to find the three angles.

When the vectors are placed tail to tail, the angles between them will be the supplements of the angles in the triangle.

cos x = 72 + 42 – 102

2 × 7 × 4 = 0.625x ≈ 128°41'

Angle between vectors is 180° – 128°41' = 51°19'

cos y = 102 + 42 – 72

2 × 10 × 4 = 0.8375y ≈ 33°7'

Angle between vectors is 180° – 33°7' = 146°53' z ≈ 180° – 128°41' – 33°7'

= 18°12'

Angle between vectors is 180° – 18°12' = 161°48' 9 a

Draw the triangle of forces.

T = 15 sin 30°

= 7.5 kg wt b The situation will be the same, except

that the 30° angle will now be 40°.

T = 15 sin 40° ≈ 9.64 kg wt

c The angle between T and N is now 80°.

Draw the triangle of forces.

Use the sine rule. T

sin 30° = 15sin 100°

T = 15 × 0.5sin 100°

≈ 7.62 kg wt

10

Draw the triangle of forces for point B.

Use the sine rule to find T1 and T2.

T1

sin 110° = 12sin 20°

T1 = 12 × sin 110°sin 20°

≈ 32.97 kg wtT2

sin 50° = 12sin 20°

T2 = 12 × sin 50°sin 20°

≈ 26.88 kg wt


Now draw the triangle of forces for point C.

Use the sine rule to find T3.

T3

sin 70° = T2

sin 40°

T3 = 26.88 × sin 70°sin 40°

≈ 39.29 kg wt Since the triangle is isosceles, W = T3 ≈ 39.29 kg wt

The mass of W is 39.29 kg.


Exercise 20B Solutions 1 F cos 40° = 10 kg wt

F = 10cos 40°

≈ 13.05 kg wt

2 Resolve in the direction of F. F – 10 cos 55° = 0

F = 5.74 kg wt

3 First resolve vertically to find N. N cos 25° – 8 = 0

N = 8cos 25°

≈ 8.83 kg wt

Keep the exact value of N in your calculator.

Resolve horizontally. F – N sin 25° = 0

F = N sin 25° ≈ 3.73 kg wt

sin 25 0sin 25 3.73 kg wt

F NF N− ° == ° ≈

4 Resolve parallel to the plane, i.e.

perpendicular to N. F is at an angle of 34° to the plane. F cos 34° – 10 sin 20° = 0

F = 10 sin 20°cos 34°

≈ 4.13 kg wt

5

Resolve vertically: T cos 30° – 12 = 0

T = 12cos 30°

Resolve horizontally: F – T sin 30° = 0

F = T sin 30°

= 12 sin 30°cos 30°

≈ 6.93 kg wt

6

Resolve parallel to the plane. 20 – W sin 40° = 0

W = 20sin 40°

≈ 31.11 kg wt

The force W exerts on the plane is the part of its weight resolved perpendicular to the plane.

F = W cos 40°

= 20 cos 40°sin 40°

= 23.84 kg wt


9 First find the angle between the string and the vertical.

7

sin x = 915 + 9

= 0.375x = 22.024°

Resolve vertically. T cos x – 3 = 0

T = 3cos 22.024°

≈ 3.24 kg wt

First resolve horizontally so only one

unknown is involved. 30 sin x – 20 sin 35° = 0

sin x = 20 sin 35°30

= 0.382x ≈ 22°29'

Keep the exact value in your calculator and resolve vertically. 0 = W – 20 cos 35° – 30 cos 22.481°W = 20 cos 35° + 30 cos 22.481° ≈ 44.10

8 Pressure of body on plane = 1 0 cos 50°

≈ 6.43 kg wt Resolve parallel to the plane. T – 10 sin 50° = 0

T = 10 sin 50° ≈ 7.66 kg wt

Resolve parallel to the second plane. T – W sin 40° = 0

W = Tsin 40°

= 10 sin 50°sin 40°

≈ 11.92 kg wt


Solutions to Multiple-choice Questions 1 50 cos 60° = 50 × 1

2

= 25 N E 2 Resolve perpendicular to the plane. N – 20 cos 30° = 0

N = 20 cos 30°

= 20 × 32

= 10 3 kg wt E 3 Resolve parallel to the plane. F – 20 sin 30° = 0

F = 20 sin 30°

= 20 × 12

= 10 kg wt A 4 Use Pythagoras’ theorem.

Resultant = 52 + 42 = 41 kg wt C 5 For the particle to be in equilibrium, B must equal the sum of the forces on

A and C. ∴ B = A cos 60° + C cos 30° (since 180 – 150 = 30). As this is true, C cannot be true. C 6

Since the forces are perpendicular,

a2 + 72 = 92

a2 = 81 – 49 = 32

a = 32 = 4 2 B 7 The angle between the forces when they

are head to tail will be 120°. Use the cosine rule. F2 = 202 + 202 – 2 × 20 × 20 × cos 120°

= 400 + 400 – 800 × – 12

= 1200F = 1200

= 20 3 kg wt B 8 The angle between the forces when they

are head to tail will be 120°. Use the cosine rule. F2 = 3002 + 2002 – 2 × 300 × 200 × cos 120°

= 90 000 + 40 000 – 120 000 × – 12

= 190 000F = 190 000

= 100 19 kg wt A

9 R = 162 + 302

= 1156

= 34 kg wt C 10 The forces will be at right angles to each

other. a2 + 82 = 122

a2 = 144 – 64 = 80

a = 80 = 4 5 B


Solutions to Short-answer Questions 1 Note that the two strings form a 3-4-5

triangle. Draw the triangle of forces.

Note: sin x = 6

10= 3

5; cos x = 8

10= 4

5

T1 = 15 sin x = 15 × 3

5= 9 kg wt

T2 = 15 cos x = 15 × 4

5= 12 kg wt

2 Draw the triangle of forces.

Use the cosine rule. F2 = 102 + 102 – 2 × 10 × 10 × cos 120°

= 100 + 100 – 200 × – 12

= 300F = 300 = 10 3 kg wt

Since the triangle is isosceles, x = 180 – 120

2 = 30°

10 3 kg wt, at 150° to each 10 kg wt force.

3 The force exerted on the body by the plane will be perpendicular to the plane. Resolve parallel to the plane, so the component of this force will be zero.

The hypotenuse of the marked triangle is

h = 122 + 62

= 180 = 6 5 cm

If x is the angle of the plane to the horizontal, sin x = 6

6 5= 1

5cos x = 12

6 5= 2

5

Resolving, T – 70 sin x = 0

T = 70 sin x = 70 × 1

5

= 70 55

= 14 5 kg wt

Resolving perpendicular to the plane,

N – 70 cos x = 0N = 70 cos x = 70 × 2

5

= 140 55

= 28 5 kg wt

4 The force exerted on the body by the

plane will be perpendicular to the plane. Resolve parallel to the plane, so the component of this force will be zero. F cos 30° – 15 sin 30° = 0

F 32

= 15 × 12

F = 153

= 15 33

= 5 3 kg wt


5 Draw the triangle of forces and use the cosine rule.

cos x = 122 + 52 – 82

2 × 15 × 5 = 105

120= 7

8

Since the required angle is 180° – x, the cosine is – 7

8.

6 F cos 30° = 20

F 32

= 20

F = 20 × 23

= 40 33

kg wt

7 Resolve parallel to the plane. F – 15 sin 45° = 0

F = 15 sin 45°

= 15 22

kg wt

8

Resolve parallel to the plane. W sin 30° – 14 = 0

W = 14sin 30°

= 140.5

= 28 kg wt

Resolve perpendicular to the plane.

N – 28 cos 30° = 0N = 28 cos 30°

= 28 32

= 14 3 kg wt 9

Calculate F by resolving parallel to the plane. F cos 30° – 12 sin 30° = 0

F 32

= 12 × 12

F = 6 × 23

= 12 33

= 4 3 kg wt


Chapter 21 – Revision of chapters 19 and 20 Solutions to Multiple-choice Questions 1 x = at2 + bt + c c = –12 D

2 v = dxdt

= 4t – 5

When t = 0, v = –5 cm/s E

3 a = dvdt

= 4

When t = 0, a = 4 cm/s2 B 4 v = 0

4t – 5 = 0 t = 5

4= 1.25 s B

5 x = 0

2t2 – 5t – 12 = 0(2t + 3)(t – 4) = 0

t = 4 s C 6 t = 3

x = 2 × 32 – 5 × 3 – 12

= –9 cm C

7 Average velocity = change in positionchange in time

= –9 – –123

= 1 cm/s A 8 The direction of velocity changes at t = 1.25. Position at t = 1.25 = 2 × 1.252 – 5 × 1.25 – 12

= –15.125 cm Distance travelled from t = 0 to t = 1.25 = –12 – –15.125 = 3.125 cm Distance travelled from t = 1.25 to t = 3 = –9 – –15.125 = 6.125 cm Distance travelled in the first 3 seconds = 3.125 + 6.125 = 9.25 cm D

9 Average speed = distance travelledchange in time

= 9.253

= 3 112

cm/s C

10 v = u + at

= 15 – 10 × 3 = –15 cm/s B 11 v = u + at

0 = 15 – 10t t = 15

10= 1.5 s D

12 Maximum height occurs when v = 0, i.e. when t = 1.5 s s = ut + 1

2 at2

= 15 × 1.5 – 12× 10 × 1.52

= 11.25 m D 13 s = ut + 1

2 at2

0 = 15t – 5t2

5t(t – 3) = 0

After 3 s, since t = 0 is the initial projection. C

14 Distance = area of trapezium = 1

2× (14 + 6) × 20

= 200 m E 15 For constant acceleration, acceleration = change in velocity

change in time= 20

5

= 4 m/s2 D


16 Resolve perpendicular to F2. The angle between F1 and F2 extended

back is 100 + 120 – 180 = 40°. F1 sin 40° – 8 sin 60° = 0

F1 = 8 sin 60°sin 40°

≈ 10.78 A kg wt 17 Resolve perpendicular to F1. The angle between F2 and F1 extended

back is 100 + 120 – 180 = 40°. The angle between the 8 kg wt force and

F1 extended back is 120 – 40 = 80°. F2 sin 40° – 8 sin 80° = 0

F2 = 8 sin 80°sin 40°

≈ 12.26 D kg wt 18 Resolve perpendicular to the plane.

N – 10 cos 25° = 0

N = 10 cos 25° ≈ B 9.06 kg wt 19 Resolve parallel to the plane. F – 10 sin 25° = 0

F = 10 sin 25°

≈ A 4.23 kg wt

20 Draw the triangle of forces. β = 180° – α

Use the cosine rule to find β.

cos β = 122 + 102 – 142

2 × 12 × 10 = 0.2β ≈ 78°

C α ≈ 180 – 78 = 102° 21 Draw the triangle of forces. φ = 180° – θ

Use the cosine rule to find φ.

cos φ = 122 + 17.52 – 152

2 × 12 × 17.5 = 0.536φ ≈ 57°

θ ≈ C 180 – 57 = 122° 22 E 60 cos 30° ≈ 51.96 kg wt 23 B 60 sin 30° = 30 kg wt


Chapter 22 – Describing the distribution of a single variable Exercise 22A Solutions 1 a 'Type of toy' is categorical data. b 'Number of students' is numerical data. c 'Favourite colour' is categorical data. d 'Weight' is categorical data e Although derived from numbers, 'light',

'medium' and 'heavy' are not numerical values, so this is categorical data.

f 'Level of enthusiasm' is categorical data.

2 a IQ is a numerical value, so this is

numerical data. b While it uses number for labelling, the

survey is actually checking attitudes, so this is categorical data.

3 a Discrete – the number of pages can only

be an integer. b Discrete – the cost can only be a whole

number of cents. c Continuous – the volume could take any

value, including fractional values. d Continuous – any amount of time could

pass, including seconds and fractions of seconds.

e Discrete – the number of tosses can only

be an integer.


Exercise 22B Solutions 1 a

b Hamburgers are the most popular, with

23 students selecting them. 2 a

b From the table, 21 + 11 = 32

respondents.

3 a

b Music is the least popular type of film,

with only 15 rentals. 4 a

b Watching TV was the most common

activity, chosen by 42% of students.



Number 0 1 2 3 4 5 6 Frequency 4 4 4 4 3 2 1

2 a 'Only child' equates to family size = 1. Reading from the graph, the answer is 4 students. b The most common number is the

column with the greatest height. 2 children (height = 10). c Add up the heights of the columns for family size = 6 and higher. 2 + 1 + 0 + 2 + 0 = 5 students d Add the heights of all the columns. 4 + 10 + 5 + 4 + 0 + 2 + 1 + 0 + 2 + 0 = 28 students 3 a Read the height of the 10-19 column. Zero (0) students. b Add all the heights of the columns. 2 + 3 + 3 + 7 + 9 + 10 + 8 + 4 + 2 = 48 students c The modal class is the column with the

greatest height. 60–69 students. d Count all the columns to the right of 50. 9 + 10 + 8 + 4 + 2 = 33 students

4 a, b Divide each frequency by the total

number of cities, 32, to get the relative frequency

Temp. (°C) Freq. Relative Freq. 0 – 1 0.03 5 – 0 0 10 – 1 0.03 15 – 9 0.28 20 – 4 0.13 25 – 5 0.16 30 – 7 0.22 35 – 4 0.13 40 – 0 0 45 – 1 0.03

c

d Count these cities using the values in the

table. 1 + 1 + 9 + 4 = 15 out of a total 32 cities. 15

32× 100 ≈ 47%

5 a

b The modal class is the column with the

greatest height. $5.00–$5.99


c Prices ($) Cumulative

Freq. less than 5 3 less than 10 9 less than 15 12 less than 20 15 less than 25 19 less than 30 19 less than 35 20 less than 40 20 less than 45 21

6 a

b

Measurement (cm) CumulativeFreq.

less than 28 0 less than 29 1 less than 30 3 less than 31 11 less than 32 20 less than 33 27 less than 34 28

c The students' estimates ranged from 28.9 cm to 33.3 cm, with most students (86%) over-estimating the 30 cm measure.

7 a

b

Marks Cumulative Freq.

less than 10 0 less than 20 0 less than 30 1 less than 40 3 less than 50 8 less than 60 16 less than 70 22 less than 80 26 less than 90 27 less than 100 30

c The students' marks ranged from 21 to 99,

with most students (over 70%) scoring more than 50% on the test.

8 a


b Length of hole (m) Cumulative

Freq. less than 240 0 less than 260 1 less than 280 6 less than 300 8 less than 320 17 less than 340 30 less than 360 33 less than 380 45 less than 400 47 less than 420 50

c i From the graph, there are about 8 out of 50 = 4

25

ii From the graph, there are about 17 holes longer than 360 m = 17

50

iii 90% of 50 holes is 45 holes. From the graph, 90% of holes are longer than 270 m.


Exercise 22D Solutions 1 a Both distributions have the same spread,

but the one on the left has a lower centre than the one on the right.

Differ in centre. b Both distributions have the same spread

and the same centre. Differ in neither. c The distribution on the right has a higher

centre and a wider spread than the one on the left.

Differ in both. 2 a The histogram has a short tail to the left

and a long tail to the right, so it is positively skewed.

b The histogram has a long tail to the left

and a short tail to the right, so it is negatively skewed.

c While not perfectly symmetric, this

histogram is still approximately symmetric.

3 While not perfectly symmetric, this histogram is still approximately symmetric.

4 While not perfectly symmetric, this


5 While not perfectly symmetric, this



Exercise 22E Solutions 1 a

b There are 2 leaves on the '6' stem, so

two months. 2 a

b There are 7 leaves on the '3' stem, and 2

leaves on the '4' stem. 7 + 2 = 9 batteries. 3 a

b There are 2 leaves on the '6' stem and

one leaf on the '7' stem, so 3 students. c As a histogram, this would have a short

tail to the left and a long tail to the right, so the distribution is positively skewed.

4 a

b The distribution is approximately

symmetric. There are outliers at the high end, but these can be ignored.

5 a

b Both distributions are approximately

symmetric and have a similar spread. Fathers tend to have ages centred in the

late 40s, compared to mothers who tend to have aged centred in the early 40s.


6 a

b Marks less than 50% are on stems

1 through 4. Six students from Class A and 2

students from Class B scored less than 50%.

c Class B performed better, as more

students scored in the higher 70% to 90% levels.


Exercise 22F Solutions 1 a Mean = 29 + 14 + 11 + 24 + 14 + 14 + 28 + 14 + 18 + 22 + 14

11= 202

11≈ 18.36

Arrange the data in order: 11 14 14 14 14 14 18 22 24 28 29 The median is the middle number: 14. b Mean = (Sum of numbers) ÷ n = 147

16≈ 9.19

Arrange the data in order: 3 3 5 5 6 6 7 9 11 12 12 12 13 13 15 15 The median is the average of the two middle numbers: 9 + 11

2= 10

c Mean = (Sum of numbers) ÷ n = 74.1

10= 7.41

Arrange the data in order: 5.6 6.5 7.0 7.1 7.5 7.8 7.9 8.2 8.2 8.3 The median is the average of the two middle numbers: 7.5 + 7.8

2= 7.65

d Mean = (Sum of numbers) ÷ n = 38.9

24≈ 1.62

Arrange the data in order: 0.1 0.2 0.2 0.2 0.5 0.5 0.6 0.7 0.7 0.9 1.0 1.1 1.2 1.3 1.5 1.7 1.7 2.0 2.7 3.1 3.2 3.4 4.6 5.8 After putting the data in order, the two middle numbers are 1.1 and 1.2. Median = 1.1 + 1.2

2= 1.15

2 Mean = (Sum of numbers) ÷ n a Sum of numbers

= (1 × 6) + (2 × 3) + (3 × 10) + (4 × 7) + (5 × 8) = 110

Total data (n) = 6 + 3 + 10 + 7 + 8 = 34 Mean = 110

34≈ 3.24

The middle two data would be the 17th and 18th scores, which are both be 3.

Median = 3

b Sum of numbers = (–2 × 5) + (–1 × 8) + (0 × 11) + (1 × 3) + (2 × 2) = –11 Total data (n) = 5 + 8 + 11 + 3 + 2 = 29 Mean = – 11

29≈ – 0.38

The median is the 15th score, which is 0.


3 Mean = (Sum of numbers) ÷ n = $2 707 400

14≈ $193 386

The median is the average of the two middle numbers.

After putting the data in order, the two middle numbers are both $140 000.

Median = $140 000 The median is a better measure, as it is

more like a typical price for a house. 4 To find the mean, add a third row to

calculate the products, or partial sums.

No. of days missed

0 1 2 3 4 5 6 9 21

No. of students

4 2 14 10 16 18 10 2 1

Product 0 2 28 30 64 90 60 18 21

Mean = 0 + 2 + 28+ 30+ 64+ 90+ 60+ 28+ 214 + 2 + 14+ 10+ 16+ 18+ 10+ 2 + 1

= 31377

= 4.06

The median is the 39th score, which is 4. Both measures are reasonable. 5 In each case, first put the data in order. a 510 560 630 715 718 750 1002 1085 1093 1112 Range = 1112 – 510 = 602 The median of the first five points is 630. The median of the second five points is 1085. Interquartile range = 1085 – 630 = 455 b –1.6 –1.2 –1.0 0.2 0.7 0.8 3.4 3.7 Range = 3.7 – –1.6 = 5.3 The median of the first four points is –1.2 + –1.0

2= –1.1.

The median of the second four points is 0.8 + 3.4

2= 2.1.

Interquartile range = 2.1 – –1.1 = 3.2

c 8.39 8.51 8.51 8.54 8.56 8.58 8.62 8.82 8.96 Range = 8.96 – 8.39 = 0.57 Exclude the median. The median of the first four points is 8.51. The median of the last four points is 8.62 + 8.82

2= 8.72.

Interquartile range = 8.72 – 8.51 = 0.21 d 15 16 16 17 18 18 19 19 20 20 21 22 Range = 22 – 15 = 7 The median of the first six points is 16 + 17

2= 16.5.

The median of the last six points is 20. Interquartile range = 20 – 16.5 = 3.5 6 First put the data in order. 159 161 189 190 192 193 196 199 203 206

209 209 224 225 231 238 244 248 276 304 a Range = 304 – 159 = 145 b The median of the first ten points is 192 + 193

2= 192.5.

The median of the last ten points is 231 + 238

2= 234.5.

Interquartile range = 234.5 – 192.5 = 42 7 a Range = 4.5 – 2.1 = 2.4 kg b There are 20 data points here. The median of the first ten points is 2.9 + 3.1

2= 3.0 kg.

The median of the second ten points is 3.9 + 4.1

2= 4.0 kg.

Interquartile range = 4.0 – 3.0 = 1.0 kg


8 a Mean = 276

12= 23

Construct a table. xi xi – x (xi – x)2

30 7 49 16 –7 49 22 –1 1 23 0 0 18 –5 25 18 –5 25 14 –9 81 56 33 1089 13 –10 100 26 3 9

9 –14 196 31 8 64

Σxi = 276 Σ(xi – x) = 1688

The standard deviation s = 168811

≈ 12.39

b Mean = $73.21

18= $4.07


2.52 –1.55 2.4025 4.38 0.31 0.0961 3.60 –0.47 0.2209 2.30 –1.77 3.1329 3.45 –0.62 0.3844 5.40 1.33 1.7689 4.43 0.36 0.1296 2.27 –1.80 3.2400 4.50 0.43 0.1849 4.32 0.25 0.0625 5.65 1.58 2.4964 6.89 2.82 7.9524 1.98 –2.09 4.3681 4.60 0.53 0.2809 5.12 1.05 1.1025 3.79 –0.28 0.0784 4.99 0.92 0.8464 3.02 –1.05 1.1025

Σxi = 73.21 Σ(x –1 x) = 29.8503

The standard deviation s = 29.850317

≈ $1.33

c Mean = 374010

= 374


200 –174 30 276 300 –74 5476 950 576 331 776 200 –174 30 276 200 –174 30 276 300 –74 5476 840 466 217 156 350 –24 576 200 –174 30 276 200 –174 30 276

Σxi = 3740 Σ(xi – x) = 711 840

The standard deviation s 711 8409

=

≈ 281.24 d Mean = 1493

19≈ 78.58


86 7.42 55.0564 74 –4.58 20.9764 75 –3.58 12.8164 77 –1.58 2.4964 79 0.42 0.1764 82 3.42 11.6964 81 2.42 5.8564 75 –3.58 12.8164 78 –0.58 0.3364 79 0.42 0.1764 80 1.42 2.0164 75 –3.58 12.8164 78 –0.58 0.3364 78 –0.58 0.3364 81 2.42 5.8564 80 1.42 2.0164 76 –2.58 6.6564 77 –1.58 2.4964 82 3.42 11.6964

Σxi = 1493 Σ(x1 – x) = 166.6316

The standard deviation s 166.631618

=

≈ 3.04


9 a i Mean = 634

36≈ 17.61


41 23.39 547.0921 16 –1.61 2.5921

6 –11.61 134.7921 21 3.39 11.4921

1 –16.61 275.8921 21 3.39 11.4921

5 –12.61 159.0121 31 13.39 179.2921 20 2.39 5.7121 27 9.39 88.1721 17 –0.61 0.3721 10 –7.61 57.9121

3 –14.61 213.4521 32 14.39 207.0721

2 –15.61 243.6721 48 30.39 923.5521

8 –9.61 92.3521 12 –5.61 31.4721 21 3.39 11.4921 44 26.39 696.4321

1 –16.61 275.8921 56 38.39 1473.7921

5 –12.61 159.0121 12 –5.61 31.4721

3 –14.61 213.4521 1 –16.61 275.8921

13 –4.61 21.2521 11 –6.61 43.6921 15 –2.61 6.8121 14 –3.61 13.0321 10 –7.61 57.9121 12 –5.61 31.4721 18 0.39 0.1521 64 46.39 2152.0321

3 –14.61 213.4521 10 –7.61 57.9121

Σxi = 634 Σ(x1 – x) = 8920.5556

The standard deviation

s = 8920.555635

≈ 15.96

ii Mean = 468724

≈ 195.29


141 –54.29 2947.4041 260 64.71 4187.3841 164 –31.29 979.0641 235 39.71 1576.8841 167 –28.29 800.3241 266 70.71 4999.9041 150 –45.29 2051.1841 255 59.71 3565.2841 168 –27.29 744.7441 245 49.71 2471.0841 258 62.71 3932.5441 239 43.71 1910.5641 152 –43.29 1874.0241 141 –54.29 2947.4041 239 43.71 1910.5641 145 –50.29 2529.0841 134 –61.29 3756.4641 150 –45.29 2051.1841 237 41.71 1739.7241 254 58.71 3446.8641 150 –45.29 2051.1841 265 69.71 4859.4841 140 –55.29 3056.9841 132 –63.29 4005.6241

Σxi = 4687 Σ(x1 – x) = 64 394.9584

The standard deviation

s = 64 394.958423

≈ 52.91

b i Interval = [17.61 – 2 × 15.96 , 17.61 + 2 × 15.96] = [0, 49.53] 34 out of 36 = 34

36× 100 = 94%

ii Interval = [195.29 – 2 × 52.91 , 195.29 + 2 × 52.91] = [89.47, 301.11]

24 out of 24 = 100%


10 a

Age Cmlt. Freq. Cmlt. Rel. Freq.

17 or less 7 0.194 18 or less 23 0.639 19 or less 26 0.722 20 or less 29 0.806 21 or less 30 0.833 24 or less 31 0.861 25 or less 32 0.889 31 or less 33 0.917 41 or less 34 0.944 44 or less 35 0.972 45 or less 36 1.0

Read off the 50%, 25% and 75% mark

from the vertical axis. From the data, the median is 18. The middle of the top half is the first

quartile, also 18. The middle of the bottom row is 20. Interquartile range = 20 – 18 = 2 b Mean = 755

36≈ 20.97

Using a table or calculator,

s = 190135

≈ 7.37

c Interval = [20.97 – 2 × 7.37, 20.97 + 2 × 7.37] = [6.23, 35.71] 33 out of 36 = 33

36× 100 = 92%

11 a i Mean = 54.3

8≈ 6.79

Sort the data from lowest to highest. 5.8 5.9 6.0 6.2 7.3 7.4 7.4 8.3 The median is 6.2 + 7.3

2= 6.75

ii Using a table or calculator,

s = 6.028757

≈ 0.93

The median of the first four points is 5.9 + 6.0

2= 5.95

The median of the last four points is 7.4 Interquartile range = 7.4 – 5.95 = 1.45 b i Mean = 108.3

8≈ 13.54

Sort the data from lowest to highest. 5.8 5.9 6.2 7.3 7.4 7.4 8.3 60 The median is 7.3 + 7.4

2= 7.35

ii Using a table or calculator,

s = 2472.478757

≈ 18.79

The median of the first four data is 5.9 + 6.2

2= 6.05

The median of the last four data is 7.4 + 8.3

2= 7.85

Interquartile range = 7.85 – 6.05 = 1.7

c The error has little effect on the median

or interquartile range. It doubles the mean and increases the

standard deviation by a factor of 20. 12 This interval would be two standard

deviations each side of the mean. Interval = [$50 – 2 × $3, $50 + 2 × $3] = [$44, $56] 95% of the shares lie between $44 and $56. 13 Note that $150 = $550 – 2 × $200 and

$950 = $550 + 2 × $200. As these values are two standard

deviations each side of the mean, it would be on 95% of days.


Exercise 22G Solutions 1 Write the heights in order from lowest to

highest. a 123 123 132 133 135 140 140

143 143 145 149 150 154 154 154 156 157 157 157 159 160

163 165 167 167 168 176 180 Median m = 154 Q1 = 140 + 143

2= 141.5

Q3

= 160 + 1632

= 161.5

Minimum = 123 Maximum = 180 b

c The distribution is slightly negatively

skewed, centred at 154 cm, with the middle 50% of heights ranging from 141.5 cm to 161.5 cm.

2 Write the numbers in order from lowest

to highest. a 0 0 0 0 0 0 0 0 0 0 0

1 1 1 1 1 2 2 2 4 4 5 7 8 10 11 11 12 13 13 14 15 18 27 28 28 38 52

Median m = 3 = 0 1Q = 13 3Q Minimum = 0 Maximum = 52 b Interquartile range = 13 – 0 = 13 Outliers can only be greater than 13 + 1.5 × 13 = 32.5 There are two outliers, 38 and 52. The new maximum for the boxplot

becomes 28. c

d The distribution is positively skewed, centred at 3.

While 75% of people borrowed 13 books or less, one student borrowed 38 books and another borrowed 52 books.

3 a The winnings are already in order. Median m = $854 533 Q1 = $748 662 + $755 795

2 = $752 228.50

Q3 = $1 639 171 + $1 697 1552

= $1 668 163

Minimum = $704 105

Maximum = $6 357 547

Interquartile range = $1 668 163 – $752 228.50 = $915 934.50 Outliers could only be greater than $1 697 155 + 1.5 × $915 934.50 = $3 071 057 The new maximum for the boxplot

becomes $2 766 051. Use this information to draw the

boxplot.

b The distribution is extremely positively

skewed, with a median value of $854 533. The middle 50% of players won from $75 228.50 to $1 573 674. There is one clear outlier, Roger Federer,

winning $6 357 547. Lleyton Hewitt is a marginal outlier, winning $2 766 051.


4 Write the amounts in order from lowest to highest.

a $4.75 $6.25 $6.75 $7.90 $8.40 $8.50 $8.50 $8.89 $9.00 $10.00

$10.00 $10.80 $10.90 $11.65 $11.69 $12.00 $12.34 $12.46 $13.00 $17.23

Median m = $10.00 Q1 = $8.40 + $8.50

2= $8.45

Q3 = $11.69 + $12.002

= $11.85

Minimum = $4.75 Maximum = $17.23

Interquartile range = $11.85 – $8.45 = $3.40 Outliers can be greater than $11.85 + 1.5 × $3.40 = $16.95 or less than $8.45 – 1.5 × $3.40 = $3.35 There is one outlier, $17.23 The new maximum for the boxplot

becomes $13.00. Use this information to draw the

boxplot.

b The distribution is symmetric, centred at

$10.00. The middle 50% of students earn $8.15–$11.85 per hour.

5 The circulations are already in order, just read them from the bottom to the top.

a Median m = 212 770 Q1 = 77 500 + 98 158

2= 87 879

Q3 = 258 700 + 273 2482

= 265 974

Minimum = 17 398 Maximum = 570 000

Interquartile range = 265 974 – 87 879 = 178 095 Outliers can only be greater than 265 974 + 1.5 × 178 095 = 533 116. There is one outlier, 570 000. The new maximum for the boxplot

becomes 327 654. Use this information to draw the

boxplot.

b The distribution is approximately

symmetric, centred at roughly 210 000, with an outlier at 570 000.

The middle 50% of papers have circulations of about 88 000 to 270 000.


Exercise 22H Solutions 1 a In both cases, put the data in order and

find the five figure summary. Before: 12 17 20 21 22 22 23 24 25 26 26 26 28 29 29 30 30 31 34 46 Median m = 26 Q1 = 22

Q3 = 29 + 302

= 29.5

Minimum = 12 Maximum = 46 Interquartile range = 29.5 – 22 = 7.5 In this case, outliers can be greater than

29.5 + 1.5 × 7.5 = 40.75 or less than 22 –1.5 × 7.5 = 10.75 There is one outlier, 46. The new maximum for the boxplot

becomes 34.

After: 15 19 21 24 25 25 26 28 29 30 30 32 33 34 34 35 36 43 50 54 Median m = 30 Q1 = 25

Q3 = 34 + 352

= 34.5

Minimum = 15 Maximum = 54 Interquartile range = 34.5 – 25 = 9.5 Outliers can be greater than 34.5 + 1.5 × 9.5 = 48.75 or less than 25 –1.5 × 9.5 = 10.75 There are two outliers, 50 and 54. The new maximum for the boxplot

becomes 43.

b The distribution is negatively skewed

before the course, centred on 26. After the course, the distribution is more

symmetric, centred on 30, which suggests that the course was effective.

2 In both cases, put the data in order and find the five figure summary.

Year 8: 0 1 1 1 1 1 1 2 2 2 2 3 3 3 3 3 3 3 4 4 4 4 4 4 5 7 7 7 Median m = 3 Q1 = 1 + 2

2= 1.5

Q3 = 4

Minimum = 0 Maximum = 7 Interquartile range = 4 – 1.5 = 2.5 Outliers can only be greater than 4 + 1.5 × 2.5 = 7.75 There are no outliers. Year 12: 1 1 1 1 2 2 2 3 3 3 4 4 5 5 6 6 6 7 7 7 7 7 7 7 8 8 8 9 Median m = 5 + 6

2= 5.5

Q1 = 2 + 32

= 2.5

Q3 = 7

Minimum = 1 Maximum = 9 Interquartile range = 7 – 2.5 = 4.5 Outliers can only be greater than 7 + 1.5 × 4.5 = 13.75 There are no outliers.

a From the boxplot, it can be seen that Year

12 students do the most homework. b From the boxplot, it can be seen that the

Year 12 group has a larger variance than the Year 8 group.


3 a In both cases, put the data in order and

find the five figure summary. 1970: 16 17 18 18 19 19 20 20 21 21 21 21 21 22 22 22 23 24 24 24 25 25 25 26 26 26 27 27 29 30 31 31 32 33 34 36 36 37 37 39 Median m = 24 + 25

2= 24.5

Q1 = 21Q3 = 30 + 31

2= 30.5

Minimum = 16 Maximum = 39 Interquartile range = 30.5 – 21 = 9.5 Inspecting the data, it is clear there are

no outliers.

1990: 17 18 18 19 19 20 22 23 23 23 24 24 24 25 26 26 27 28 28 28 28 28 29 29 29 30 31 32 32 33 33 34 35 35 38 39 41 43 44 46 Median m = 28 Q1 = 23 + 24

2= 23.5

Q3 = 33

Minimum = 17 Maximum = 46 Interquartile range = 33 – 23.5 = 9.5 Outliers can be greater than 33 + 1.5 × 9.5 = 47.25, or less than 23.5 – 1.5 × 9.5 = 9.25 There are no outliers.

b The distributions of ages in both groups

are slightly positively skewed, with the mothers in 1970 (median = 24.5) generally younger than the mothers in 1990 (median = 28).

The variability in both groups is the same (IQR = 10 for both groups).


Solutions to Multiple-choice Questions 1 'Level of Exercise' is not numerical, so

therefore it must be categorical. D 2 The vertical scale is 2% for each dotted line. There is 2% in the 20–30 column. B 3 The median is in the interval that

contains 50%, as you add the percentages.

4% + 2% + 33% = 39% of the employees work less than 40 hours per week.

39% + 39% = 78% % of the employees work less than 50 hours per week.

∴ the median will be in the interval 40–50. D 4 There are 19 employees, listed in order.

The median will be the 10th in this list, which corresponds to 7 meetings. C

5 Ignore the median, and find the medians

of the first 9 and the last 9 data points. This will be the 5th and 15th employee,

corresponding to 4 and 14 meetings respectively.

The interquartile range is 14 – 4 = 10. D

6

Reading from the graph, the number is

closest to 150. (It is a little less than 150.) C

7 The boxplot has a short tail to the left

and a long tail to the right, so it is positively skewed. A

8 Reading from the boxplot, the

interquartile range is about 17 – 5 = 12. D 9 The typical wage is shown by the black

bar in each boxplot. Company 3 has the lowest wage,

approximately $30 000 per year. C 10 Variation is shown by the length of the

boxplot, not including outliers. Company 1 has the largest variance. A


Solutions to Short-answer Questions 1 a This can be expressed as a number, so it

is numerical data. b Although the scale is numerical, the

actual data is not, so it is categorical. 2 a This is not expressed as a number, so it

is categorical data. b From the graph, the 'self-employed'

group is roughly halfway between 5% and 10%, or 7.5%.

3 Given the spread of data, an appropriate

class interval width would be 5. Count the number in each class interval, and draw the graph.

4 a

b Use the stem and leaf plot from part a. There are 22 data points, so the mean is

the average of the 11th and 12th times, which are both 52.

The median is 52 minutes. The first quartile will be the middle

score of the first 11 scores, and the second quartile will be the middle score of the second 11 scores.

These are the 6th and 17th times respectively, i.e. 47 min and 57 min.

5 Mean = 397014

≈ $283.57

Sort the data from lowest to highest. 185 190 210 210 215 245 265

270 280 285 300 315 320 680 Median = 265 + 270

2= $267.50

6 a Interval = [179 – 2 × 14, 179 + 2 × 14] = [151, 207] By counting, 26 out of 28 = 26

28× 100 ≈ 92.9%

b Since about 95% of data are usually

within two standard deviations of the mean, this is the kind of percentage you would expect.

7 Using the information provided:

8 First list the numbers from lowest to

highest. 0 0 2 3 3 4 7 7 7 8 10 11 12 12 13 13 14 14 15 15 16 16 16 16 17 18 21 21 21 22 23 23 23 30 31 44 Median m = 14 + 15

2= 14.5

Q1 = 7 + 82

= 7.5

Q3 = 21

Minimum = 0 Maximum = 44 Interquartile range = 21 – 14.5 = 6.5 In this case, outliers could only be

greater than 21 + 1.5 × 6.5 = 30.75 There is one outlier, 44. 31 is technically

an outlier, but is so close to the limit it can be considered normal data.

The new maximum for the boxplot becomes 31.


Chapter 23 – Investigating the relationship between two numerical variables Exercise 23A Solutions 1 a

b Larger values of y are associated with

smaller values of x, so the association is negative.

c The scatterplot does not show any

significant outliers. 2 a


larger values of x, so the association is positive.

c The scatterplot does not show any

significant outliers.

3 a


larger values of x, so the association is positive.

c The scatterplot shows an outlier at (122, 378), which suggests an aircraft

much slower than its size would suggest. 4 a


smaller values of x, so the association is negative.

c The scatterplot shows an outlier at (10, 8700), which suggests a caravan

much more expensive than its age would suggest.


Exercise 23B Solutions 1 Read from the table to determine the

correlation. a q = 0.20 = no correlation b q = –0.30 = weak negative correlation c q = –0.85 = strong negative correlation d q = 0.33 = weak positive correlation e q = 0.95 = strong positive correlation f q = –0.75 = strong negative correlation g q = 0.75 = strong positive correlation h q = –0.24 = no correlation i q = –1 = strong negative correlation j q = 0.25 = weak positive correlation k q = 1 = strong positive correlation l q = –0.50 = moderate negative correlation 2 a Count the points. There are 9. Draw a vertical line through the 5th

from the left, and through the 5th from the bottom.

q = (a + c) – (b + d)

a + b + c + d = 6 – 1

7≈ 0.71

b Count the points. There are 11. Draw a vertical line through the 6th from

the left, and through the 6th from the bottom.

q = (a + c) – (b + d)

a + b + c + d = 8 – 1

9≈ 0.78

c Count the points. There are 15. Draw a vertical line through the 8th


Ignore the extra points on the vertical

dotted line. q = (a + c) – (b + d)

a + b + c + d = 11 – 1

12≈ 0.82


d Count the points. There are 27. Draw a vertical line through the 14th


Ignore the extra point on the horizontal

dotted line.

q = (a + c) – (b + d)a + b + c + d

= 23 – 124

≈ 0.92

3 a There are 10 points. Draw a vertical line midway between

the 5th and 6th from the left, and midway between the 5th and 6th from the bottom.

q = (a + c) – (b + d)

a + b + c + d = 2 – 8

10= –0.6

b q = –0.6 = moderate negative

relationship


the 6th and 7th from the left and midway between the 6th and 7th from the bottom.

q = (a + c) – (b + d)

a + b + c + d = 10 – 2

12≈ 0.67

b q = 0.67 = moderate positive

relationship 5 a There are 16 points. Draw a vertical line midway between


q = (a + c) – (b + d)a + b + c + d

= 16 – 016

= 1

( ) ( ) 16 0 1

16a c b dqa b c d+ − + −

= = =+ + +

b q = 1 = strong positive relationship.




q = (a + c) – (b + d)

a + b + c + d = 4 – 10

14≈ – 0.43

( ) ( ) 4 10 0.43

14a c b dqa b c d+ − + −

= =+ + +

= −

b q = –0.43 = weak negative relationship


Exercise 23C Solutions 1 Read from the table to determine the

relationship. a r = 0.20 = no linear relationship b r = –0.30 = weak negative linear

relationship c r = –0.85 = strong negative linear

relationship d r = 0.33 = weak positive linear

relationship e r = 0.95 = strong positive linear

relationship f r = –0.75 = strong negative linear

relationship g r = 0.75 = strong positive linear

relationship h r = –0.24 = no linear relationship i r = –0.50 = moderate negative linear

relationship j r = 0.25 = weak positive linear

relationship k r = 1 = perfect positive linear

relationship l r = –1 = perfect negative linear

relationship 2 a This scatterplot is nearly a mirror-image

of the plot of 'age convicted' against 'testosterone'.

That scatterplot has r = –0.814, so we can estimate that r = 0.8 for this plot.

b This scatterplot is roughly a mirror-

image of the plot of 'age convicted' against 'testosterone', although not as close a match as the plot in part a.


c This scatterplot is reasonably similar to the scatterplot of 'mortality' against 'smoking ratio', although with fewer points.

That scatterplot has r = 0.716, so we can estimate that r = 0.7 for this plot.

d This scatterplot is roughly a mirror-

image of the plot of 'age convicted' against 'testosterone', although not as close a match as the plot in part a.


e This scatterplot is roughly a mirror-image

of the scatterplot of 'mortality' against 'smoking ratio', although with more points.

That scatterplot has r = 0.716, so we can estimate that r = –0.7 for this plot.

f This scatterplot is roughly a mirror-image

of the scatterplot of verbal' against 'mathematics'.

That scatterplot has r = 0.275, so we can estimate that r = –0.2 for this plot.

3 Use your calculator to determine r in

this question. a r = –0.86 b This is a strong negative linear

relationship. 4 Use your calculator to determine r in

this question. a r = 0.95 b This is a strong positive linear

relationship. 5 Use your calculator to determine r in

this question. a r = 0.77 b This is a strong positive linear

relationship.


6 Use your calculator to determine r in this question.

a r = –0.77 b This is a strong negative linear

relationship. 7 a

There appears to be a strong positive

linear relationship. b Yes; the data is numeric and the

relationship is linear. There are no outliers.

8 a

There appears to be a strong positive

linear relationship. b Yes; the data is numeric and the

relationship is linear.

c There are 9 points. Draw a vertical line through the 5th


q = (a + c) – (b + d)

a + b + c + d = 6 – 1

7≈ 0.71

Using a calculator, r = 0.87

d q = 0.71 = moderate positive linear relationship

r = 0.87 = strong positive linear relationship

e i The point will move down (shown on

the horizontal axis, although it will be well below this axis.) The median of Test 1 will be unaffected. The median of Test 2 will move down to the next point below.

ii q = (a + c) – (b + d)

a + b + c + d = 5 – 2

7≈ 0.43

Using a calculator, r = –0.004 The error in the data has a much

greater effect on Pearson’s correlation coefficient.


Exercise 23D Solutions Note: Answers will vary for lines drawn by eye. 1

2

3 a

b One possible line goes through (2000, 2012) and (5200, 6040).

y – y1

x – x1= y2 – y1

x2 – x1

y – 2000x – 2012

= 5200 – 20006040 – 2012

= 32004028

≈ 0.794y – 2012 = 0.794(x – 2000)

y = 424 + 0.794x

Other lines are possible and will give different answers.

c The positive slope indicates that districts

with high rates in Year 1 also had high rates in Year 2.

4 a

b One possible line goes through (40, 88)

and (60, 96).

y – y1

x – x1= y2 – y1

x2 – x1

y – 88x – 40

= 96 – 8860 – 40

= 820

= 0.4y – 88 = 0.4(x – 40)

y = 72 + 0.4x


c The intercept (72 cm) is the predicted

height at age 0. The slope predicts an increase of 0.4 cm in height each month.

d If your equation differs slightly from the

one given here, so will your answers. i y = 7 2 + 0.4x

= 72 + 0.4 × 42 ≈ 89 cm

ii 18 years = 18 × 12 = 216 months. y = 72 × 0.4x

= 72 + 0.4 × 216 ≈ 158 cm

e Part i is reasonable as it is a value close

to the data. Part ii is not reliable as the relationship may no longer be linear at this age.


5 a


y – y1

x – x1= y2 – y1

x2 – x1

y – 163x – 159

= 183 – 163181 – 159

= 2022

≈ 0.91y – 163 = 0.91(x – 159)

= 18.3 – 0.91x


c y = 18.3 + 0.91x

= 18.3 × 0.91 × 170 = 173.0 cm

If your equation differs slightly from the one given here, so will your answer.

6 a


y – y1

x – x1= y2 – y1

x2 – x1

y – 2600x – 100

= 3900 – 2600200 – 100

= 1300100

= 13y – 2600 = 13(x – 100)

y = 1300 + 13x


c The fixed cost is the constant in y = 1300 + 13x, or $1300. d The cost of production is the constant

multiplier in y = 1300 + 13x, or $13. 7 a Draw a scatterplot to assist you.

0

10

20

30

40

50

60

70

80

0 1 2 3 4 5

Drug dose (mg)

Res

pons

e tim

e (m

in)

6

One possible line goes through (4.0, 14) and (0.5, 63).

y – y1

x – x1= y2 – y1

x2 – x1

y – 14x – 4.0

= 63 – 140.4 – 4.0

= 49–3.5

= –14y – 14 = –14(x – 4.0)

y = 70 – 14x


b The intercept is the predicted time taken

to experience pain relief if no drug is given. From the slope we predict a reduction of 14 minutes in time taken to experience pain relief for each mg of drug administered.

c y = 70 – 14x

= 70 – 14 × 6 = –14 min

This isn't realistic, as relief would be experienced 14 minutes before the drug was given!


8 a Draw a scatterplot to assist you.

0

2000

4000

6000

8000

10000

12000

14000

0 100 200 300 400 500 600 700

Advertising ($)

Busi

ness

($)

One possible line goes through (160, 2900) and (600, 10 900).

y – y1

x – x1= y2 – y1

x2 – x1

y – 2900x – 160

= 10 900 – 2900600 – 160

= 8000440

≈ 18.2y – 2900 = 18.2(x – 160)

≈ 18.2x


b The intercept predicts zero sales if

nothing is spent on advertising. The slope means that on average, each $1 spent on advertising is associated with an increase of $18.20 in sales.

c i y = 18.2 × 1000

= $18 200 ii y = 18.2 × 0

= $0


Exercise 23E Solutions Use your CAS/graphics calculator in this exercise to determine the least squares line. Methods will differ depending on the calculator used. 1 a Using a calculator, y = 68.2 + 0.46x b The y-intercept is the predicted height at

birth. From the slope, we predict an increase in height of 0.46 cm each month.

c i y = 68.2 + 0.46x

= 68.2 + 0.46 × 42 ≈ 88 cm

ii 18 years = 18 × 12 = 216 months y = 68.2 + 0.46x

= 68.2 + 0.46 × 216 ≈ 168 cm

d The height at 42 months is reliable,

since this is within the range of data given (interpolation).

The height at 18 years is less reliable since this is outside the range of data given (extrapolation).

2 Using a calculator, y = 487.6 + 0.77x 3 a Using a calculator, y = 50.2 + 0.72x b An increase of 1 cm in the mother’s

height is associated with an increase of 0.72 cm in the daughter’s height, on average.

c y = 50.2 + 0.72x

= 50.2 + 0.72 × 170 ≈ 172 cm

4 a Using a calculator, y = 1330 + 12x b The fixed cost is the constant in y = 1330 + 12x, or $1330 c The cost of production is the constant

multiplier in y = 1330 + 12x, or $12. 5 a Using a calculator, response time = 57.0 – 10.2 × drug dose b The intercept of 57.0 minutes is the

predicted time for pain relief when no drug is given. From the slope, we predict a 10.2 minute decrease in response time for each 1 mg of drug given.

c y = 57.0 – 10.2x

= 57.0 – 10.2 × 6 = –4.4 min

This isn't realistic, as relief would be experienced 4.4 minutes before the drug was given.

6 a Using a calculator, business = 1123.8 + 18.9 × advertising b Intercept is the volume of business with

no advertising, From the slope we predict an increase in business of $18.90 for every dollar spent on advertising.

c i y = 1123.8 + 18.9x

= 1123.8 + 18.9 × 1000 = $20 044

ii y = 1123.8 + 18.9x = 1123.8 + 18.9 × 0 = $1123.80


Solutions to Multiple-choice Questions 1 Both variables must be numeric. This

applies to D only. 'Year 9' and 'Year 11', while appearing to be numerical, are really categorical. D

2 The data should be linear with no

outliers. In B, there is no relationship between the variables; C has an outlier, and A and D are not linear. E

3 Refer back to the table; q = 0.32 is a

weak positive relationship. A 4 There are 18 points. Draw a vertical line

midway between the 9th and 10th from the left and midway between the 9th and 10th from the bottom.

q = (a + c) – (b + d)

a + b + c + d = 2 – 16

18

= – 1418

= – 79

B

5 Looking at the graph, it's clear that there is a strong positive correlation, but not a perfect positive correlation.

0.75 ≤ r ≤ 1 is the range for a strong positive correlation, so r = 0.8 is the closest fit. B 6 The data suggest a strong positive

correlation, since food expenditure generally increases with weekly income. r = 0.8 is the closest fit (this can be confirmed using a calculator). E

7 Using a calculator to determine the least

squares regression line, it is closest to –42.864 + 0.482 × weekly income. C 8 y = 40 + 0.10x

= 40 + 0.10 × 600

= $100 C 9 The amount spent on entertainment is

the constant modifier of the x-value. This is 0.10, so for every dollar (x), 10 cents are spent on entertainment. A 10 First, note that the slope is negative. Taking the ends of the data, slope = rise

run = 75 – 225

21 – 10 = – 150

11 ≈ – 13.6

The slope is closest to –10. D



b Larger values of y are generally

associated with larger values of x, so the association is positive.

2 There are 12 points. Draw a vertical line

midway between the 6th and 7th from the left and midway between the 6th and 7th from the bottom.

q = (a + c) – (b + d)

a + b + c + d = 6 – 6

12= 0

3 r = 1n – 1Σ

i = 1

n

⎝⎜⎛ xi – x

sx ⎠⎟⎞⎝⎜⎛ yi – y

sy ⎠⎟⎞

Although r can be calculated by hand, it is much easier and faster to use a calculator, which shows that r = 0.927.

4 One possible line goes through (160, 120) and (190, 180)

Gradient = riserun

≈ 180 – 120190 – 160

= 6030

= 2

Run from (160, 120) = 160, so rise is 160 × 2 = 320

Intercept with vertical axis = 120 – 320 = –200 Equation is approximately weight = – 200 + 2 × height


5 a Using a calculator or computer, Errors = 14.9 – 0.533 × time 6 a Intercept: there's no sensible

interpretation, since no-one could complete the task in negative time. Slope: For each additional second taken to complete the task, the number of errors is reduced by about 1

2 an error.

b Errors = 14.9 – 0.533 × time

= 14.9 – 0.533×10 ≈ 9.6

Since the number of errors must be a whole number, you would predict 10 errors, maybe 9.


Chapter 24 – Revision of chapters 22 and 23 Solutions to Multiple-choice Questions 1 Bar charts are used to display frequency

distributions of categorical data. The only categorical data in this question is option C. C

2 Histograms are used to display

frequency distributions of numerical data. The only numerical data in this question is weight measured in kg. A

3 The histogram has a short tail to the left

and a long tail to the right, so it is positively skewed. It has an outlier at

x = 14. E 4 17 students travel 20 or less 20 km to

school, while 16 travel more than 20 km, so option D is true. D

5 Adding up the number of ‘leaves’ on the

right gives 30 students. C 6 There are 7 'leaved' in the 2, 3 and 4

'stems', so 7 students failed. B 7 10% of 30 students is 3 students. The

top three marks are 86, 90 and 93, of which the lowest is 86. D

8 If turned on its side, the plot would

be a histogram with a short tail to the right and a slightly longer tail to the right, so it is negatively skewed. C

9 Most scores on the left are on the 2

'stem', while most scores on the right are on the 2 and 3 stems. This shows that, in general, scores improved after the test. B

10 The scores on the left are spread more widely up and down the range, while the scores on the left are more tightly centred on the 6 and 7 stems.

This suggests that the pulse rates of those who exercise are less variable than those who don't exercise. D

11 The new mark will go next to the

median, or could replace the median, which will still be 50. B

12 Neither the maximum nor the minimum

will change, so the range will still be 70 – 20 = 50. B 13 Since the mean is 50, it will not be changed by adding another score of 50. B 14 Old variance = 12.0 × 12.0 = 144.0 Old sum of squared deviations from

mean = 144.0 × (99 – 1) = 14 112 Variation of 50 from the mean = 0 ∴new sum of squared deviations from mean = 14 112 + 0 = 14 112

Standard deviation = 14 112100 – 1

= 142.545

≈ 11.9 B 15 The width of each boxplot shows its

variation. Since Y is wider than X, Class Y scores are more variable than

Class X scores. D


16 If the median is less than the mean, the ages in the second half of the data are likely to be higher than those in the first half. This suggests the data may be negatively skewed. C

17 The boxplot shows that there is

significantly more data above the mean than below it. This means the distribution is positively skewed. B

18 The histogram is positively skewed,

which generally means that the mean is larger than the median. E 19 The boxplot shows a negatively skewed

distribution with much more variability towards the left.

Option B is the closest fit to this description. B

20 The third quartile is the median of the

upper half of the distribution. If the woman is in this quartile, it

means that about 75% of women are shorter than her. B

21 Scatterplots are used to display bivariate

numerical data. Both sets must be numerical, and this is only the case in B and D. However, D is not bivariate, as the data cannot realistically be arrayed as a unique set of points. B

22 The first scatterplot shows a strong non-linear relationship. The q-correlation is only used to test for a

linear relationship, so it would not be useful here. A

23 Gradient = rise

run= 20 – 0

0 – 10= –2

This is closest to –1. A

24 Calculate the gradient. Using (1, 10) and (10, 30), m ≈ 30 – 10

10 – 1 ≈ 2

The y-intercept is around 10. Therefore a good approximation could be y = 2x + 10. C 25 r = 0.7 is a moderate positive linear

relationship, so an increase in the first variable will usually mean an increase in the second variable.

Therefore we can say that countries with high smoking rates also tend to have high rates of heart disease. C

26 r = 0.35 is a weak positive linear

relationship. The scatterplot is a curve rather than a

line, so there probably is a relationship, but since r is low then it is unlikely to be linear. C

27 A negative correlation means that as

values of x increase, values of y tend to decrease. C

28 Use typical points such as (98, 100) and

(78, 80). m ≈ 98 – 78

100 – 80 = 20

20= 1

Extending the line back to the y-axis gives a y-intercept of about 0.

The equation is roughly y = x. A 29 Since the equation uses positive

numbers and addition, the cost of production will increase as the number of units produced increases.

Since N is the number of units, the total cost of production increases by $35 for each unit produced. C

30 The least squares regression line is used

to minimise vertical deviations from the data (deviations along the y-axis). D


Chapter 25 – Proof and number Exercise 25A Solutions 1 a If p is even, it is a multiple of 2, and pq

will also be even. If pq is even and q is even, then p may

be odd. p is even ⇒ pq is even b If p + q is odd, p and q cannot both be

odd, so pq will be even. If pq is even, p and q could both be

even, meaning p + q could be even. p + q is odd ⇒ pq is eve c If x = 0, then xy = 0 If xy = 0, x could be non-zero if y = 0. x = 0 ⇒ xy = 0 d If ab = ac and a = 0, then b may not be

equal to c. If b = c, then ab = ac, whether or not a = 0. ab = ac ⇐ b = c 2 a This is only true if A and B have no

common elements. If there is any overlap, the union will be smaller. For instance, if A = {2, 4, 6, 8, 10} and B = {6, 8, 10}, n (A ∪ B) = 5. Since it's not always true, the statement is false.

b Consider the element x, which is an

element of A. Since A is a subset of B, x is also an element of B. This is true for every x, so the set of

elements that are in both A and B will be the same as every element in A.

x ∈ A ⇒ x ∈ B x ∈ A ∩ B ⇒ x ∈ A ∴ and x ∈ A ⇒ x ∈ A ∩ B The statement is therefore true.

c A ∩ B = Ø is actually true whenever A and B have no elements in common.

For instance, if A = {2, 4, 6} and B = {8, 10}, A ∩ B = Ø. The statement is therefore false. d If A' = ξ, then it contains all the elements

under consideration. A' therefore contains no elements, as all the elements are in A, so A' = Ø.

The statement is therefore true. 3 a n (A ∪ B) = 8 ⇒ n (A) = 5 and n (B) = 3 This would mean that if the union of A and B has 8 elements, A would have 5 elements and B would have 3 elements. However, there are many other possible

combinations that would fit; for instance, if A had 8 elements and B = Ø, their union would also have 8 elements.

The statement is therefore false. b A ∩ B = A ⇒ A ∈ B This would mean that if the intersection

of A and B is equal to A, then A is an element of B.

Since the intersection contains all the elements the two sets have in common, and it is equal to A, then all the elements of A are elements of B, which means

A is an element of B. The statement is therefore true. c A = Ø or B = Ø ⇒ A ∩ B = Ø The empty set contains no elements, so

it therefore has no elements in common with any other set. This means that the intersection of an empty set and any other set will also be an empty set.

The statement is therefore true.


d A = ξ ⇐ A' = Ø If A' = Ø, then it has no elements, which

means A contains all the elements. The identity set ξ contains all the elements, so A = ξ.

The statement is true. 4 a Swap the sides to find the converse: n2 is odd implies n is odd. Odd numbers can never have even

factors, so the root of an odd number must always be an odd number.

Therefore the statement is true (assuming that n is an integer).

b Swap the sides to find the converse: N

2 is divisible by 9 ⇒ N is divisible by 3. Since 32 = 9, this is the same as saying N

2 is divisible by 32 ⇒ N is divisible by 3. A number's factors are squared if the

original number is squared. Therefore the statement is true

(assuming that N is an integer). c Swap the sides to find the converse: x2 > 4 ⇒ x < –2 x2 will be greater than 4 if x is less than

–2, but it will also be true if x is greater than 2.

Since the first part does not only lead to the second part, the statement is false.

5 a Let the three odd numbers be 2n + 1, 2n + 3 and 2n + 5. Their sum = 2n + 1 + 2n + 3 + 2n + 5 = 6n + 9, which will be

divisible by 3 if n is an integer. We can therefore say "The sum of three

consecutive odd numbers is divisible by 3." b Let the four odd numbers be 2n + 1, 2n + 3, 2n + 5 and 2n + 7. Their sum = 2n + 1 + 2n + 3 + 2n + 5 + 2n + 7 = 8n + 16, which will be

divisible by 8 if n is an integer. We can therefore say "The sum of four

consecutive odd numbers is divisible by 8."

6 The sum of the squares is

(n – 2)2 + (n – 1)2 + n2 + (n + 1)2 + (n + 2)2

= n2 – 4n + 4 + n2 – 2n + 1 + n2 + n2 + 2n + 1 + n2 + 4n + n = 5n2 + 10 = 5(n2 + 2), which is divisible by 5.

7 There are many possible counter-

examples for these statements. a The order of operation changes from

multiplication → addition to multiplication → addition, so this is false.

32 + 42 = 52

(3 + 4)2 = 72

b Any even number multiplied by 7 will

give an even number divisible by 7, so this is false.

2 × 7 = 14. c The order of operation changes from

addition → division to division → addition, so this is false.

1 + 4 = 51 + 4 = 3

d The order of operations is different, so

this will be false if a, b and c have different values.

12

⎝⎜⎛1 + 2 + 4

2 ⎠⎟⎞ = 2

12

⎝⎜⎛1 + 2

2+ 4

⎠⎟⎞ = 11

4

e All prime numbers (other than 2) are odd,

and the sum of two odd numbers is an even, non-prime number, so this is false.

3 + 7 = 10 (The result could be a prime number if one

of the original numbers is 2, e.g. 2 + 3 = 5, but you only need one counter-example to prove that the rule is false.)

f The order of operation changes from

addition → division to division → addition, so this is false.

1100 + 10

= 1110

1100

+ 110

= 1100

+ 10100

= 11100


8 a The converse is: a > b implies a – b is positive. Subtracting a smaller number from a

larger number always gives a positive result, e.g. 10 – 3 = 7, so this is true.

b The converse is: x = y implies x = 0 and y = 0. x and y could be any numbers, not just

zero, e.g. 10 = 10, so this is false. c The converse is: x = –y implies x + y = 0. Subtracting any number from itself

always gives a result of zero, e.g. 10 – 10 = 0, so this is true. d The converse is: xy is even implies x is even and y is odd. If x and y were both even, xy would also

be even, e.g. 8 × 6 = 48, so this is false. e The converse is:

A perfect square being even implies its square root is even. Multiplying two even numbers always produces an even number, while multiplying two odd numbers always produces an odd number. An even perfect square will thus always be created by multiplying an even number by the same even number, e.g. 64 = 8 × 8, so this is true.

9 a For any natural number N, if a is a

divisor, there will be a corresponding natural number divisor N

a. This will be

distinct from a in every case except N = a2. Thus if N is not a perfect square, there will

be an even number of factors. If N is a perfect square, there will be a set of pairs of factors as above, and an additional factor equal to the square root of N.

Thus N will have an odd number of divisors if and only if it is a perfect square.

The statement is true.

b Consider the powers of 2. The only factors of 2N are 1, 2, 22, . . . 2N – 1, 2N.

Thus 2N will have exactly N + 1 factors, and 2N – 1 will have exactly N factors.

Thus numbers can be found with any number of factors, including 1, since

20 = 1 has exactly one factor. The statement is true for all odd

numbers. c The same argument used in part b

applies to all numbers, not just odd numbers, so the statement is true.

d Since the above argument works for any

prime number, and there is an infinite number of prime numbers, there will be an infinite number of numbers with exactly N divisors, for all natural number values of N.

The statement is true. 10 a (a – b)2 ≥ 0

a2 – 2ab + b2 ≥ 0 Add 2ab to both sides of the inequality: a2 + b2 ≥ 2ab b If u ≥ 0 and v ≥ 0, then u and v can be

written as respectively and a2 and b2

a = u , b = v .

u + v2

= a2 + b2

2a2 + b2

2≥ 2ab

2 ≥ ab ≥ u v ≥ uv

∴ u + v2

≥ uv


11 (a + b)⎝⎜⎛1

a+ 1

b⎠⎟⎞ = (a + b) b + a

ab

= (a + b)2

ab

= a2 + 2ab + b2

ab

= 2abab

+ a2 + b2

ab

= 2 + 2(a2 + b2)2ab

In Q.10a we showed that a2 + b2 ≥ 2ab.

∴ 2 + 2(a2 + b2)2ab

> 2 + 2

(a + b)⎝⎜⎛1

a+ 1

b⎠⎟⎞ > 4

12 a If a and b are even, then they can be

written as a = 2n and b = 2m. a + b = 2n + 2m = 2(m + n), which must be an even number. b If a and b are odd, then they can be

written as a = 2n + 1 and b = 2m + 1. a + b = 2n + 1 + 2m + 1 = 2n + 2m + 2 = 2(m + n + 1), which must be an

even number. c If a is even and b is odd, then they can

be written as a = 2n and b = 2m + 1. a + b = 2n + 2m + 1 = 2(m + n) + 1, which must be an

odd number. d If a and b are odd, then they can be

written as a = 2n + 1 and b = 2m + 1. ab = (2n + 1)(2m + 1) = 2mn + 2m + 2m + 1 = 2(mn + m + n) + 1, which must be

an odd number.

e If a + b is even, then a + b = 2n. b = 2n – a If a – b is even, then a – b = 2n. b = a – 2n a + b = b + 2n + b = 2b + 2n = 2(b + n), which must be an even number. This proves the conjecture. f If a + b is even, then a + b = 2n. If a – b is even, then a – b = 2n. n = a + b

2 is an integer and m = a – b

2 is also an integer.

n + m = a + b2

+ a – b2

= 2aa

= a

n – m = a + b2

– a – b2

= 2b2

= b

∴ ab = (n + m)(n – m)

= n2 – m2, the difference of squares. 13 The total number of tiles is mn. The number of tiles on the perimeter is n + n + (m – 2) + (m – 2) = 2n + 2m – 4 We are told 2n + 2m – 4 = 1

2mn

∴4n + 4m – 8 = mn Make n the subject. mn – 4n = 4m – 8

n(m – 4) = 4m – 8n = 2m – 8

m – 4 = 4m – 16 + 8

m – 4 = 4 + 8

m – 4

n will only be an integer if m – 4 is a factor of 8.

∴m – 4 = 1, 2, 4 or 8 m = 5, 6, 8 or 12 These values give n = 12, 8, 6 and 5

respectively. You should notice that the last two solutions are equivalent to the first two solutions. Since m ≤ n, there will be two solutions – m = 5, n = 12 and m = 6, n = 8.


14 This is one proof. There may be others. Any number (except 1) that consists of

all 1s can be written as 100n + 11, where n is an integer.

All integers can be written as 10m + a, where m is any integer and a is any integer from 0 to 9 inclusive.

(10m + a)2 = 100m2 + 20am + a2

Since this is a multiple of 10 plus , the last digit of a must be 1.

a2

2

The only values of a that satisfy this requirement are a = 1 and a = 9.

If a = 1,

(10m + a)2 = 100m2 + 20m + 1

= 100m2 + 10 × 2m + 1 In this case, the second last digit must be

even and cannot equal 1. If a = 9,

(10m + a)2 = 100m2 + 180m + 81 = 100m2 + 100 + 10 × 8m + 81

Again, the second last digit must be even and cannot equal 1.

Therefore, if any number ending in 1 is a perfect square, its second last digit must be even, and it therefore cannot consist of all 1s, unless it is 1.


Exercise 25C Solutions 1 a If a solution is not readily seen, use trial

and error on the variable with the largest coefficient, as you will expect fewer trials until you find a multiple of the other variable.

Try x = 0: 3y = 1 has no integral solutions.

Try x = 1: 11 + 3y = 1 has no integral solutions.

Try x = 2: 22 + 3y = 1 has the solution y = –7. The HCF of this solution is 1. The general solution will be x = 2 + 3t, y = –7 + 11t, t ∈ Z b An obvious solution is x = 1, y = 0. The HCF of this solution is 1. The general solution will be x = 1 + 7t, y = –2t, t ∈ Z Any even value of y will give a solution.

If y = –2, x = 8. x = 8 + 7

2 t, y = –2 – 2

2 t, t ∈ Z

To get integer solutions, replace t by 2t. x = 8 + 7t, y = –2 – 2t, t ∈ Z

c This equation is equivalent to 8x + 21y = 33. It is also obvious that y must be odd, and x must be a multiple of 3. y = 5 gives the solution x = –9. x = –9 + 21t, y = 5 – 8t, t ∈ Z d This is the same equation as in part a,

hence the solution will be the same. x = 2 + 3t, y = –7 + 11t, t ∈ Z

e Any even value of y will give a solution. If y = 2, x = 4. The HCF of this solution is 2. The general solution will be x = 4 + 7

2 t, y = 2 – 2

2 t, t ∈ Z

To get integer solutions, replace t by 2t. x = 4 + 7t, y = 2 – 2t, t ∈ Z f This is the same equation as in part e,

hence the solution will be the same. x = 4 + 7t, y = 2 – 2t, t ∈ Z 2 a If x and y are both positive, then the left

side must be at least 11 + 3 > 1. There are no solutions such that x and y

are both positive. c If x and y are both positive, then the left

side must be at least 24 + 63 = 87. Any increase in either x or y will make

the left side too large. There are no solutions such that x and y

are both positive. e Write the equation as 2x = 22 – 7y

x = 11 – 72y

If y = 2, x = 4. If y = 4, x = –3, and any larger values of

y will make the left side of the equation negative.

There is one solution: x = 4, y = 2. 3 Let h be the highest common factor of a and b. a, b and c can be written as a = hp, b = hq, c = hr + k, where 0 < k < h. The equation becomes hpx + hqy = hr + k. For all integer values of x and y, the left

side of the equation will be a multiple of h, while the right side will not. Therefore the equation can have no integral solutions.


4 a Let s be the number of spiders and b the

number of beetles. Equating the numbers of legs gives 8s + 6b = 54. b This equation simplifies to 4s + 3b = 27. 4s = 27 – 3y

= 3(9 – b)s = 3(9 – b)

4

Solutions will only exist when 9 – b is a multiple of 4, and b > 0, 9 – b > 0.

This occurs when b = 1, s = 6 and when b = 5, s = 3.

The answer could be written '3 spiders and 5 beetles, or 6 spiders and 1 beetle.'

5 Equating the value of the coins,

20x + 50y = 5002x + 5y = 50

5y = 50 – 2x = 2(25 – x)y = 2(25 – x)

5

= 2 ⎝⎜⎛5 – x

5⎠⎟⎞

This gives the results as in the table below.

6 All solutions can be given by x = 100 + 83t, y = 1 – 19t 100 + 83t > 0

83t > – 100t > – 100

83

1 – 19t > 0–19t > – 1

t < 119

Since t is an integer, –1 ≤ t ≤ 0. The second solution occurs when t = –1. x = 100 – 83

= 17y = 1 + 19 = 20

For 19x + 98y = 1998, one obvious solution is x = 100, y = 1.

x = 100 + 98t, y = 1 – 19t 100 + 98t > 0

98t > – 100t > – 100

98

1 – 19t > 0–19t > – 1

t < 119

Since t is an integer, –1 ≤ t ≤ 0. The second solution occurs when t = –1. x = 100 – 98

= 2y = 1 + 19 = 20

7 Equating the value of the notes,

10x + 50y = 500x + 5y = 50

x = 50 – 5y = 5(10 – y)

This gives the results as in the table below.

8 Total number of pieces of fruit = 63x + 7. y = 63x + 7

23 = 7(9x + 1)

23

63 7 7(9 1)

23 23x xy + +

= =

9x + 1 must be a multiple of 23. 9x + 1 = 23n

9x = 23n – 1x = 23n – 1

9

If n = 2, x = 5 and y = 14.


11 Let x be the number of 5-litre jugs used and y the number of 3-litres jugs used.

If n = 9t + 2,

x = 23n – 19

= 23(9t + 2) – 19

= 23t + 23 × 2 – 19

= 23t + 5y = 7(9x + 1)

23 = 7((23n – 1) + 1)

23 = 7n = 7(9t + 2)

5x + 3y = 75x = 7 – 3y

x = 7 – 3y5

Solutions will only exist when 7 – 3y is a multiple of 3.

This occurs when y = –1: x = 7 + 35

= 2

To measure exactly 7 litres, you would pour two full 5-litre jugs into a container and then remove one 3-litre jugful.

The next solution will be x = 28, y = 112. 12 Obviously the post office can't sell 1c or

2c worth of postage. Nor can it sell 4c or 7c worth, because there's no way to arrange 3c and 5c to get those values.

The general solution is x = 5 + 23t, y = 14 + 63t; t ≥ 0 and t ∈ Z. 9 Consider the value of the two types of

cattle. It can sell 6c worth (3 + 3 = 6) and 8c

worth (3 + 5 = 8). 410x + 530y = 10 000

41x + 53y = 1000

So the problem can be rephrased as 3x + 5y = n, n ≥ 8 Using a graphics calculator, a

spreadsheet, or trial and error, where x is the number of 3c stamps and y the number of 5c stamps. x = 5, y = 15.

5 of the $410 cattle and 15 of the $530 cattle.

If n = 8, 3x + 5y = 8; the obvious solution is x = 1, y = 1.

If n = 9, 3x + 5y = 9; the obvious solution is x = 3, y = 0. 10 Let the required number be x.

If it leaves a remainder of 6 when divided by 7, then x = 7n + 6.

If n = 10, 3x + 5y = 10; the obvious solution is x = 0, y = 2.

If it leaves a remainder of 9 when divided by 11, then x = 11m + 9.

Since this set of three can be made using 3x + 5y, the next set of three amounts (11, 12, 13) can be made as 3x + 5y + 3, or by adding another 3c stamp.

7n + 6 = 11m + 97n – 11m = 3

One solution is n = 2, m = 1.

The general solution is n = 2 + (–11)t, Similarly, every set of three consecutive amounts can be made by adding an additional 3c stamp.

m = 1 – 7t. Replacing t with –t gives n = 2 + 11t, m = 1 + 7t. Therefore it's possible to create all

amounts in excess of 3c, except for 4c and 7c.

t = 0 gives n = 2, m = 1, x = 7 × 2 + 6 = 20. The smallest positive number is 20. The general form is x = 7n + 6

= 7(2 + 11t) + 6 = 77t + 20 for t ∈ N ∪ {0}

13 Consider total cost. 1.7a + b = 29.6

17a + 10b = 296 Using a graphics calculator, a

spreadsheet, or trial and error, a = 8, b = 16. 8 of type A and 16 of type B.


Exercise 25D Solutions 1 a b = aq + r, 0 ≤ r < a

43 = 5q + r = 5 × 8 + 3

(5, 43) = (5, 3) by theorem 25 = 3 × 1 + 2

(3, 5) = (3, 2) by theorem 23 = 2 × 1 + 1

(2, 3) = (2, 1) by theorem 22 = 2 × 1 + 0

∴ (43, 5) = (5, 3) = 1

b

b = aq + r, 0 ≤ r < a39 = 13q + r

= 13 × 3 + 0

c

∴ (39, 13) = (13, 0) = 13

b = aq + r, 0 ≤ r < a37 = 17q + r

= 17 × 2 + 3

(17, 37) = (17, 3) by theorem 217 = 3 × 5 + 2

(3, 17) = (3, 2) by theorem 23 = 2 × 1 + 1

(2, 3) = (2, 1) by theorem 22 = 2 × 1 + 0

∴ (37, 17) = (17, 3) = 1

d b = aq + r, 0 ≤ r < a

128 = 16q + r = 16 × 8 + 0

∴ (128, 16) = (16, 0) = 16 2 If d is a common factor of a and b, then a = nd and b = md. a + b = nd + md

= d(n + m) d is a divisor of a + b. ∴

If d is a common factor of a and b, then a = nd and b = md.

a – b = nd – md = d(n – m)

d is a divisor of a – b. ∴

3 a 9284 = 4361 × 2 + 562

(4361, 9284) = (4361, 562)4361 = 562 × 7 + 427

(4361, 562) = (427, 562)562 = 427 × 1 + 135

(427, 562) = (135, 427)427 = 135 × 3 + 22

(135, 427) = (22, 135) This process could continue, but at this

point it is quicker and easy to notice that the two numbers have no common factor other than 1, so (4361, 9284) = 1.

b 2160 = 999 × 2 + 162

(999, 2160) = (162, 999)999 = 162 × 6 + 27

(162, 999) = (27, 162)162 = 27 × 6 + 0

(999, 2160) = 27 c (–372, 762) = (372, 762)

762 = 372 × 2 + 18(372, 762) = (372, 18)

372 = 18 × 20 + 12(372, 18) = (12, 18)

18 = 12 × 1 + 6(12, 18) = (6, 12)

12 = 6 × 2 + 0(–372, 762) = 6

d 716 485 = 5255 × 136 + 1805

(716 485, 5255) = (1805, 5255)5255 = 1805 × 2 + 1645

(1805, 5255) = (1805, 1645)1805 = 1645 × 1 + 160

(1805, 1645) = (160, 1645)1645 = 160 × 10 + 45

(160, 1645) = (45, 160) This process could continue, but at this

point it is quicker and easy to notice that the two numbers have a highest common factor of 5, so

(716 485, 5255) = 5.


4 a Apply the division algorithm to 804 and 2358. 2358 = 804 × 2 + 750

804 = 750 × 1 + 54750 = 54 × 13 + 48

54 = 48 × 1 + 648 = 6 × 8

Working backwards with these results, 6 = 54 – 48 × 1

6 = 54 – (750 – 54 × 13)6 = 54 – 750 + 54 × 136 = 54 × 14 – 750 6 = (804 – 750 × 1) × 14 – 7506 = 804 × 14 – 750 × 14 – 7506 = 804 × 14 – 750 × 156 = 804 × 14 – (2358 – 804 × 2) × 156 = 804 × 14 – 2358 × 15 + 804 × 306 = 804 × 44 – 2358 × 15

A solution is x = 44, y = –15.

The general solution is x = 44 + 2358

6 t

= 44 + 393ty = –15 – 804

6 t

= –15 – 134t, t ∈ Z

b This is equivalent to 3x + 4y = 1. The algorithm is still useful. 4 = 3 × 1 + 1

1 = 3 × –1 + 4 A solution is x = –1, y = 1. The general solution is x = –1 + 4t

y = 1 – 3t, t ∈ Z

c 478 = 3 × –478 + 4 × 478 A solution is x = –478, y = 478. The general solution is x = –478 + 4t

y = 478 – 3t, t ∈ Z

d The algorithm is still useful. –5 = 3 × –2 + 1

1 = 3 × 2 + –51 = 3 × 2 – 5 × 1

38 = 3 × 76 – 5 × 38

A solution is x = 76, y = 38

The general solution is x = 76 + 5t

y = 38 + 3t

This can be simplified. If t – 15 is used instead of t, then

x = 76 + 5(t – 15) = 1 + 5ty = 38 + 3(t – 15) = –7 + 3t, t ∈ R

e Apply the division algorithm to 804 and 2688. 2688 = 804 × 3 + 276

804 = 276 × 2 + 252276 = 252 × 1 + 24252 = 24 × 10 + 12

24 = 12 × 2


12 = 252 – (276 – 252 × 1) × 1012 = 252 – 276 × 10 + 252 × 1012 = 252 × 11 – 276 × 10 12 = (804 – 276 × 2) × 11 – 276 × 1012 = 804 × 11 – 276 × 22 – 276 × 1012 = 804 × 11 – 276 × 3212 = 804 × 11 – (2688 – 804 × 3) × 3212 = 804 × 11 – 2688 × 32 + 804 × 9612 = 804 × 107 – 2688 × 32

A solution is x = 107, y = –32.

The general solution is x = 107 + 2688

12 t

= 107 + 224ty = –32 – 804

12 t

= –32 – 67t, t ∈ R


f Apply the division algorithm to 1816 and 2688.

2688 = 1816 × 1 + 8721816 = 872 × 2 + 72

872 = 72 × 12 + 872 = 8 × 9


8 = 872 – (1816 – 872 × 2) × 128 = 872 – 1816 × 12 + 872 × 248 = 872 × 25 – 1816 × 12 8 = (2688 – 1816 × 1) × 25 – 1816 × 128 = 2688 × 25 – 1816 × 25 – 1816 × 128 = 2688 × 25 – 1816 × 37

A solution is x = –37, y = 25.

The general solution is x = –37 + 2688

8 t

= –37 + 336ty = 25 – 1816

8 t

= 25 – 227t, t ∈ R


Solutions to Multiple-choice Questions 1 The square of an even number is always

an even number, and an even number plus an odd number always gives an odd result.

This means that m2 + n will be odd, so option E is false. E

2 If x – 3 > 0, then x – 2 > 0.

The product of two numbers greater than zero will also be greater than zero.

Therefore option D is true. D 3 If p and q are both positive, and p + q = 1, then p and q are both less than 1, and 1

p and 1

q will be greater than 1.

Since p > q, 1pq

> 1p and 1

pq> 1

q.

The largest quantity will be 1pq

. C

4 If pq < 0, and p > q, then p will be

positive and q will be negative, in which case 1

p would be positive and greater

than 1q

, which would be negative.

Therefore pq must be greater than zero (i.e. p and q are both positive) for 1

p to

be less than 1q

. E

5 You can choose any number of 2s from

0 to p in (p + 1) ways. For each of these, you can choose any number of 3s from 0 to q in (q + 1) ways, and for each of these combinations you can choose any number of 5s from 0 to r in (r + 1) ways.

The total number of ways = (p + 1)(q + 1)(r + 1) D

6 Let the digits be a and b, so the number n = 10a + b. The sum of its digits is a + b. 10a + b = k(a + b)

= ka + kb10a – ka = kb – b(10 – k)a = b(k – 1)

a = k – 110 – k

b

The number formed by interchanging the digits is m = 10b + a.

Adding the two numbers,

10a + n + 10b + 1 = 11a + 11b = 11(a + b)

k(a + b) + m = 11(a + b)m = (11 – k)(a + b)

The new number is the sum of the digits multiplied by 11 – k. C

7 m + n = mn

n = mn – m = m(n – 1)

m = nn – 1

This will only be integral if n = 2, m = 2 or n = 0, m = 0.

There are two solutions. B 8 Except for option D, the statements will

be affected by either a, b or c being negative. D

9 Solutions can be written in terms of

another variable, and the number will be infinite. E 10 (n – 1), (n – 2) and (n – 3) are

consecutive numbers. Thus at least one of the factors of (n – 1)(n – 2)(n – 3) will be a multiple of 3, and at least one will be even. Thus the number will be divisible by 1, 2, 3, and 6, but not always by 5. B


Solutions to Short-answer Questions 1 1885 = 365 × 5 + 60

(1885, 365) = (60, 365)365 = 60 × 6 + 5

(60, 365) = (60, 5)60 = 5 × 12 + 0

(1885, 365) = 5 2 a Apply the division algorithm to 43 and 9. 43 = 9 × 4 + 7

9 = 7 × 1 + 27 = 2 × 3 + 12 = 2 × 1

Working backwards with these results,

1 = 7 – 2 × 31 = 7 – (9 – 7 × 1) × 31 = 7 – 9 × 3 + 7 × 31 = 7 × 4 – 9 × 3 1 = (43 – 9 × 4) × 4 – 9 × 31 = 43 × 4 – 9 × 16 – 9 × 31 = 43 × 4 – 9 × 19

A solution to 9x + 43y = 1 is x = –19, y = 4. A solution to 9x + 43y = 7 is x = –19 × 7 = –133, y = 4 × 7 = 28.

The general solution is x = –133 + 43t

y = 28 – 9t, t ∈ R

Other solutions are possible. t = 4 gives a specific solution of x = 39,

y = –8, leading to a general solution of x = 39 + 43t

y = –8 – 9t, t ∈ R

b If x > 0, 39 + 43t > 0

t > – 3943

If y > 0, –8 – 9t > 0t < – 8

9

These two inequations cannot both be true if x is an integer.

There is no solution for x ∈ Z+ , y ∈ Z+ .

3 If a and b are odd, they may be written as 2n + 1 and 2m + 1 respectively, where n and m are integers.

ab = (2n + 1)(2m + 1) = 4mn + 2n + 2m + 1 = 2(2mn + n + m) + 1

This will be an odd number since 2mn + n + m is an integer. 4 12 121 = 10 659 × 1 + 1462

(12 121, 10 659) = (1462, 10 659)10 659 = 1462 × 7 + 425

(1462, 10 659) = (1462, 425)1462 = 425 × 3 + 187

(1462, 425) = (187, 425)425 = 187 × 2 + 51

(187, 425) = (187, 51)187 = 51 × 3 + 34

(187, 51) = (51, 34)51 = 34 × 1 + 17

(51, 34) = (34, 17)34 = 17 × 2 + 0

(12 121, 10 659) = 17 5 a The algorithm is still useful. 7 = 5 × 1 + 2

5 = 2 × 2 + 11 = 5 – 2 × 21 = 5 – (7 – 5 × 1) × 21 = 5 – 7 × 2 + 5 × 21 = 5 × 3 – 7 × 2

A solution is x = 3, y = –2. The general solution is x = 3 + 7t

y = –2 – 5t, t ∈ R

b If 1 = 5 × 3 – 7 × 2, then 100 = 5 × 300 – 7 × 200. A solution is x = 300, y = –200. The general solution is x = 300 + 7t

y = –200 – 5t, t ∈ R

300 7 , 200 5x t y t= + = − −


c If y ≥ x, –2 – 5t ≥ 3 + 7t

–12t ≥ 5t ≤ – 5

12

Since t is an integer, t ≤ 0. The solution is x = 3 + 7t

y = –2 – 5t, t ≤ 0, t ∈ R

6 First, let Tom’s age be t and Fred’s age

be f. Since it appears Tom is older than Fred,

and we must look at the time when Tom was Fred’s age, we will define d as the difference in ages, specifically how many years Tom is older than Fred.

t = f + dt + f = 63

∴(f + d) + f = 632f + d = 63

When Tom was Fred’s age, d years ago, Fred was aged f – d.

Tom is now twice that age, 2(f – d). ∴ t = 2(f – d) 2( )t f d∴ = − Since t = f + d, f + d = 2(f – d)

= 2f – 2d3d = f

Substitute f = 3d into 2f + d = 63. 6d + d = 63

7d = 63d = 9f = 3d = 27

t + f = 63t = 36

Tom is 36 and Fred is 27.


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Advanced General Mathematics Worked Answers Enhanced Edition NO COPY WRITE INFRINGEMENT INTENDED

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