MATHEMATICS

Advanced Higher

Revision Materials

Revision Materials

Geometry Unit

Geometry, Proofs and Systems of Equations

NATIONAL

QUALIFICATIONS

Advanced Higher Mathematics – Geometry, Proof and Systems of Equations Unit Assessment Preparation - Further Practice Questions

Page 2

Advanced Higher Mathematics – Geometry, Proof and Systems of Equations Unit Assessment Preparation - Further Practice Questions

Page 3

1.1 Applying Algebraic skills to matrices and systems of equations

Using Gaussian elimination to solve a 3X3 system of linear equations

1. Expand the following using the Binomial Theorem:

(a) (b) (c) (d)

(e) (f) (g) (h)

1.1 Applying Algebraic skills to matrices and systems of equations

Performing matrix operations of addition, subtraction and multiplication

2. Given the matrices

, , ,

, , ,

find the matrix:

(a) (b) (c)

(d) (e) (f)

(g) (h) (i) (j) (k) (l) (m) (n) (o) (p) (q) (r) (s) (t) (u) (v) (w)

1.1 Applying Algebraic skills to matrices and systems of equations

Calculating the determinant of a matrix

Finding the inverse of a matrix

3. Given the matrices

, , , ,

, , , ,

,

,

(a) Find

(b) Determine the value(s) of for which each matrix is singular.

x + y + z = 10

2x + 3y + z = 21

x + 2y + 4z = 19

x + y + 2z = 4

3x + 4y + z = 19

2x + y + 4z = 6

x + 2y - z = 8

x + 3y + 2z = 27

2x - 2y + 5z = 2

x + y- z = 11

2x + 3y + 2z = 16

-x + 2y + 2z = 4

x + 2y + z = 4

2x + 5y - 3z = -9

4x - 2y - z = 21

x + y - z = 3

x + 2y + z = 1

2x - 3y + 2z = 2

x - 3y + z = 9

2x - 5y + 3z = 13

3x + 2y - 2z = -12

-x + y + z = 0

x + 2y + z = 5

2x + y - z = 7

A =1

2

-1

1

0

2

æ

èç

ö

ø÷B =

0

3

1

2

-1

1

æ

èç

ö

ø÷C =

4

-5

1

0

2

2

æ

èç

ö

ø÷D =

1

3

3

3

2

1

æ

è

çç

ö

ø

÷÷

E =2 0

4 1

æ

èç

ö

ø÷F =

0 -2

1 2

æ

èç

ö

ø÷G =

-1 3 2

-2 0 1

4 2 3

æ

è

çç

ö

ø

÷÷

H =

0 -1 2

3 2 0

1 -1 4

æ

è

çç

ö

ø

÷÷

2A- 3B-C 3A- 2B+C A- 2B+ 3C2(A+ B+C) 4A- B- 2C 4B- A+ 2C

AD BD CD EA EB ECFA FB FC EF DE DFAG BG CG GH HG

P =2 0

4 a

æ

èç

ö

ø÷Q =

1 -2

0 b

æ

èç

ö

ø÷R =

-2 3

1 c

æ

èç

ö

ø÷S =

-3 4

1 d

æ

èç

ö

ø÷

T =3 1

-2 e

æ

èç

ö

ø÷U =

12 3

8 f

æ

èç

ö

ø÷ V =

5 -2

-1 g

æ

èç

ö

ø÷ W =

-4 -5

2 h

æ

èç

ö

ø÷

A =

k 3 2

-2 0 1

4 2 3

æ

è

çç

ö

ø

÷÷

B =

0 -1 2

m 2 0

1 -1 4

æ

è

çç

ö

ø

÷÷C =

-1 3 2

-2 0 1

p +1 2 3

æ

è

çç

ö

ø

÷÷

D =

2 3 -2

1 2 + q 4

-3 1 2

æ

è

çç

ö

ø

÷÷

P-1,Q-1,R-1,S-1,T -1,U-1,V -1,W -1

a,b,c,d,e, f ,g,h,k,m, p,q

Advanced Higher Mathematics – Geometry, Proof and Systems of Equations Unit Assessment Preparation - Further Practice Questions

Page 4

1.2 Applying algebraic and geometric skills to vectors

Calculating the vector product

4. (a) Given the vectors and , calculate

(b) Given the vectors and , calculate

(c) Given the vectors and , calculate

(d) Given the vectors and , calculate

(e) Given the vectors and , calculate

(f) Given the vectors and , calculate

1.2 Applying algebraic and geometric skills to vectors

Finding the equation of a line in three dimensions

5. Obtain the equation of each line in Parametric form, Symmetric form and Vector form. Each line passes through the pair of points as detailed below:

(a) and (b) and

(c) and (d) and

(e) and (f) and

(g) and (h) and

6. Obtain the equation of each line in Parametric form, Symmetric form and Vector form.

(a) A line passes through the point and is parallel to the line

(b) A line passes through the point and is parallel to the line

a =

3

2

-1

æ

è

çç

ö

ø

÷÷

b =

4

0

-2

æ

è

çç

ö

ø

÷÷

a´b

c =

4

0

-1

æ

è

çç

ö

ø

÷÷

d =

6

-2

3

æ

è

çç

ö

ø

÷÷

c´ d

e =

1

1

1

æ

è

çç

ö

ø

÷÷

f =

-2

-3

4

æ

è

çç

ö

ø

÷÷

e´ f

g = i - k h = 6 j + 5k g´ h

m = -2i - 3 j + k n = -i + 3 j + 4k m´ n

p = i + j + 2k r = 2 j - k p´ r

A 0, 1, 3( ) B -1, 2, -4( ) C 5, -1, 0( ) D 6, 2, -7( )E 3, 11, -2( ) F 6, -1, 0( ) G 2, 1, 0( ) H 3, 7, 10( )J 0, 0, 0( ) K 1, 2, 3( ) L 1, -2, -1( ) M 2, 3, 1( )N 3, -1, 6( ) P 0, -3, -1( ) R 1, 2, -1( ) S -1, 0, 1( )

S(1,-2,3)

0

1

3

æ

è

çç

ö

ø

÷÷

+ t

2

1

-1

æ

è

çç

ö

ø

÷÷

T (-1,2,-2)

5

2

0

æ

è

çç

ö

ø

÷÷

+ t

1

-1

1

æ

è

çç

ö

ø

÷÷

Advanced Higher Mathematics – Geometry, Proof and Systems of Equations Unit Assessment Preparation - Further Practice Questions

Page 5

(c) A line passes through the point and is parallel to the line

(d) A line passes through the point and is parallel to the line

(e) A line passes through the point and is parallel to the line

(f) A line passes through the point and is parallel to the line

1.2 Applying algebraic and geometric skills to vectors

Finding the equation of a plane

7. Obtain the equation of each plane in Parametric form, Cartesian form and Vector form.

(a) The plane has normal vector and passes through the point .

(b) The plane has normal vector and passes through the point .

(c) The plane has normal vector and passes through the point .

(d) The plane which passes through the point and is perpendicular to

(e) The plane which passes through the point and is perpendicular to .

(f) The plane which passes through the point and is perpendicular to .

(g) The plane which is parallel to the vectors and and passes through the

point .

(h) The plane which is parallel to the vectors and and passes through the

point .

(i) The plane which is parallel to the vectors and and passes through

the origin.

U(4,2,-1)

3

-2

1

æ

è

çç

ö

ø

÷÷

+ t

3

1

3

æ

è

çç

ö

ø

÷÷

V(1,-1,1)x - 3

2=y- 7

2=z -10

1

W (3,4,5)x +1

2=y- 2

1=z + 3

-3

Z(-1,3,0)x - 2

-2=y+1

-1=z - 2

4

2

3

1

æ

è

çç

ö

ø

÷÷

A(0,2,6)

5

4

-3

æ

è

çç

ö

ø

÷÷

B(2,1,-1)

2

-3

1

æ

è

çç

ö

ø

÷÷

C(5,3,-2)

D(-4,6,7) -4i + 6 j + 7k

E(-1,2,1) i - 3 j + 2k

F(1,-3,1) i + 2 j - 2k

3i + 2 j - k 4i - 2k

G(1,1,0)

i + j + 2k 2 j - k

H(1,2,-1)

i + j + k -2i - 3 j + 4k

Advanced Higher Mathematics – Geometry, Proof and Systems of Equations Unit Assessment Preparation - Further Practice Questions

Page 6

1.3 Applying geometric skills to complex numbers

Performing geometric operations on complex numbers

8. (a) Given z = 3 + i

(i) Find the modulus and argument of z using exact values.

(ii) Write in polar form.

(b) Given z = -1+ 3i (c) Given z = -3+ 3i (d) Given z = 5 3 + 5i

(e) Given z = 2 3 + 6i (f) Given z = -1- i (g) Given z = 2(1+ i)

9. (a) The complex number w has modulus 3 and argument p

4.

(i) Write w in Cartesian form using exact values.

(ii) Plot w on an Argand diagram.

(b) The complex number w has modulus 5 and argument p

4.

(c) The complex number w has modulus 2 and argument 2p

3.

(d) The complex number w has modulus 10 and argument -p

5.

(e) Given w = 5(cosp

6+ isin

p

6) (f) Given w = 2(cos

p

2+ isin

p

2) (g) Given w = 3(cos

4p

9+ isin

4p

9)

1.4 Applying algebraic skills to number theory

Using Euclid’s algorithm to find the greatest common divisor of two positive integers

10. Use the Euclidean algorithm to find the greatest common divisor of:

(a) 231 and 17 (b) 149 and 139 (c) 280 and 117 (d) 132 and 424 (e) 140 and 252 (f) 1365 and 770 (g) 1696 and 1504 (h) 2093 and 1679

z

Advanced Higher Mathematics – Geometry, Proof and Systems of Equations Unit Assessment Preparation - Further Practice Questions

Page 7

1.5 Applying algebraic and geometric skills to methods of proof

Disproving a conjecture by providing a counter-example

11. (a) For any real numbers a,b,c and d , it is conjectured that

a < b and c < dÞ ac < bd

Find a counter-example to disprove this conjecture. (b) For any real numbers a and b it is conjectured that

a

b<1Þ a < b

Find a counter-example to disprove this conjecture. (c) For any real numbers a and b it is conjectured that

a2 > b2 Þ a > b

Find a counter-example to disprove this conjecture.

1.5 Applying algebraic and geometric skills to methods of proof

Using direct and indirect proof in straightforward examples 12. (a) Prove by contradiction, that if 3n is even then n is also even, where

n is a natural number.

(b) Prove by contradiction, that if 5n is even then n is also even, where

n is a natural number.

(c) Prove by contradiction, that if n2 is odd then n is also odd, where

n is a natural number.

(d) Prove by contradiction, that if 2n2 is odd then n is also odd, where

n is a natural number.

(e) Use direct proof to show that if 3 is added to the square of an odd integer

the answer is divisible by 6.

(f) Use direct proof to show that the product of any two odd integer is an even integer.

(g) Use direct proof to show that if 5 is taken away from any odd number the answer is a

multiple of 2.

Advanced Higher Mathematics – Geometry, Proof and Systems of Equations Unit Assessment Preparation - Further Practice Questions

Page 8

Answers:

1 (a) (b) (c)

(d) (e) (f)

(g) (h)

2 (a) (b) (c)

(d) (e) (f)

(g) (h) (i)

(j) (k) (l)

(m) (n) (o)

(p) (q) (r)

(s) (t) (u)

(v) (w)

3 (a)

(b)

x = 5,y = 3,z = 2 x = 4,y = 2,z = -1 x = -2,y = 7,z = 4

x = 2,y = 6,z = -3 x = 5,y = -2,z = 3 x = 2,y = 0,z = -1

x = -2,y = -4,z = -1 x =1,y = 3,z = -2

-2

18

0

8

-5

5

æ

èç

ö

ø÷7

-5

-4

-1

4

6

æ

èç

ö

ø÷13

-19

0

-3

8

6

æ

èç

ö

ø÷

10

0

2

6

2

10

æ

èç

ö

ø÷-4

15

-7

2

-3

3

æ

èç

ö

ø÷7

0

7

7

0

6

æ

èç

ö

ø÷

-2

11

-1

10

æ

èç

ö

ø÷0

0

1

14

æ

èç

ö

ø÷13

1

16

-13

æ

èç

ö

ø÷

2

6

-2

-3

0

2

æ

èç

ö

ø÷0

3

2

6

-2

-3

æ

èç

ö

ø÷8

11

2

4

4

10

æ

èç

ö

ø÷

-4

5

-2

1

-4

4

æ

èç

ö

ø÷-6

6

-4

5

-2

1

æ

èç

ö

ø÷10

-6

0

1

-4

6

æ

èç

ö

ø÷

0

1

-4

-6

æ

èç

ö

ø÷

14

14

10

3

2

1

æ

è

çç

ö

ø

÷÷

3

2

1

4

-2

-4

æ

è

çç

ö

ø

÷÷

1

4

3

10

1

11

æ

èç

ö

ø÷-6

-3

-2

11

-2

11

æ

èç

ö

ø÷2

13

16

-11

15

-4

æ

èç

ö

ø÷

11 6 6

1 1 0

9 -3 20

æ

è

çç

ö

ø

÷÷

10 4 5

-7 9 8

9 11 13

æ

è

çç

ö

ø

÷÷

P-1 =1

2a

a

-4

0

2

æ

èç

ö

ø÷Q-1 =

1

b

b

0

2

1

æ

èç

ö

ø÷R-1 =

1

-2c- 3

c

-1

-3

-2

æ

èç

ö

ø÷

S-1 =1

-3d - 4

d

-1

-4

-3

æ

èç

ö

ø÷T -1 =

1

3e+ 2

e

2

-1

3

æ

èç

ö

ø÷U -1 =

1

12 f - 24

f

-8

-3

12

æ

èç

ö

ø÷

V -1 =1

5g - 2

g

1

2

5

æ

èç

ö

ø÷ W -1 =

1

10 - 4h

h

-2

5

-4

æ

èç

ö

ø÷

a = 0 b = 0 c = -3

2d = -

4

3e = -

2

3f = 2 g =

2

5h =

5

2

detA = 22 - 2k, k =11 detB = 2m- 4, m = 2 detC =19 + 3p, p =19

3

detD = -2q- 49, q = -49

2

Advanced Higher Mathematics – Geometry, Proof and Systems of Equations Unit Assessment Preparation - Further Practice Questions

Page 9

4 (a) (b) (c) (d)

(e) (f)

5 Equation of each line in Parametric Form:

(a) (b)

OR OR

(c) (d)

OR OR

(e) (f)

OR OR

(g) (h)

OR OR

5 Equation of each line in Symmetric Form:

(a) (b) (c)

OR OR OR

(d) (e) (f)

OR OR OR

(g) (h)

OR OR

5 Equation of each line in Vector Form:

(a) OR (b) OR

(c) OR (d) OR

-4i + 2 j - 8k -2i -18 j - 8k 7i - 6 j - k 6i - 5 j + 6k

-15i + 7 j - 9k -5i + j + 2k

x = t -1, y = -t + 2, z = 7t - 4 x = t + 5, y = 3t -1, z = -7t

x = t, y = -t +1, z = 7t + 3 x = t + 6, y = 3t + 2, z = -7t - 7

x = 3t + 3, y = -12t +11, z = 2t - 2 x = t + 2, y = 6t +1, z =10t

x = 3t + 6, y = -12t -1, z = 2t x = t + 3, y = 6t + 7, z =10t +10

x = t, y = 2t, z = 3t x = t +1, y = 5t - 2, z = 2t -1

x = t +1, y = 2t + 2, z = 3t + 3 x = t + 2, y = 5t + 3, z = 2t +1

x = -3t + 3, y = -2t -1, z = -7t + 6 x = -2t +1, y = -2t + 2, z = 2t -1

x = -3t, y = -2t - 3, z = -7t -1 x = -2t -1, y = -2t, z = 2t +1

x +1

1=y- 2

-1=z + 4

7

x - 5

1=y+1

3=z

-7

x - 3

3=y-11

-12=z + 2

2x

1=y-1

-1=z - 3

7

x - 6

1=y- 2

3=z + 7

-7

x - 6

3=y+1

-12=z

2x - 2

1=y-1

6=z

10

x

1=y

2=z

3

x -1

1=y+ 2

5=z +1

2x - 3

1=y- 7

6=z -10

10

x -1

1=y- 2

2=z - 3

3

x - 2

1=y- 3

5=z -1

2x - 3

-3=y+1

-2=z - 6

-7

x -1

-2=y- 2

-2=z +1

2x

-3=y+ 3

-2=z +1

-7

x +1

-2=y

-2=z -1

2

0

1

3

æ

è

çç

ö

ø

÷÷

+ t

1

-1

7

æ

è

çç

ö

ø

÷÷

-1

2

-4

æ

è

çç

ö

ø

÷÷

+ t

1

-1

7

æ

è

çç

ö

ø

÷÷

5

-1

0

æ

è

çç

ö

ø

÷÷

+ t

1

3

-7

æ

è

çç

ö

ø

÷÷

6

2

-7

æ

è

çç

ö

ø

÷÷

+ t

1

3

-7

æ

è

çç

ö

ø

÷÷

3

11

-2

æ

è

çç

ö

ø

÷÷

+ t

3

-12

2

æ

è

çç

ö

ø

÷÷

6

-1

0

æ

è

çç

ö

ø

÷÷

+ t

3

-12

2

æ

è

çç

ö

ø

÷÷

2

1

0

æ

è

çç

ö

ø

÷÷

+ t

1

6

10

æ

è

çç

ö

ø

÷÷

3

7

10

æ

è

çç

ö

ø

÷÷

+ t

1

6

10

æ

è

çç

ö

ø

÷÷

Advanced Higher Mathematics – Geometry, Proof and Systems of Equations Unit Assessment Preparation - Further Practice Questions

Page 10

(e) OR (f) OR

(g) OR (h) OR

6 Equation of each line in Parametric Form:

(a) (b)

(c) (d)

(e) (f)

6 Equation of each line in Symmetric Form:

(a) (b) (c)

(d) (e) (f)

6 Equation of each line in Vector Form:

(a) (b) (c)

(d) (e) (f)

7 (a) Cartesian = (b) Cartesian =

Parametric = Parametric =

Vector = Vector =

0

0

0

æ

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+ t

1

2

3

æ

è

çç

ö

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÷÷

1

2

3

æ

è

çç

ö

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÷÷

+ t

1

2

3

æ

è

çç

ö

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÷÷

1

-2

-1

æ

è

çç

ö

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÷÷

+ t

1

5

2

æ

è

çç

ö

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÷÷

2

3

1

æ

è

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ö

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÷÷

+ t

1

5

2

æ

è

çç

ö

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÷÷

3

-1

6

æ

è

çç

ö

ø

÷÷

+ t

-3

-2

-7

æ

è

çç

ö

ø

÷÷

0

-3

-1

æ

è

çç

ö

ø

÷÷

+ t

-3

-2

-7

æ

è

çç

ö

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÷÷

1

2

-1

æ

è

çç

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÷÷

+ t

-2

-2

2

æ

è

çç

ö

ø

÷÷

-1

0

1

æ

è

çç

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÷÷

+ t

-2

-2

2

æ

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çç

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÷÷

x = 2t +1, y = t - 2, z = -t + 3 x = t -1, y = -t + 2, z = t - 2

x = 3t + 4, y = t + 2, z = 3t -1 x = 2t +1, y = 2t -1, z = t +1

x = 2t + 3, y = t + 4, z = -3t + 5 x = -2t -1, y = -t + 3, z = 4t

x -1

2=y+ 2

1=z - 3

-1

x +1

1=y- 2

-1=z + 2

1

x - 4

3=y- 2

1=z +1

3

x -1

2=y+1

2=z -1

1

x - 3

2=y- 4

1=z - 5

-3

x +1

-2=y- 3

-1=z

4

1

-2

3

æ

è

çç

ö

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÷÷

+ t

2

1

-1

æ

è

çç

ö

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÷÷

-1

2

-2

æ

è

çç

ö

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÷÷

+ t

1

-1

1

æ

è

çç

ö

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÷÷

4

2

-1

æ

è

çç

ö

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÷÷

+ t

3

1

3

æ

è

çç

ö

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÷÷

1

-1

1

æ

è

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ö

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÷÷

+ t

2

2

1

æ

è

çç

ö

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÷÷

3

4

5

æ

è

çç

ö

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÷÷

+ t

2

1

-3

æ

è

çç

ö

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÷÷

-1

3

0

æ

è

çç

ö

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÷÷

+ t

-2

-1

4

æ

è

çç

ö

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÷÷

2x+ 3y+ z =12 5x+ 4y- 3z =17

x = t, y =2 + 3t

3, z = 6 + t x =

2 + 5t

5, y =

1+ 4t

4, z =

1+ 3t

3

0

2

6

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ö

ø

÷÷

+

2

3

1

æ

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çç

ö

ø

÷÷t

2

1

-1

æ

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ö

ø

÷÷

+

5

4

-3

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÷÷t

Advanced Higher Mathematics – Geometry, Proof and Systems of Equations Unit Assessment Preparation - Further Practice Questions

Page 11

(c) Cartesian = (d) Cartesian =

Parametric = Parametric =

Vector = Vector =

(e) Cartesian = (f) Cartesian = x+ 2y- 2z = -7

Parametric = Parametric =

Vector = Vector =

(g) Cartesian = -4x+ 2y- 8z = -2 OR 2x- y+ 4z =1

Parametric =

Vector =

1

1

0

æ

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çç

ö

ø

÷÷

+

3

2

-1

æ

è

çç

ö

ø

÷÷s +

4

0

-2

æ

è

çç

ö

ø

÷÷t (h) Cartesian =

Parametric =

Vector =

1

2

-1

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çç

ö

ø

÷÷

+

1

1

2

æ

è

çç

ö

ø

÷÷s +

0

2

-1

æ

è

çç

ö

ø

÷÷t

(i) Cartesian = 7x- 6y- z = 0

Parametric =

Vector =

1

1

1

æ

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çç

ö

ø

÷÷s +

-2

-3

4

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÷÷t

8 (a) modulus = 2 , argument = p

6, polar form = 2(cos

p

6+ isin

p

6)

(b) modulus = 2 , argument = 2p

3, polar form = 2(cos

2p

3+ isin

2p

3)

(c) modulus = 2 3 , argument = 5p

6, polar form = 2(cos

5p

6+ isin

5p

6)

(d) modulus = 10 , argument = p

6, polar form = 10(cos

p

6+ isin

p

6)

(e) modulus = 4 3, argument = p

3, polar form = 4 3(cos

p

3+ isin

p

3)

2x- 3y+ z = -1 -4x+ 6y+ 7z =101

x =5 + 2t

2, y =

2 - 3t

-3, z = 6 + t x =1+ t, y =1+ t, z =1+ t

5

3

-2

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ø

÷÷

+

2

-3

1

æ

è

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ö

ø

÷÷t

-4

6

7

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ø

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+

-4

6

7

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x - 3y+ 2z = -5

x = -1+ t, y =2 - 3t

-3, z =

1+ 2t

2x =1+ t, y =

-3+ 2t

2, z =

1- 2t

-2

-1

2

1

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ø

÷÷

+

1

-3

2

æ

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1

-3

1

æ

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ö

ø

÷÷

+

1

2

-2

æ

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ö

ø

÷÷t

x =1+ 3s+ 4t, y =1+ 2s, z = -s- 2t

-5x + y+ 2z = -5

x =1+ s, y = 2 + s+ 2t, z = -1+ 2s- t

x = s- 2t, y = s- 3t, z = s+ 4t

Advanced Higher Mathematics – Geometry, Proof and Systems of Equations Unit Assessment Preparation - Further Practice Questions

Page 12

8 (f) modulus = 2 , argument = 3p

4, polar form = 2(cos

3p

4+ isin

3p

4)

(g) modulus = 2 , argument = p

4, polar form = 2(cos

p

4+ isin

p

4)

10 (a) GCD = 1 (b) GCD = 1 (c) GCD = 1 (d) GCD = 4 (e) GCD = 28 (f) GCD = 37 (g) GCD = 32 (h) GCD = 23

11 (a) Choose suitable values, eg a = -3, b =1, c = -5 and d = -2

Show result is false, eg ac =15 and bd = -2 Communicate result, Since ac > bd conjecture is not true (b) Choose suitable values, eg a = -3, and b = -4

Show result is false, eg a

b=

3

4<1

Communicate result, Since a > b conjecture is not true (c) Choose suitable values, eg a = -5 and b = -3

Show result is false, a2 = 25 and b2 = 9 Communicate result, Since a < b conjecture is not true 12 (a) State opening conjecture, Assume that there is an odd number n such that 3n is even

Start process, Let n = 2k -1, kÎN

Complete process, 3n = 3(2k -1) = 6k - 3 = 2(3k -1)-1, which is odd since 3k -1ÎN

Contradiction statement and conclusion, This contradicts the initial statement so if n is even then 3n is also even. 12 (b) State opening conjecture, Assume that there is an odd number n such that 5n is even

Start process, Let n = 2k -1, kÎN

Complete process, 5n = 5(2k -1) =10k - 5 = 2(5k - 2)-1, which is odd since 5k - 2 ÎN

Contradiction statement and conclusion, This contradicts the initial statement so if n is even then 5n is also even. 12 (c) State opening conjecture,

Assume that there is an even number n such that n2 is odd

Start process, Let n = 2k, kÎN

Complete process, n2 = (2k)2 = 4k2 = 2(2k2 ), which is even since 2k2 ÎN

Contradiction statement and conclusion,

Advanced Higher Mathematics – Geometry, Proof and Systems of Equations Unit Assessment Preparation - Further Practice Questions

Page 13

This contradicts the initial statement so if n2 is odd then n is also odd. 12 (d) State opening conjecture,

Assume that there is an even number n such that 2n2 is odd

Start process, Let n = 2k, kÎN

Complete process, 2n2 = 2(2k)2 = 8k2 = 2(4k2 ), which is even since 4k2 ÎN

Contradiction statement and conclusion,

This contradicts the initial statement so if 2n2 is odd then n is also odd.

12 (e) Start process, odd number 2n-1, where nÎN

Complete process, (2n-1)2 = 4n2 - 4n+1+ 3 = 4n2 - 4n+ 4

Communicate result, = 4(n2 - n+1) is divisible by 4 since n2 - n+1 is clearly an integer

(f) Start process, odd integers 2n-1, where nÎN

Complete process, (2n-1)(2n-1) = 4n2 - 4n+1= 4(n2 - n)+1

Communicate result, 4(n2 - n)+1 is odd since 4(n2 - n) is clearly even

(g) Start process, odd number (2n-1)- 5 = 2n- 6, where nÎN

Complete process, 2n- 6 = 2(n- 3)

Communicate result, is a multiple of 2