- 1 -

MATHEMATICS HIGHER

Units 1, 2 and 3

2002

Paper II

Questions

&

Worked Solutions

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1) Triangle ABC has vertices A(-1, 6),

B(-3, -2) and C(5, 2).

Find

(a) the equation of the line p, the

median from C of triangle ABC.

(b) the equation of the line q, the

perpendicular bisector of BC.

(c) the co-ordinates of the point of

intersection of the lines p and q.

Soln

(a) The line p, the median from C, goes through the point C and the mid-point of AB.

Find the mid-point of AB.

A(-1, 6) Let P be the mid-point of AB.

B(-3, -2) P( )

= ( )

= ( )

P = (-2, 2)

Find the gradient of CP Equation of CP i.e the line p

C(5, 2) mCP = y2 − y1 C(5, 2) mCP = 0

P(-2, 2) x2 − x1 y − b = m(x − a)

= 2 − 2 y − 2 = 0(x − 5)

-2 − 5 y − 2 = 0

= 0 y = 2

-7

= 0

B(-3, -2)

C(5, 2)

x1 + x2 , y1 + y2

2 2

-1 + -3 , 6 + -2

2 2

-4 , 4

2 2

A(-1, 6)

p

q

x

y

- 3 -

(b) The perpendicular bisector of BC, q bisects BC

meets BC at right angles i.e. mq • mBC = -1

Find the mid-point of BC Find the gradient of BC

B(-3, -2) Let Q be the mid-point of BC. mBC = y2 − y1

x2 − x1

C (5, 2) Q( ) B(-3, -2)

= ( ) C (5, 2)

= ( )

Q = (1, 0)

Gradient of q,

the perpendicular bisector of BC = -2

i.e. mq = -2

Equation of q,the perpendicular bisector of BC

C(1, 0) mq = -2

y − b = m(x − a)

y − 0 = -2(x − 1)

y = -2x + 2

(c) Find the point of intersection of the lines p and q.

Find the point of intersection of y = 2

y = -2x + 2

2 = -2x + 2

2x = 0

x = 0

The point of intersection of the lines p and q is (0, 2)

-3 + 5 , -2 + 2

2 2

2 , 0

2 2

x1 + x2 , y1 + y2

2 2 = 2 − (-2)

5 − (-3)

= 2 + 2

5 + 3

= 4

8

= 1

2

For perpendicular lines

m1 x m2 = -1

BC and q are

perpendicular,

mBC x mq = -1

1 x –2 = -1

2

The gradient of q is -1

- 4 -

6

0

0

3

3

8

6 - 3

0 - 3

0 - 8

3

-3

-8

6

0

0

6 - 3

6 - 3

0 - 8

3

-3

-8

3

3

8

cos ∠ADB = DA . DB

DADB

= (3 x 3) + (-3 x 3) + (-8 x -8)

√(32 + (-3)

2 + (-8)

2 x √(3

2 + (3)

2 + (-8)

2

= 9 + (-9) + 64

√ 9 + 9 + 64 x √ 9 + 9 + 64

= 64

√ 82 x √ 82

cos ∠ADB = 64 82

= 0.780487804….

∠ADB = 38.69473992…..

∠ADB = 38.70 ( 1 decimal place )

6 units

6 units

6 units

6 units

2) The diagram shows a square-based pyramid

of height 8 units.

Square OABC has a side length of 6 units.

The co-ordinates of A and D are (6, 0, 0) and

(3, 3, 8).

C lies on the y-axis.

(a) Write down the co-ordinates of B.

(b) Determine the components of DA and DB.

(c) Calculate the size of angle ADB.

Soln

(a) B is 6 units on the x-axis, 6 units on the y-axis and 0 units on the z-axis.

B is the point (6, 6, 0).

(b) DA = a - d

= ( ) − ( )

= ( )

= ( )

DB = b - d (c)

= ( ) − ( )

= ( )

= ( )

From the above:

DA . DB = (3 x 3) + (-3 x 3) + (-8 x -8)

= 9 + (-9) + 64

= 64

DA= √ 32 + (-3)

2 + (-8)

2 = √ 9 + 9 + 64 = √ 82

DB= √ 32 + (3)

2 + (-8)

2 = √ 9 + 9 + 64 = √ 82

z

6

6

0

3

3

-8

C B

A (6,0,0) x

y

z

O

D(3,3,8)

C B

A (6,0,0) x

y

z

O

D (3,3,8)

(6,6,0)

- 5 -

. . 0

y = f(x)

A

(2, 0)

.

positive zero positive negative zero

3) The diagram shows part of the graph of the

curve with equation y = 2x3 − 7x

2 + 4x + 4.

(a) Write down the x-coordinate of the maximum

turning point.

(b) Factorise 2x3 − 7x

2 + 4x + 4.

(c) State the coordinates of the point A and

hence find the values of x for which

2x3 − 7x

2 + 4x + 4 < 0.

Soln

(a) For stationary points dy = 0

dx

Differentiate the curve and set equal to zero y = 2x3 − 7x

2 + 4x + 4

dy = 6x2 − 14x + 4

dx

Set dy = 0

dx

6x2 − 14x + 4 = 0

2(3x2 − 7x + 2) = 0

2(3x− 1)(x− 2) = 0

(3x− 1) = 0 or (x− 2) = 0

x = 1 or x = 2

3

dy = 0 when x = 1 or x = 2 i.e. stationary points occur at x = 1 or x = 2

dx 3 3

Nature of stationary points

x 0 1 1 2 3

3

dy = 2(3x− 1)(x− 2) 2(0 – 1)(0 – 2) 2(1– 1)(1 – 2) 2(3 – 1)(1 – 2) 2(6 – 1)(2 – 2) 2(9 – 1)(3 – 2)

dx = 2(-1)(-2) 3 = 2(2)(-1) = 2(5)(0) = 2(8)(1)

= 4 = 2(0)(-5 ) = -4 = 0 = 16

3

= 0

At x = 1 there is a maximum turning point. The x-coordinate of the maximum turning point is 1

3 3

x

y

- 6 -

.

(b) Factorise 2x3 − 7x

2 + 4x + 4.

From the diagram x = 2 is one solution.

(x − 2) is a factor of 2x3 − 7x

2 + 4x + 4.

Using the factor theorem

2 2 -7 4 4

4 -6 -4

2 -3 -2 0

y = (x − 2)(2x2 − 3x − 2)

2x2 − 3x − 2 = (2x + 1)(x − 2)

y = (x − 2)(2x + 1)(x − 2)

y = (x − 2)2(2x + 1)

y = 2x3 − 7x

2 + 4x + 4 = (x − 2)

2(2x + 1)

(c) Set y = 0 (x − 2) = 0 or (2x + 1) = 0

x = 2 or x = -1

2

The points (2, 0) and ( -1 , 0)

2

are on the curve y = 2x3 − 7x

2 + 4x + 4.

A is the point ( -1 , 0)

2

y = 2x3 − 7x

2 + 4x + 4

2x3 − 7x

2 + 4x + 4 < 0 when x < -1

2

x

y

(2, 0)

. x

y

(2, 0)

A

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A

(4) A man decides to plant a number of fast-growing trees as a boundary between

his property and the property of his next door neighbour. He has been

warned, however, by the local garden centre that, during any

year the trees are expected to increase in height by 0.5 metres. In response to

this warning he decides to trim 20% off the height of the trees at the start of any year.

(a) If he adopts the “20% pruning policy”, to what height will he expect the

trees to grow in the long run?

(b) His neighbour is concerned that the trees are growing at an alarming rate

and wants assurances that the trees will grow no taller than 2 metres.

What is the minimum percentage that the trees will need to be trimmed

each year so as to meet this condition?

Soln

(a) At the start of any year, after pruning 20% off the trees, the man is left with

80% of the height of the trees.

80% = 0.80 = 0.8

Also, during the previous year, the trees have grown 0. 5 metres.

Recurrence relation u n+1 = 0.8u n + 0.5 u n+1 = au n + b

L = 0.5 a = 0.8 b = 0.5

1 − 0.8

= 0.5 -1 < 0.8 < 1 , ∴∴∴∴a limit exists

0.2

= 5 L = b

2 1 − a

= 2.5

If the man adopts a “20% pruning policy”

he will expect the trees, in the long run,

to grow to a height of 2.5 metres.

(b) The neighbour wants the trees to grow to a height of

no more than 2 metres.

Set L = 2

The trees are still expected, during any year, to

increase in height by 0.5 metres.

b = 0.5

Require the minimum percentage that the trees need

to be trimmed each year for a Limit of 2 metres.

Find the value of a.

For the trees to grow no taller than 2 metres, the recurrence relation is u n+1 = 0.75u n + 0.5. The man is left with 75% of the height of the trees.

To not exceed 2 metres, the minimum % for trimming each year is 25%.

L = b

1 − a

L = 2, b = 0.5

2 = 0.5

1 − a

(x by 1 − a both sides)

2(1 − a) = 0.5

(divide by 2 both sides)

1 − a = 0.25

1 − 0.25 = a

a = 0.75

- 8 -

y

x

y = 1 + 10x – 2x2

y = 1 + 5x – x2

. . o

5) Calculate the shaded area enclosed

between the parabolas with equations

y = 1 + 10x – 2x2 and

y = 1 + 5x – x2.

Soln

Find the x-coordinates of the pints

of intersection of

y = 1 + 10x – 2x2 and y = 1 + 5x – x

2.

Set equal

i.e. 1 + 10x – 2x2 = 1 + 5x – x

2

10x – 2x2 = 5x – x

2

0 = x2

– 5x

x2

– 5x = 0

x(x – 5) = 0

x = 0 or x – 5 = 0

x = 5

5

∫1 + 10x – 2x2 − (1 + 5x – x

2 ) dx 0

5

= ∫1 + 10x – 2x2 − 1 – 5x + x

2 dx

0

5

= ∫5x – x2

dx 0

5

= 5x2

– x3

= 5(5)2

– 53 – 5(0)

2 – 0

3

2 3 0 2 3 2 3

= 125 – 125 – 0

– 0

2 3 2 3

= 375 – 250

6 6

= 125

6

= 20 units2

5

6

- 9 -

6) Find the equation of the tangent to the curve y = 2sin(x − ) at the point where

x =

Soln

x = y = 2sin(x − )

= 2sin( )

= 2sin( )

= 2sin( )

= 2 . 1

2

= 1

The point ( ) lies on both the curve and the tangent to the curve.

Find the gradient of the tangent to the curve at any point.

i.e. differentiate y = 2sin(x − )

dy = 2cos(x − )

dx

x =

= 2cos( )

= 2cos( )

= 2cos( )

= 2. √3

2

= √3

At the point ( ) the gradient of the tangent to the curve is √3

π 6

π .

3

π 6

π 3

π − π 3 6

2π − π

6 6

π 6

sin π = 1 = 0.5

6 2

π

6

π 3 1

2

√3

π ,

3 1

π 6

π 6

π 3

π − π 3 6

2π − π

6 6

π 6

cos π = √3

6 2

π

6

π

3 1

2

√3

π ,

3 1

- 10 -

Find the equation of the tangent. y − b = m( x − a )

point ( ) y − 1 = √3 ( x − )

mtan = √3 y − 1 = √3x − √3.

( x by 3 both sides )

3y − 3 = 3.√3x − √3π

3y = 3.√3x − √3π + 3

( divide by 3 both sides )

y = √3x − √3π + 1

3

y = √3x + 1 − π

√3

π ,

3 1 π

3

π 3

- 11 -

7) Find the x-coordinate of the point where the graph of the curve with equation

y = log3( x − 2) + 1 intersects the x-axis.

Soln

The graph of y = log3( x − 2) + 1 intersects the x-axis when y = 0.

log3( x − 2) + 1 = 0

log3( x − 2) = -1

From above, the graph of y = log3( x − 2) intersects the x-axis at the point ( )

( x − 2) = 3-1

( x − 2) = 1

3

x = 1 + 2

3

x = 2 or x = 7

3

1

3

7 , 0

3

- 12 -

1

2

3

2

3

2

8) A point moves in a straight line such that its acceleration a is given by

a = 2(4 − t) , 0 ≤ t ≤ 4. If it starts at rest, find an expression for the velocity v

where a = dv .

dt

Soln

v = ∫ dt

= 2 ∫ dt

= 2 . (4 − t) . (-1) + C

= 2 . 2 . (-1) . (4 − t) + C

3

v = -4(4 − t) + C

3

Find the value of C

v = -4(4 − t) + C

3

When t = 0, v = 0

0 = -4(4 − 0) + C

3

0 = -4(4 ) + C

3

4(√4 )3 = C

3

C = (4)(2)3

3

C = (4)(8) v = -4(4 − t) + C

3 3

C = 32 v = -4(4 − t) + 32 v = -4 (4 − t) − 8

3 3 3 3

2(4 − t) 1

2

(4 − t) 1

2

3

2

3

2

3

2

3

2

3

2

Change side, change sign

3

2

3

2

3

2

- 13 -

9) Show that the equation (1 − 2k)x2 − 5kx − 2k = 0 has real roots for all integer

values of k.

Soln

(1 − 2k)x2 − 5kx − 2k = 0 has real roots given b

2 − 4ac ≥ 0.

a = 1 − 2k b2 − 4ac = (-5k)

2 − 4(1− 2k)(-2k)

b = -5k = 25k2− (4 − 8k)(-2k)

c = -2k = 25k2

+ 2k (4 − 8k)

= 25k2

+ 8k − 16k2

= 9k2

+ 8k

9k2

+ 8k ≥ 0 for all integer values of k.

∴b2 − 4ac ≥ 0 for all integer values of k.

∴(1 − 2k)x2 − 5kx − 2k = 0 has real roots for all integer values of k.

9k2

+ 8k ≥ 0 for all integer values of k

Find the zeros of the function ( i.e. set y = 0) Find the stationary value

9k2

+ 8k = 0 dy = 18k + 8

k(9k + 8) = 0 dk

Set dy = 0

k = 0 or 9k + 8 = 0 dk

k = -8 18k + 8 = 0

9 k = -8 = -4

18 9

(0,0) and ( ) lie on the curve y = 9k2

+ 8k.

Stationary value when k = -4

9

Nature of stationary point

k -1 -4 0

9

dy -18+8 0 0+8

dk = -10 = 8

= 18k + 8

From the above diagram

9k2

+ 8k < 0 for –8 < k < 0

9 y = 9k2

+ 8k has a minimum turning point

k is not an integer when –8 < k < 0 when k = -4

9 9

∴ 9k2

+ 8k < 0 when k is not an integer.

∴ 9k2

+ 8k ≥ 0 for all integer values of k.

y = 9k2

+ 8k is a parabola

-8 , 0

9

neg. zero pos.

x

y

-4

9

-8

9

-1

0

y = 9k2

+ 8k

- 14 -

6m

8m

Village Hall

Manse Lane

Th

e V

enn

el

o

y

x

(0, 6)

(a, 0) (8, 0)

l

b

.

. .

8

10 6

a

l

multiply by a

8

∗ ∗

10) The shaded rectangle on this map

represents the planned extension to the

village hall. It is hoped to provide the

largest possible area for the extension.

The coordinate diagram represents the

right angled triangle of ground behind the

hall. The extension has length l metres

and breadth b metres, as shown. One

corner of the extension is at the point

(a, 0).

(a) (i) Show that l = 5 a.

4

(ii) Express b in terms of a and hence deduce that the area, A m2, of the

extension is given by A = 3 a(8 − a).

4

(b) Find the value of a which produces the largest area of the extension.

Soln

(a) (i) From the 2nd

diagram above

Similar Triangles

By Pythagoras

10 =√62

+ 82

=√36 + 64

=√100

RSF (Reduction Scale Factor) = small = a l = 10 • a = 5 a

big 8 8 4

l = 5 a

4

- 15 -

8

10 6

b

8 − a

8 − a

b

8

10 6

Rotate

Clockwise

b 8 − a

multiply by (8 −a)

10

∗ ∗

(a) (ii) From the 2nd

diagram

Similar Triangles

The area of the extension, A m2, is given by the formula A = lb

= 5 a • 3 (8 − a)

4 5

= 3 a (8 − a)

4

A = 3 a (8 − a)

4

Reflect

RSF (Reduction Scale Factor) = small = (8 − a) b = 6 • (8 − a) = 3 • (8 − a)

big 10 10 5

b = 3 (8 − a)

5

l = 5 a

4

b = 3 (8 − a)

5

- 16 -

(b) Find the value of a which produces the largest area of the extension.

A = 3 a (8 − a)

4

= 24a − 3a2

4 4

= 6a − 3a2

4

A = 6a − 3 a2

4

Differentiate

dA = 6 − 6 a Set dA = 0 6 − 3 a = 0

da 4 da 2

= 6 − 3 a (x by 2 both sides)

2 12 − 3a = 0

12 = 3a

3a = 12

a = 4

For A = 6a − 3 a2

, a stationary point occurs when a = 4.

4

Check nature of the stationary point

a 3 4 5

dA 6 − 9 6 − 12 6 − 15

da 2 2 2

= 6 − 3 a = 6 − 4.5 = 6 − 6 = 6 − 7.5

2 = 1.5 = 0 = -1.5

positive zero negative

A = 6a − 3 a2 has a maximum turning point at a = 4

2

i.e. a = 4 is the value of a which produces the largest area of the extension.

- 17 -