year 2013 is work hard, try hard regret!! remembering! · year 2013 is going to be a year worth...

Year 2013 is

going to be a

year worth

remembering!

Work hard, try hard and NO regret!!

Additional

Mathematics

4038

Six Years Series

2007 – 2012

Questions and

Full solutions

Dear Students

This is a compilation of the latest past six years Additional Mathematics O level 4038.

It is meant for you as a reference and personal use. The material is copyrighted and it is only for

your personal use.

Good luck.

Mr Ang June 2013

Solutions to O Level Add Math paper 1 2007

By KL Ang, Jun 2013 Page 1

1. Find the set of values of the constant k for which the 1y k x intersects the curve

2 6y x x k at two distinct points.

Solution :

21 6k x x x k

2 6 2 0x k x k

Discriminant, 2

6 4 1 2k k

2 20 36k k

Let 0 to get two distinct roots,

2 20 36 0k k

18 2 0k k

2k or 18k

Alternative solution:

21 6k x x x k

2 6 2 0x k x k --------- (1)

For equation (1) to have two distinct solutions, we just need the minimum point of this “smiley”

curve to be below the x-axis, or negative.

Let 2 6 2y x k x k

Line of symmetry,

6 6

2 1 2

k kx

[Analysis]

To solve simultaneous equations that will give two distinct solutions. After combining the two

equation, the quadratic equation will have to have a positive discriminant.

k

w 2 20 36w k k

2 18

In Summary:

This is a rather routine question. Knowing the discriminant of quadratic

equation well, particularly nature of roots, will always be important. Solving

quadratic inequality occurs frequently in this paper.



When 6

2

kx

,

2

min

6 66 2

2 2

k ky k k

21

6 24

k k

2

5 94

kk

For two distinct roots, min 0y

2

5 9 04

kk

2 20 36 0k k

18 2 0k k

2k or 18k

An alternative approach to solve quadratic inequality

Given a “Smiley” quadratic curve, like 2 20 36 0w k k , it is easier to consider the

complementary set of 2 20 36 0w k k .

Consider 2 20 36 0w k k ,

18 2 0k k

2 18k This is the range of values that we need to exclude!

So, the solution to 2 20 36 0k k , is

2k or 18k

Using this approach, we always turn any quadratic inequality into a “Smiley” and then with the

knowledge of as 2 0ax bx c , where 0a . The solution to this kind of inequality is always

1 2x x x ., where 2

1 2ax bx c a x x x x and 1 2x x .

k

w 2 20 36w k k

2 18



2. Given that 1 3s , express 2 2

1

s

s

in the form 3a b , where a and b are integers.

Solution :

By long division,

2 2 3

11 1

ss

s s

Given that 1 3s ,

2 2 31 3 1

1 1 3 1

s

s

33

2 3

3 2 33

2 3 2 3

2

2

6 3 33

2 3

3 6 3 3

6 2 3

[Analysis]

This is about rationalizing denominator where

1 a

aa and

2

11 a b a b

a ba b a b a b

.




Given that 1 3s ,

2

2 1 3 22

1 1 3 1

s

s

2

1 2 3 3 2

2 3

6 2 3

2 3

6 2 3 2 3

2 3 2 3

2

2

2 6 2 3 3 6 2 3

2 3

12 4 3 6 3 6

4 3

6 2 3

In Summary:

Take note of the first solution. After simplifying the algebraic expression before

ratinalising the denominator can usually make the question easier to do.



3. The value, V dollars, of a house is given by 0

ktV V e , where 0V dollars is the original value

of the house when built, t is the time in years since it was built and k is a constant.

Calculate

(i) the value of k if, after 10 years, the value of the house has doubled,

(ii) the value of t when the value of the house is three times its original value.

Solution :

(i)

Given that when 10t , 02V V

10

0 02k

V V e ,

102 ke

ln 2 10k

1

ln 2 0.069310

k

(ii)

When 03V V ,

0 03 ktV V e

3 kte

ln3 kt

ln 3 ln 3

15.8ln 2

10

tk

[Analysis]

A straight forward question on exponential(indices) and logarithm.

In Summary:

Answer for the part (ii) may vary a little, depending on the value used in k.



4.

(i) Find the integer which satisfies the equation 3 2 11 3 0x x x .

(ii) Find, in the form 3a b , a and b are integers, the other values which satisfy the

equation.

Solution :

(i)

Let 3 2f 11 3x x x x ,

when 3x , 3 2

f 3 3 3 11 3 3 0

therefore 3x is a factor of 3 2f 11 3x x x x

By long division,

2f 3 4 1x x x x .

23 4 1 0x x x

3x or 2 4 1 0x x

Since the discriminant of 2 4 1 0x x ,

Discriminant, 2

4 4 1 1 12 0 is not a square number, the roots will

not be a rational root. So the only integer root is 3x .

Alternatively,

f 1 0 , f 1 0 , f 3 0 , so the only integer root is 3x .

(ii)

When 2 4 1 0x x , by quadratic formula,

24 4 4 1 1

2 1x

2 3x

[Analysis]

This question involves solving a polynomial equation. So, factor theorem maybe a productive

tool. In part (i), the possible integer zeros(roots) are 1, 3 . Part (ii) is to be completed with

the application of quadratic formula.



Alternatively,

When 2 4 1 0x x , by completing the square method,

2 4 1 0x x

2 22 2 2 2 2 1 0x x

2

2 3 0x

2

2 3x

2 3x

2 3x


3 2 11 3 0x x x

3 2 23 4 11 3 0x x x x

2 23 4 11 3 0x x x x

2 3 4 1 3 0x x x x

23 4 1 0x x x

In Summary:

Take note that in part (i), it is sufficient to apply remainder theorem with non-

zero remainders for all possible roots to find all integer(rational) roots.

Finding roots with completing the square method, may be easier in this case.

Long division of polynomial is really a very useful technique.



5.

(i) Differentiate ln sin x with respect to x .

(ii)

The diagram shows part of the curve coty x , cutting the x -axis at ,02

. The line

3y intersects the curve at P . A line is drawn from P , parallel to the y -axis, to

meet the x -axis at Q . Use your result from part (i) to find the area of the shaded

region.

Solution:

(i) Given that ln siny x ,

Let sinu x , cosdu

xdx

lny u , 1dy

du u

By chain rule,

dy dy du

dx du dx

1

cosdy

xdx u

cos

sin

dy x

dx x

cotdy

xdx

[Analysis]

Part (i) is to apply chain rule. Part (ii) is to integrate the area under the curve, which will

require the x-coordinate of the point Q to be found.

x

y coty x

3 P

Q

2



(ii)

Given that coty x , when 3y , where 02

x

3 cot x

1

3tan x

1

tan3

x

1 1tan

3x

6

x

2

2

6

6

cot ln sinxdx x

ln sin ln sin2 6

1

ln 1 ln2

ln 2 0.693 square units

In Summary:

Part (i) solution shows the full application of chain rule. Part (ii) provides an

interesting twist, make sure that you understand the range of x that is acceptable in

this question. Additional solution will need to be rejected.



6. The matrices A and B are such that 2

1A= B . Given that

2 1

2 1

B= , find

(i) the matrix A ,

(ii) the value of the constant k for which 1 4k B A I , where I is the identity matrix.

Solution:

(i) Given that 2 1

2 1

B= ,

det B=2 1-2 1 4

11 11

2 24

B =

2

21

1 1 1 1 1 1 1 31 1 1

2 2 2 2 2 2 6 24 16 16

B

2

11 31

6 216

A= B

(ii) For 1 4k B A I

11 1

2 24

kk

B =

1 3 1 0 3 3 1 14 1 3

46 2 0 1 6 6 2 216 4 4

A I

1 1 1 13

2 2 2 24 4

k

Therefore , 3k

[Analysis]

Consider 1A A=I

, given that A is a 2 2 matrix, 1 0

0 1I

. 2AA=A .




(i) Given that 2

1A= B ,

1 1 BA=BB B

1BA=IB

1BBA=BB

2B A=I

Let a b

c d

A

2 1 2 1 1 0

2 1 2 1 0 1

a b

c d

2 3 1 0

6 1 0 1

a b

c d

2 3 2 3 1 0

6 6 0 1

a c b d

a c b d

1

2 3 1 2 1 3 16

6 0 6 6

16

aa c a c

a c a cc

3

2 3 1 2 1 3 16

6 0 6 2

16

bb d b d

b d b dd

1 3

1 3116 16

6 2 6 216

16 16

A

(ii) Given that 1 4k B A I ,

1 4k B B A I B

4k I AB IB

1 14k I B B B B

14k I B I B

14k I B B ------------ (1)



det B=2 1-2 1 4

11 11

2 24

B =

Substitute B and 1B into (1),

1 0 1 1 2 11

40 1 2 2 2 14

k

1 0 3 0

0 1 0 3k

1 0 1 0

30 1 0 1

k

Therefore , 3k

In Summary:

Learn to apply some matrix pre- or post multiplications can be helpful as

shown in part (ii) of the alternative solution. Beware that AB BA , in general.



7. A curve has the equation 2

2

xey

x

, where 2x .

(i) Find dy

dx.

(ii) Show that the y -coordinate of the turning point is 3

2

e.

(iii) By considering the sign of dy

dx, or otherwise, determine whether the turning point is a

maximum or a minimum.

Solution:

(i) Given 2

2

xey

x

, for 2x

2 2

2

2 2 1

2

x xe x edy

dx x

2

2

2 3

2

xe xdy

dx x

(ii) When 0dy

dx ,

2

2

2 30

2

xe x

x

As 2 0xe for all real x , and 2

2 0x , therefore only

2 3 0x .

3

2x 2

[Analysis]

Apply quotient rule in part (i). The tangent at turning point is at zero gradient, 0dy

dx . The

real challenge is in part (iii).



So, when 3

2x ,

32

23

3

22

32

2

ey e

e

(iii)

x 1.5 ( 1.8) 1.5 1.5 ( 1.2)

dy

dx 0 0 0

The turning point at 3

2x is a minimum point.


(i) Given 2

2

xey

x

, for 2x

2

ln ln2

xey

x

2ln ln ln 2xy e x

ln 2 ln 2y x x

2 ln 2ln d x xd y

dx dx

1 1

22

dy

y dx x

2 3

2

dy xy

dx x

2

2

2 3

2

xe xdy

dx x

x

y

2

2

xey

x

1.2 1.8 1.5



(iii)

2 3

2

dy xy

dx x

2

2

2 3

2 3 2

2

xd

d y dy x xy

dx dx x dx

,product rule

Let 2 3

2

xu

x

,

ln ln 2 3 ln 2u x x

1 2 1

2 3 2

du

u dx x x

2 1

2 3 2

duu

dx x x

2 3 2 1

2 2 3 2

du x

dx x x x

Therefore,

2

2

2 3 2 3 2 3 2 1

2 2 2 2 3 2

d y x x xy y

dx x x x x x

2

2

2 3 2 2 2

2 2 2 3

d y x xy

dx x x x

2 2

2

2 3 2 2 2

2 2 2 2 3

xd y e x x

dx x x x x

2 2

22

2 3 2 2 2

2 22

x x xd y e

dx x xx

When 3

2x ,

23

22 0 4 0

d ye

dx

Therefore the turning point at 3

2x is a minimum.

In Summary:

Clearly, doing a 2nd

derivative on this function is not going to be any easy.

Observe the technique of taking logarithm before differentiation help to get the job

done.



9. The function f : 2cos3x x is defined for 0 x .

(i) State the period f .

(ii) State the amplitude of f .

(iii) Sketch the graph of fy x .

(iv) On the diagram drawn in part (iii), sketch the graph of 2

1x

y

for 0 x .

(v) State the number of solutions, for 0 x , of the equation 2 cos3 2x x .

Solution:

(i) The period of f is 2

3

.

(ii) The amplitude of f is 2 .

(iii) Sketch the graph of fy x .

x 0 6

3

2

2

3

5

6

y 2 0 2 0 2 0 2

[Analysis]

Apply quotient rule in part (i). The tangent at turning point is at zero gradient, 0dy

dx . The

real challenge is in part (iii).

0 x

y

2

2

6

3

2

2

3

5

6

2cos3y x

21

xy



(iv) For the graph of 2

1x

y

for 0 x .

(v) State the number of solutions, for 0 x , of the equation 2 cos3 2x x .

2 cos3 2x x

2

2cos3 1x

x

The number of solutions is the number of points of intersections between the curve,

2cos3y x , and the line, 2

1x

y

,is 3.

x 0

y 1 1

In Summary:

This question may see repeated visits in this exam. A good ability to sketch curve,

trigo curves in this case, will be important.



10. (a) Solve the inequality 2 3 5x .

(b) The graph of 2y x c passes through the point 1,5 .

(i) Find the possible values of the constant c .

(ii) Find, for each value of c , the x -coordinate of the point where the graph meets

the x -axis.

Solution:

(a)

When 3

2 3 02

x x , 2 3 2 3x x .

2 3 5x

2 3 5x

4x

Therefore, 3

42

x

When 3

2 3 02

x x , 2 3 2 3x x .

2 3 5x

2 3 5x

1x

Therefore, 3

12

x

Combining the two inequality, we get 1 4x

[Analysis]

This is a simple question on linear modulus. 1st, is to make a sketch of 2 3y x . 2

nd , then

place a line of 5y . We can then clearly infer from the sketch the desire solution.

Continue to sketch the V-shape curve that passes through 1,5 . It will be self-evident there

the two ‘c’ values are to be.

O x

y

3

2

5

2 3y x 3



(b) When 2y x c passes through the point 1,5 , substitute 1x and 5y into the

modulus equation.

(i) 5 2 c

When 2 0 2c c , 2 2c c .

5 2 c

3c

When 2 0 2c c , 2 2c c .

5 2 c

7c

(ii)

When 3c , 0y ,

0 2 3x

0 2 3x

3

2x

When 7c , 0y ,

0 2 7x

0 2 7x

7 1

32 2

x

In Summary:

Modulus is a key topic to understand as it will be use in ‘A’ level. It is best by

going through this solution very carefully so as to enhance your understanding.

O x

y

5

1 2

c 2

c




(a) Solve the inequality 2 3 5x .

Since 2

2 3 2 3x x ,

2

2 3 5x

2 22 3 5x

2 22 3 5 0x

2 3 5 2 3 5 0x x

2 2 2 8 0x x

1 4 0x x

1 4x

(b) When 2y x c passes through the point 1,5 , substitute 1x and 5y into the

modulus equation.

(i) 5 2 c

2

5 2 c

225 2 c

2 22 5 0c

2 5 2 5 0c c

3 7 0c c

3c or 7c

In Summary:

The alternative solution is actually easier to understand than those as shown in

textbook.



12. Answer only one of the following two alternatives.

EITHER

The diagram shows a car badge ATBUCVDWA , in which ABCD is a square of side 8cm .

Points P , Q , R and S are the mid-points of AB , BC , CD and DA respectively. The arc

BUC is part of a circle with centre S . Arcs CVD , DWA and ATB have centres P , Q and

R respectively.

(i) Show that angle BSC is 0.927 radians, correct to 3 significant figures.

(ii) Find the perimeter of the car badge.

(iii) Find the area of the car badge.

Solution:

(i) 2 2 2 2 24 8BS AB AW

4 5BS

By cosine rule,

2 2 2

2 cosBC BS CS BS CS BSC

2

28 2 4 5 2 4 5 4 5 cos BSC

160 64 160cos BSC

[Analysis]

The content of this question has been moved to Mathematics syllabuses.

A

B C

D

T

U

V

W

S

P

Q

R

A

B C

D

T

U

V

W

S

P

Q

R



3cos

5BSC

1 3cos

5BSC

0.927BSC radians

(ii)

4 5 0.927BC

The perimeter 4 4 5 0.927 33.2 cm

(iii)

Area of sector SBUC 21

0.927 4 5 37.082

Area of the car badge 24 37.08 8 84.3 2cm

In Summary:

This question will now appear in General Mathematics paper.

5

3

4



OR

The table shows experimental values of two variables x and y .

It is known that x and y are related by the equation 10 xy Ab , where A and b are

constants.

(i) Using graph paper, draw the graph of lg 10y against x and use your graph to

estimate the value of A and of b .

(ii) By drawing a suitable line on your graph, solve the equation 210x xAb .

Solution:

(i)

10 xy Ab

10 xy Ab

lg 10 lg xy Ab

lg 10 lg lgy A x b

lg A is the vertical axis intercept, lgb is the gradient of the line.

x 0.5 1.0 1.5 2.0

y 15.9 19.1 23.4 30.2

x 0.5 1.0 1.5 2.0

y 15.9 19.1 23.4 30.2

10y 5.9 9.1 13.4 20.2

lg 10y 0.77 0.96 1.13 1.31

[Analysis]

Part (i) effectively tell us what we should be doing with the given curve. Copy the table and

update it very carefully.



From the graph,

lg 0.6A

3.98A

0.65

lg 0.37141.75

b

2.35b

(ii)

Given that 210x xAb ,

lg lg 2A x b x

The line to add is lg 10 2y x , from the graph, 0.35x .

x

ln 10y

O 0.5 1 1.5 2

0.5

1

1.5

1.75

0.65

0.35

lg 10 2y x lg 10 lg lgy A x b



In Summary:

Be careful that it is lg not ln in this question. Students do have to expect this

type of question to appear regularly in this exam. Your answers may vary, depending

on the drawing. It is advisable to draw the curve as large as possible.



1. Prove the identity 2

11 sin 1 sin

1 tanA A

A

.

Solution :

L.H.S. 2

1

1 tan A

2

1

sec A

2cos A

21 sin A

1 sin 1 sinA A

= R.H.S.

Alternative solution :

R.H.S. 1 sin 1 sinA A

21 sin A

2cos A

2

1

sec A

2

1

1 tan A

= L.H.S.

In Summary:

These identities are useful not only in proving questions about identities, but

also in solve trigo equations and some trigo applications.

[Analysis]

Recognise that 2 21 tan sec , 2 2sin cos 1 , 1

seccos

.



2. A company supplies 4 garden centres – Allseed, Budwise, Croppers and Digwell – with

bags of compost, which are sold in 3 sizes – large 145 litres, medium 75 litres and small 20

litres. The number of bags of compost supplied to each garden centre in one delivery is

shown in the following table.

Over a six-month period Allseed received 5 such deliveries, Budwise 6, Croppers 8 and

Digwell 7. Write down three matrices such that matrix multiplication will give the total

amount of compost supplied over the six-month period and hence find this total.

Solution :

4,3 3,1 4,1

200 500 200 Allseed145

300 600 0 Budwise75 per delivery

0 400 300 Croppers20

0 700 0 Digwell

,

We will need to construct a delivery matrix to pre- or post- multiply the result of the above

matrix. In order to get a (1,1) matrix from (4,1), we will perform 1,4 4,1 1,1 . So we are

to pre-multiply a (1,4) delivery matrix. The three matrices are,

200 500 200

300 600 0

0 400 300

0 700 0

,

145

75

20

, 5 6 8 7

Large Medium Small

Allseed 200 500 200

Budwise 300 600 -

Croppers - 400 300

Digwell - 700 -

[Analysis]

The first challenge is to figure out what is in the question. The total compost ordered =

(frequent of delivery) x (Amounts per delivery) x (capacity of each size). Next, it is to figure

out what matrices will permit this to happen.



200 500 200145

300 600 05 6 8 7 75

0 400 30020

0 700 0

70500

885005 6 8 7

36000

52500

1539000

The total amount of compost supplied over the six-month period is 1539000 litres

In Summary:

The difficulty in this question is in multiplying out the matrices correctly.



3. The line 2 3 12x y meets the curve 2 4 8y x at the points P and Q . Find the length

of the line PQ .

Solution :

To solve 2

2 3 12

4 8

x y

y x

,

Let 2 12 3x y

4 24 6x y --------- (1)

Substitute (1) into 2 4 8y x ,

2 24 6y y

2 6 16 0y y

2 8 0y y

8y or 2y

When 8y , 18x .

When 2y , 3x .

2 2

18 3 8 2PQ

325

5 13 units


Given that 2 3 12x y ,

3

62

yx

3336 6

2 2 2

p qp

q p

y yyx x

2 2

2 9 9

4 4

p q q p

q p

y y y yx x

2 2

q p q pPQ x x y y

2

29

4

q p

q p

y yy y

2

13

4

q py y

13

2q py y

To solve 2

2 3 12

4 8

x y

y x

,

Let 2 12 3x y

4 24 6x y --------- (1)

Substitute (1) into 2 4 8y x ,

2 24 6y y

2 6 16 0y y

2 8 0y y

8y or 2y

[Analysis]

Points of intersections of a line and a curve, is found by solving simultaneous equations. It is

easier to replace variable x . The length 2 2

q p q pPQ x x y y .




Let 3 12 2y x --------- (1)

29 36 72y x

2

3 36 72y x --------- (2)

Substitute (1) into (2),

2

12 2 36 72x x

2144 48 4 36 72x x x

24 84 216 0x x

2 21 54 0x x

3 18 0x x

3x or 18x

When 18x , 8y .

When 3x , 2y .

13 13

2 8 10 5 132 2

PQ units

In Summary:

Students should check the answers against the given equations to guard against

mistakes. Avoid substituting fraction into equation.



4. Find the coefficient of 3x in binomial expansion of

(i) 7

1 2x ,

(ii) 721 7 1 2x x .

Solution :

(i) Given that 7

1 2x ,

Each term in this expansion, 77 7

1 2 2r r r r

rT x xr r

7 0 1 2 30 1 2 3

7 7 7 71 2 2 2 2 2

0 1 2 3x x x x x

7 2 31 2 1 14 84 280x x x x

the coefficient of 3x term is 280 .

(ii)

Given that 721 7 1 2x x

2 2 31 7 1 14 84 280x x x x

2 3 2 2 31 14 84 280 7 1 14 84 280x x x x x x x

2 3 2 31 14 84 280 7 98x x x x x

2 31 14 77 182x x x



(i) Given that 7

1 2x ,

Each term in this expansion, 77 7

1 2 2r r r r

rT x xr r

When 3r , 3 3 3 3

3

7 7 6 52 8 280

3 3!T x x x




(ii)

Given that 721 7 1 2x x

7 721 2 7 1 2x x x

Expansion of the first term is as shown in part (i), 7

2r r

rT xr

Expansion of the second term, 2 27 7

7 2 7 2r rr r

rT x x xr r

When 2 3r , 1r . 1 3 3 3

1

77 2 7 14 98

1T x x x

the coefficient of 3x term is 280 98 182 .

In Summary:

A routine question, Be competent in handling the term expression, Tr . In this

way, the question can be done much quicker.



5. (i) Differentiate tan 2 1x with respect to x .

(ii) Explain why the curve tan 2 1y x has no stationary points.

(iii) Find, in terms of p , the approximate change in tan 2 1x as x increases from 1 to

1 p , where p is small.

Solution:

(i)

Let tan 2 1y x .

2dsec 2 1 2

d

yx

x

2d2sec 2 1

d

yx

x

(ii) For tan 2 1y x to have stationary points, d

0d

y

x .

22sec 2 1 0x

2sec 2 1 0x

sec 2 1 0x

Since sec 2 1 1x or sec 2 1 1x , sec 2 1 0x

Therefore d

0d

y

x , tan 2 1y x has no stationary point.

(iii) Given that d

d

y y

x x

, where f fy x x x

1x , 1 1x p p , 22sec 2 1y

xx

,

22 sec 3 2.04y p p

[Analysis]

Part (i) can be done by quotient rule of sin

tancos

. Part (ii) is to show that the first

derivative cannot be zero. Part (iii) is no longer in the current examination. It is included for

completeness of this question.



7. Solve the equation

(i) log 72 3 log 3x x ,

(ii) 5 253log log 10y y .

Solution:

(i) Given that log 72 3 log 3x x , where 0x and 1x

log 72 log 3 3x x

log 72 3 3x

327 8 x

33 3 2x

6x

(ii) Given that 5 253log log 10y y , where 0y

55

5

log3log 10

log 25

yy

55

log3log 10

2

yy

5 56log log 20y y

55log 20y

5log 4y

45y

625y

In Summary:

Students will be well advised to master solving logarithmic equation. Base

change formula is a common requirement. One cannot separate indices operations

from logarithm.

[Analysis]

Part (i) is to see that log 72x and log 3x share the same base. Part (ii) will need to apply the

base change formula.



9. (a) Solve the equation 22cos 5sin 1 0x x for 0 360x .

(b) Solve the equation tan 1 cot 2 0y y for 0 2y radians.

Solution:

(a) Given that 22cos 5sin 1 0x x , for 0 360x

22 1 sin 5sin 1 0x x

22sin 5sin 3 0x x

22sin 5sin 3 0x x

2sin 1 sin 3 0x x

2sin 1 0x or sin 3 0x

1

sin2

x sin 3x Rejected, 1 sin 1x

1 1sin

2x

, x is in 3

rd and 4

th Quadrants.

Principal angle, 30x

210 ,330x

(b) Given that tan 1 cot 2 0y y for 0 2y radians,

1

tan 1 2 0tan

yy

tan 1 2 0y

tan 3y

1tan 3y , y is in 2nd

and 4th

Quadrants

Principal angle, 1.249y

1.249, 2 1.249y

1.89, 5.03y

In Summary:

The solution is with principal angle method.

[Analysis]

To solve part (a), just need to change 2 2cos 1 sinx x , then it becomes a quadratic equation

look-a-like. Using 1

cottan

xx

to simplify the equation



11. Solutions to this question by accurate drawing will not be accepted.

The diagram, which is not drawn to scale, shows a triangle ABC in which the point A is

9,9 and the point B is 1, 3 . The point C lies on the perpendicular bisector of AB

and the equation of the line BC is 8 11y x . Find

(i) the equation of the perpendicular bisector of AB ,

(ii) the coordinates of C .

The point D is such that ACBD is a rhombus.

(iii) Find the coordinates of D .

(iv) Show that 2AB CD .

[Analysis]

A perpendicular bisector divides a line segment into two equal parts, passes through the

midpoint of AB . Use the gradient of AB to find the gradient of the normal. Use the equation

of the perpendicular bisector and the equation of BC to find the point C .

Lastly, the points D and E are the intersections of the two curves.

x

C

9,9 A

1, 3B O

y



Solution:

(i) Let the mid-point of AB be M ,

1 9 3 9

,2 2

M

5,3M

Gradient of AB 3 9 3

1 9 2

Gradient normal of AB 1 2

3 3

2

The equation of the perpendicular bisector of AB ,

2

3 53

y x

2 19

3 3

xy

(ii)

2 19

3 3

8 11

xy

y x

,

2 19

8 113 3

xx

2 19 24 33x x

52 26x

2x

When 2x , 5y , the coordinates of 2,5C

(iii) The mid-point of CD is 5,3M . Let ,D x y ,

2 5

, 5,32 2

x yM

2

52

x

53

2

y

8x 1y

the coordinates of 8,1D .

x

C

9,9 A

1, 3B O

y

M

D



(iv)

2 2

1 9 3 9 208AB

2 2

8 2 1 5 52CD

208 52

4 21352

AB

CD

Therefore, 2 CD AB


(iii) The mid-point of CD is 5,3M and 2,5C .

5 3,3 2 8,1D


(iv)

tanCD CM

AB BM

3tan

2 , gradient of the line AB

tan 8 , gradient of the line BC

tan tan

tan1 tan tan

38

23

1 82

13

2

13

x

C

9,9 A

1, 3B O

y

M

D

x

2,5C

5,3M

O

y

D

2

3 2

3



1

2

1

tan2

CD CM

AB BM

Therefore, 2 CD AB



12. Answer only one of the following two alternatives.

EITHER

A particle starts from a fixed point A and travels in a straight line. The velocity, 1msv ,

of the particle, st after leaving A , is given by 1 4 9v t t .

(i) Find the acceleration of the particle when it is at instantaneous rest.

(ii) Obtain an expression, in terms of t , for the displacement, from A , of the particle st

after leaving A .

Solution:

(i) The particle is at instantaneous rest when its velocity is zero.

Let 0v , 0t

0 1 4 9t t

4 9 1t t

Since 4 9 0t and 1 0t when 0t ,

2 2

4 9 1t t

24 9 2 1t t t

2 2 8 0t t

4 2 0t t

4t or 2t Rejected as 0t

Acceleration,

d 1

1 4d 2 4 9

v

t t

d 2

1d 4 9

v

t t

When 4t ,

2d 2 31 ms

d 54 4 9

v

t



(ii) Let the displacement from A , of the particle be s t ,

ds t v t

1 4 9 dt t t

2 3

21

4 92 6

ts t t t c where c is an integration constant.

When 0t , displacement is zero, 0 0s .

3

21

0 96

c

9

02

c

9

2c

2 3

21 9

4 92 6 2

ts t t t


(i) The particle is at instantaneous rest when its velocity is zero.

Let 0v , 0t

0 1 4 9t t

4 4 4 4 9 0t t

4 9 4 4 9 5 0t t

Let 4 9 0u t ,

2 4 5 0u u

1 5 0u u

5u or 1u , Rejected as 0u .

5 4 9t

25 4 9t

4t

Let 4 9u t ,

2 4 9u t , 2 9

4

ut



2d

4d

u

t

d

2 4d

uu

t

d 2

d

u

t u

1 4 9v t t

2 9

14

uv u

d

1d 2

v u

u

d d d

d d d

v v u

t u t

d 2

1d 2

v u

t u

d 2

1d

v

t u

d 2

1d 4 9

v

t t

When 4t ,

2d 2 31 ms

d 54 4 9

v

t

(ii) Let the displacement from A , of the particle be s t , when 0t , 0 0s .

0

0 d

t

s t s v t

0

1 4 9 d

t

t t t

2 3

2

0

14 9

2 6

t

tt t

2 3 3

2 21 1

4 9 92 6 6

tt t

2 3

21 9

4 92 6 2

tt t



O

2

161

5y

x

x

y

Q

P

A

OR

The diagram shows part of the curve

2

161

5y

x

, cutting the x -axis at Q . The

tangent at the point P on the curve cuts the x -axis at A . Given that the gradient of this

tangent is 4 , calculate

(i) the coordinates of P ,

(ii) the area of the shaded region PQA .

Solution:

(i)

2

161

5y

x

3

d 162 1

d 5

y

x x

3

d 32

d 5

y

x x

At P , d

4d

y

x ,

3

324

5 x

3

5 8x

5 2x

3x

When 3x , 3y . 3,3P

O

2

161

5y

x

x

y

Q

P

A T



(ii)

When 0y ,

2

160 1

5 x

2

5 16x

5 4x

5 4x

1x or 9x , Rejected since 3x

1,0Q

Equation of PA ,

3 4 3y x

4 9y x

When 0y , 9

4x .

9,0

4A

Let point 3,0T ,

the area of the shaded region PQA ,

3 3

2

91

4

161 d 4 9 d

5x x x

x

3

32

9

41

162 9

5x x x

x

2

216 16 9 93 1 2 3 9 3 2 9

5 3 5 1 4 4

81

5 3 98

9

28

7

8 square units



1.

The diagram shows a triangle ABC in which 4cmAB , 2cmBC and angle

120ABC . The line CB is extended to the point X where angle 90AXB .

(i) Find the exact length of AX .

(ii) Show that angle 1 3tan

2ACB

.

Solution :

(i)

60ABX OR By similar triangle with 30-60-Right triangle

4sin 60AX 4

23

AX

3

42

2 3 cmAX

2 3 cm

[Analysis]

First, recognize that ABX is a special 30-60-Right triangle. Part (i) demands “exact” value

which implies that NO rounding of numbers.

A

C

B

X

4cm

2cm

120



(ii)

4cos60BX OR By similar triangle with 30-60-Right triangle

1

42

4

1 2

BX

2cm 2cmBX

Therefore,

tanAX

ACBCX

2 3

2 2

3

2

1 3tan

2ACB

In Summary:

A very important point for students is to be able to handle Surds at ease. Any

rounding-off of answers with calculator will result in losing answer marks.



2. Solve, for x and y , the simultaneous equations

9 27 1yx ,

8 2 16 2x

y

Solution :

Given that 9 27 1yx ,

2 33 3 1x y

2 3 03 3x y

2 3 0x y ---------- (1)

Given that 8 2 16 2x

y ,

6 8

2 2 2 2y x

6 9

2 2y x

6 9y x ---------- (2)

(1)+2x(2),

15 18y

1

15

y

Substitute 1

15

y into (1), we get

9

5x .

In Summary:

Let me highlight again here that Competency in indices operations is the

foundation in doing well in this exam. Remember to check each solution.

[Analysis]

Recognize that 29 3 and 327 3 , and also 2 3 6

38 2 2 2

and

2 4 8

416 2 2 2

. Then, apply the rules of indices.




Given that 9 27 1yx ,

3 3log 9 27 log 1yx

3 3 3log 9 log 27 log 1yx

3 32 log 3 3 log 3 0x y

2 3 0x y ---------- (3)

Given that 8 2 16 2x

y ,

2 2log 8 2 log 16 2x

y

2 2 2 2log 8 log 2 log 16 log 2x

y

2 2 2 2

13 log 2 log 2 4log 2 log 2

2 2

xy

13 4

2 2

xy

93

2 2

xy --------- (4)

Solving (3) and (4) simultaneously to get the solutions.



3. Given that 7 8

1 6

A , find 1A and hence solve the simultaneous equations

8 7 11 0p q ,

6 7 0p q .

Solution :

Given that

7 8

1 6

A ,

the determinant, 7 6 8 1 50

1

6 8

6 81 50 50

1 7 1 750

50 50

A

Given that 8 7 11 0p q and 6 7 0p q , we get

7 8 11q p

6 7q p

So,

7 8 11

1 6 7

q

p

6 8

1150 50

1 7 7

50 50

q

p

[Analysis]

Question on Matrix is an on-and-off sub-topic. Students can employ Matrix operations to

solve any linear simultaneous equations. One will need to know determinant to start.

Remember, “hence” in the question implies that you MUST use 1A in your solution, no mark

will be awarded for otherwise.



6 11 8 7 1

50 5

61 11 7 7

550

q

p

Therefore, 1

15

p , 1

5q .


Let 1Aa b

c d

,

7 8 1 0

1 6 0 1

a b

c d

7 8 6 1 0

7 8 6 0 1

a b a b

c d c d

7 1

8 6 0

7 0

8 6 1

a b

a b

c d

c d

6 8,

50 50

1 7,

50 50

a b

c d

1

6 8

50 50

1 7

50 50

A

In Summary:

As a reminder, the second part MUST be done with the inverse matrix.

[Analysis]

Consider 1A A=I

, given that A is a 2 2 matrix, 1 0

0 1I

.



4.

(i) Find 3 lnd

x xdx

.

(ii) Hence find 2 ln .x xdx

Solution :

(i)

3

2 3

2 2

ln

13 ln

3 ln

dx x

dx

x x xx

x x x

(ii)

3 2 2ln 3 lnd

x x x x xdx

3 2 2ln 3 lnd

x x dx x x x dxdx

3 2 2ln 3 lnd

x x dx x xdx x dxdx

2 3 23 ln lnd

x xdx x x dx x dxdx

3 31ln

3x x x c

[Analysis]

Part (i) can be done with product rule. Part (ii) is to be completed with the result from part (i).

The method used in part (ii) is VERY important.

In Summary:

This is an excellent question on integration. Students are best to familiar

themselves with the approach to integration, as often, integration begins with

constructing a differentiation.



5.

(i) Express

8 46

5 1

x

x x

in partial fractions.

(ii) Hence, or otherwise, find the gradient of the curve

8 46

5 1

xy

x x

at the point

where 2x .

Solution(1):

(i)

8 46

5 1 5 1

x A B

x x x x

-------- (1)

(1) 5x ,

8 46

51 1

x BA x

x x

Let 5x ,

8 5 46

5 1A

1A

(1) 1x ,

8 46

15 5

x Ax B

x x

Let 1x ,

8 1 46

1 5B

9B

so,

8 46 9 1

5 1 1 5

x

x x x x

.

[Analysis]

Read careful the steps taken in part (i). This is essentially the ‘cover-up’ method. Part (ii) is to

follow from part (i) or to be taken completely as new. Gradient of a curve is dy

dx



(ii)

8 46 9 1

5 1 1 5

xy

x x x x

1 1

9 1 5y x x

2 2

9 1 1 1 5dy

x xdx

2 2

9 1

1 5

dy

dx x x

At 2x ,

2 2

9 1 8

92 1 2 5

dy

dx

Solution(2):

(i)

8 5 68 46

5 1 5 1

xx

x x x x

8 5 6

5 1 5 1

x

x x x x

8 6

1 5 1x x x

By inspection,

6 1 1

5 1 5 1x x x x

8 46 8 1 1

5 1 1 5 1

x

x x x x x

Therefore,

8 46 9 1

5 1 1 5

x

x x x x

(ii)

Given that 2

8 46 8 46

5 1 4 5

x xy

x x x x

, by quotient rule

2

22

8 4 5 8 46 2 4

4 5

x x x xdy

dx x x

2

22

8 92 224

4 5

dy x x

dx x x



At 2x ,

2

22

8 2 92 2 224 8

92 4 2 5

dy

dx

In Summary:

Part (i) solutions show that there is really no need to memorise any specific

method to decompose a fraction. However, the cover-up method is the quickest to the

solution. Part (ii) again demonstrates that the “Hence” approach is the quickest to the

solution. Since the question provide alternative approaches, students should be

reminded to apply the normal classroom practice as their response.



6. A cyclist starts from rest from a point A and travels in a straight line until he comes to rest

at a point B . During the motion, his velocity, 1msv , is given by 216

2v t t , where t is

the time in seconds after leaving A . Find

(i) the time taken for the cyclist to travel from A to B ,

(ii) the distance AB ,

(iii) the acceleration of the cyclist when 8t .

Solution:

(i) At A and B , the velocity is zero.

210 6

2t t

2 12 0t t

12 0t t

0t or 12t

the time taken is 12 seconds

(ii) The area under the velocity curve yields the displacement.

12 12

2

0 0

16

2vdx t t dx

123

2

0

36

tt

[Analysis]

Make a sketch of the velocity curve to visualize the question.

0v

A B

0v

0t

v



3

2 123 12 144m

6

(iii) Acceleration is the derivative of the velocity.

Given that 216

2v t t ,

acceleration, 6dv

tdt

When 8t , 26 8 2msdv

dt

In Summary:

When integrating velocity for displacement, one has to be mindful about the

distance travelled. For distance travelled, the integration is to be carried out by taking

the absolute value of the negative area. Question may also be asked for the velocity

when the acceleration is zero, or to find the maximum velocity.



7. The equation of a curve is sin

2 cos

xy

x

. Find the x -coordinate, where 0

2x

, of the

point at which the tangent to the curve is parallel to the x -axis.

Solution(1):

Given sin

2 cos

xy

x

,

2

cos 2 cos sin sin

2 cos

x x x xdy

dx x

2

2cos 1

2 cos

dy x

dx x

When 0dy

dx ,

2

2cos 10

2 cos

x

x

As 2 cos 0x for all real x ,

2cos 1 0x .

1cos

2x

1 1cos

2x

3x

for 0

2x

[Analysis]

The tangent parallel to the x -axis is at zero gradient, 0dy

dx .

In Summary:

Competent in using quotient rule and solve simple trigo equation are needed

here. Question is simple.



8. (i) Show that 2sin3 sin 4sin cosx x x x .

(ii) Find all the angles between 0 and which satisfy the equation

2sin3 sin 2cosx x x .

sin sin cos cos sinA B A B A B

(i) Show that 2sin3 sin 4sin cosx x x x .

Solution(1):

From L.H.S sin3 sinx x

sin 2 sin 2x x x x

2sin 2 cosx x

2 2sin cos cosx x x

24sin cosx x

=R.H.S

Solution(2):

From L.H.S sin3 sinx x

sin 2 cos cos2 sin sinx x x x x

2 22sin cos cos cos sin sin sinx x x x x x x

2 2 22sin cos sin cos sin 1x x x x x

2 2 22sin cos sin cos cosx x x x x

24sin cosx x

=R.H.S

[Analysis]

In part (i) an inspection of the identity, suggests that multiple angles formulae or factor

formulae maybe productive. eg sin sin cos cos sinA B A B A B ;

sin sin 2sin cosA B A B A B , where 2A x and B x .



Solution(3):

From R.H.S 24sin cosx x

2 2sin cos cosx x x

2 sin 2 cosx x

sin 2 sin 2x x x x

sin3 sinx x

=L.H.S

(ii) Solve, for 0 x radians, the equation

2sin3 sin 2cosx x x .

Solution(1):

Given that 2sin3 sin 2cosx x x , for 0 x ,

2sin3 sin 2cos 0x x x

2 24sin cos 2cos 0x x x

2cos 4sin 2 0x x

2cos 0x or 4sin 2 0x

2

x

1

sin2

x

5

,6 6

x

Therefore the solutions are 5

, ,6 2 6

x

In Summary:

Take note that applying the factor formulae is most time efficient (solution(1)).

There are numerous ways to answer this question, of varying length.

[Analysis]

Students can answer this part as if it is a fresh one taking no reference from the earlier part.



Solution(2):

Given that 2sin3 sin 2cosx x x , for 0 x ,

2sin3 sin 2cos 0x x x

2sin 2 cos cos2 sin sin 2cos 0x x x x x x

2 2 22sin cos cos cos sin sin sin 2cos 0x x x x x x x x

2 2 22sin cos 2cos 1 sin sin 2cos 0x x x x x x

2 24sin cos 2cos 0x x x

2cos 4sin 2 0x x

2cos 0x or 4sin 2 0x

2

x

1

sin2

x

5

,6 6

x


, ,6 2 6

x

In Summary:

Trigo questions are typically multiple approaches. Take note that solution(1) is

most efficient. The knowledge of trigo ratios of special angles and triangles are needed.



9. Ann is older than her sister Betty. Their ages in years are such that twice the square of

Betty’s age subtracted from the square of Ann’s age gives a number equal to 6 times the

difference of their ages. Given also that the sum of their ages is equal to 5 times the

difference of their ages, find the age in years of each of the sisters.

Solution:

Let Ann’s age be A , and Betty’s age be B , where A B .

2 22 6A B A B ---------- (1)

5A B A B

2 3A B

3

2A B ---------- (2)

Substitute (2) into (1), 2

3 04

BB

3 04

BB

0B or 12B , so Betty is 12 years old

(Rejected)

Substitute 12B into (2), 18A , so Ann is 18 years old.

[Analysis]

This is a very odd question. 1st, is to formulate equations, then 2

nd to solve them.

In Summary:

This is a very unusual question.



10. (a) Find the smallest value of the integer a for which 2 5 2ax x is positive for all values

of x .

(b) Find the smallest value of the integer b for which 25 2x bx is negative for all

values of x .

Solution(1):

(a)

The curve 2 5 2y ax x needs to be a “smiley”. So 0a .

As this curve should not touch the x -axis, there should not be any real root from the

equation 2 5 2 0ax x . Therefore, the discriminant is negative.

The discriminant, 25 4 2 0a

25

8a

The smallest value of the integer 4a .

(b)

The curve 25 2y x bx is a “frown”.

As this curve should not touch the x -axis, there should not be any real root from the

equation 25 2 0x bx . Therefore, the discriminant is negative.

The discriminant, 2 4 5 2 0b

2 40b

40 40b

The smallest value of the integer 6b .

Solution(2):

(a)

The curve 2 5 2y ax x needs to be a “smiley”. So 0a . The lowest point of this

curve should not touch the x -axis, therefore the minimum value of 2 5 2ax x is to be positive.

By completing the square,

[Analysis]

Consider the curve 2 5 2y ax x , this curve needs to “hang” above x -axis.

Consider the curve 25 2y x bx , this curve needs to “hang” below x -axis.



2 2

2 5 5 25 2

2 2ax x a x

a a a

2

2

5 8 25

2 4

aa x

a a

25 8 25

2 4

aa x

a a

Therefore, the minimum value 8 25

4

a

a

is positive.

8 25

04

a

a

Given that 0a ,

8 25 0a

25

8a

The smallest value of the integer 4a .

(b)

The curve 25 2y x bx is a “frown”. The highest point of the curve should not touch

the x -axis, therefore the maximum value of 25 2x bx is to be negative.

The line of symmetry of the curve,

2 5 10

b bx

The maximum value of the curve lies on the line of symmetry.

Let 10

bx ,

2

5 210 10

b bb

2 2

220 10

b b

2

220

b

The maximum value 2

2 020

b ,

2 40b

40 40b



The smallest value of the integer 6b .

Solution(3):

(a)

The turning point of the curve 2 5 2y ax x , is 2 5dy

axdx

.

When 0dy

dx ,

5

2x

a , this is the line of symmetry.

The rest of the solution follows the method in solution(2)(b).

(b)

The turning point of the curve 25 2y x bx , is 10dy

x bdx

.

When 0dy

dx ,

10

bx , this is the line of symmetry.

The rest of the solution follows that of solution(2)(b).

In Summary:

Solution (1) is the shortest. The other solutions show how other methods can also

be employed to tackle the question. Visualising the question in quadratic curves is

important.



11. (i) In the binomial expansion of

7k

xx

, where k is a positive constant, the coefficients

of 3x and x are the same. Find the value of k .

(ii) Using the value of k found in part (i), find the coefficient of 7x in the expansion of

7

21 5k

x xx

.

Solution:

7 7

7

21

k kx x

x x

The r th term of this expansion, 7 7

2

r

r r

kT x C

x

, where 0,1, ,7r

7 7 2r r

r rT C k x

For 3x term, 7 2 3r

2r

The coefficient of this 3x term,

7 2

2

2

2

7!

2!5!

21

C k

k

k

For x term, 7 2 1r

3r

The coefficient of this x term,

[Analysis]

Part (i) has a little complication with this binomial expansion of

7k

xx

, which can be

rewritten as

7

7

21

kx

x

.



7 3

3

3

3

7!

3!4!

35

C k

k

k

Therefore,

2 321 35k k

Since 0k , 3 5k

3

5k

(ii)

7

21 5k

x xx

7 7

25k k

x x xx x

The expansion of the first term above, is as given in part (i), 7 7 2r r

r rT C k x

The expansion of the second term above, 7 9 25 r r

r rT C k x

For 7x term,

from the first term, 7 2 7r

0r

from the second term, 9 2 7r

1r

coefficient from 1st term, 7 0

0 1C k

coefficient from 2nd

term, 7 1

15 21C k

The coefficient of 7x term is, 1 21 20

In Summary:

Students should be well familiar with using the general term expression to

location the coefficient of the required expansion.



12.

The variables x and y are connected by the equation xy kb , where k and b are constants.

Experimental values of x and y were obtained. The diagram above shows the straight line graph,

passing through the points 0,1.3 and 11,0.8 , obtained by plotting lg y against x . Estimate

(i) the value, to 2 significant figures, of k and of b .

(ii) the value of y when 8x .

Solution:

(i)

Given that 0,1.3 and 11,0.8 ,

Gradient, 1.3 0.8 1

lg0 11 22

b

1

2210b

0.90b

When 0x , lg 1.3y .

lg 1.3k 1.310k

20k

O x

lg y

0,1.3

11,0.8

[Analysis]

The curve xy kb is to be linearised as lg lg lgy k x b , in which lgb and lg k are gradient

and lg y -axis intercept respectively.



(ii)

The equation of the line is, lg 1.322

xy

When 8x , 8

lg 1.322

y

lg 0.936y

0.93610y

8.63y


(i)

Let the equation of the line be Y mX C , where lg y Y , lg k C , x X

Given that 0,1.3 and 11,0.8 ,

1.3 C

0.8 11 1.3m

1

22m

Therefore,

lg 1.3k 1.310k

20k

and,

1

lg22

b

1

2210b

0.90b

(ii)

20 0.9x

y

When 8x , 8

20 0.9 8.61y

In Summary:

Be careful that it is lg not ln in this question. Students do have to expect this

type of question to appear regularly in this exam.



13.

The diagram shows a glass window, PQRST , consisting of a rectangle PQST of height h cm and

width 6x cm and an isosceles triangle QRS in which 5QR RS x cm. The perimeter of the

window is 360cm.

(i) Show that the area of the window, 2cmA , is given by 21080 36A x x .

Given that x can vary,

(ii) find the stationary value of A ,

(iii) determine whether this stationary value is a maximum or a minimum.

Solution:

(i)

Given that the perimeter is 360cm,

360 6 2 10x h x

180 8h x --------- (1)

The area,

1

6 4 62

A h x x x

26 12A xh x --------- (2)

[Analysis]

This is a typical min/max problem with a fixed perimeter to find largest area.

6 cmx

Q

P T

S

R

cmh cmh

5 cmx 5 cmx




26 180 8 12A x x x

21080 36A x x

(ii)

1080 72dA

xdx

Let 0dA

dx , 0 1080 72x

15x

When 15x , the stationary value of A ,

2 21080 15 36 15 8100cmA

(iii) 2

272

d A

dx

The stationary point is a maximum point.

Alternative solution for (ii) and (iii)

21080 36A x x

236 30A x x

2 236 15 36 15A x

2

36 15 8100A x

At 15x , the value of the stationary value is 28100cm

2

36 15 8100 8100x

The stationary point is a maximum point.

In Summary:

The alternative solution is a non-calculus one.



1. A man buys a new motorcycle. After t months its value $V is given by 10000 ptV e ,

where p is a constant.

(i) Find the value of the motorcycle when the man bought it.

The value of the motorcycle after 12 months is expected to be $4000. Calculate

(ii) the expected value of the motorcycle after 18 months,

(iii) the age of the motorcycle, to the nearest month, when its expected value will be

$1000.

Solution :

(i) When 0t , 0

10000p

V e

10000V

The value of the motorcycle when the man bought it, is $1000.

(ii) When 12t , 4000V 12

4000 10000p

e

2

12 ln5

p

1

ln 2 ln5 0.0763612

p

When 18t , 18

10000p

V e

1 2

18 ln12 510000V e

3 2ln

2 510000V e

3 2ln

2 510000V e

2529.82V

the expected value of the motorcycle after 18 months is, $2529.82.

(iii) When 1000V ,

1000 10000 pte

[Analysis]

In this question, the value of the motorcycle is modeled by the formula, 10000 ptV e . The

rest of the question is straight forward which involves exponential and logarithm.



1 2ln

12 51

10

t

e

1 2

ln ln10 12 5

t

2

12ln10 ln5

t

30t

the age of the motorcycle, to the nearest month is 30 months.

In Summary:

Please make sure that you are comfortable with operations with indices and

logarithms. Depending on the features available with your electronic calculators, the

actual steps taken can vary. Answer for part (ii) may vary.



2. The roots of the quadratic equation 22 4 3 0x x are and . Find the quadratic

equation whose roots are 2 2 and 2 2 .

Solution :

Given that 22 4 3 0x x ,

Sum of roots, 4

22

Product of roots,

3

2

For a quadratic equation whose roots are 2 2 and 2 2 ,

Sum of roots, 22 2 2 22 2 4 2 4 5

Product of roots,

2 2 22 2 2 2 33

2 2 2 4 2 2 44

Therefore, 2 335 0

4x x ,

or 24 20 33 0x x

[Analysis]

A direct application of Sum of Roots(SOR), and Product of Roots(POR).

Recall that 2 2 2 2 .

In Summary:

This type of questions used to be popular a few decades ago, is making a come

back. Do expect this trend to continue. In general, 24 20 33 0k x x where 0k

will be the required solution.

[Analysis]

A direct application of Sum of Roots(SOR), and Product of Roots(POR).

Recall that 2 2 2 2 .



3.

(i) Prove the identity tan cot 2cosec2A A A .

(ii) Find all the angles between 0 and 360 which satisfy the equation tan cot 3A A .

Solution :

(i) Given that tan cot 2cosec2A A A ,

L.H.S.= tan cotA A ,

sin cos

cos sin

A A

A A

2 2sin cos

sin cos

A A

A A

2 1

2sin cosA A

2

sin 2A

=cosec2A

=R.H.S.

(ii) Given that tan cot 3A A for 0 360A ,

2cosec2 3A

3

cosec22

A

1 3

sin 2 2A

2

sin 23

A

[Analysis]

Part (i) needs to apply 1 cos

cottan sin

AA

A A . And the left hand side is a double angle, 2A

that provide a hint of possible use of double angle formulae. Take note that

1cosec2

sin 2A

A . Part (ii) has to provide solution for 0 360A which is also for

0 2 720A .



1 22 sin

3A

for 0 2 720A

Principal angle, 2 41.81A

2 41.81 , 138.19 ,401.81 ,498.19A

20.9 , 69.1 ,200.9 ,249.1A

In Summary:

Students need to recall that 1 cos

cottan sin

AA

A A ;

1cosec

sinA

A ;

1sec

cosA

A ;

2 2sin cos 1A A ; sin 2 2sin cosA A A . The principal angle is the value returned

from your calculator.




(i) 3 32 log 3 7 log 2 3x x ,

(ii) 53log log 5 2yy .

Solution :

(i)

Given that 3 32 log 3 7 log 2 3x x ,

3 3 32log 3 log 3 7 log 2 3x x

2

3 3 3log 3 log 3 7 log 2 3x x

3 3log 9 3 7 log 2 3x x

9 3 7 2 3x x

25 60 0x

12 2

25 5

x

(ii)

Given that 53log log 5 2yy ,

5

5

13log 2

logy

y

Let 5logu y ,

1

3 2uu

23 2 1 0u u

3 1 1 0u u

1

3u or 1u

5

1log

3y 5log 1y

[Analysis]

In part (i), 1

3 7 0 23

x x and 1

2 3 0 12

x x . Together, 1

23

x .

In part (ii), 0y and 1y . Apply base change formula.

Must check that the solutions are admissible with the given equation.



1

35y

5y

1

lg lg53

y

lg 0.23299y

0.2329910y

0.585y


(iii)

Given that 53log log 5 2yy ,

lg lg5

3 2lg5 lg

y

y

Let lg

lg5

yu ,

1

3 2uu

23 2 1 0u u

3 1 1 0u u

1

3u or 1u

lg 1

lg5 3

y

lg1

lg5

y

1

lg lg53

y lg lg5y

1

35y

5y

0.585y

In Summary:

Be mindful of the restrictions imposed by logarithm.



5. The term containing the highest power of x in the polynomial f x is 42x . Two of the

roots of the equation f 0x are 1 and 2 . Given that 2 3 1x x is a quadratic factor

of f x , find

(i) an expression for f x in descending powers of x ,

(ii) the number of real roots of the equation f 0x , justifying your answer,

(iii) the remainder when f x is divided by 2 1x .

Solution:

(i)

2f 2 3 1 1 2x x x x x

4 3 2f 2 8 4 10 4x x x x x

(ii) Given the 2 3 1x x ,

the discriminant 2

3 4 1 1 5 0 , this quadratic factor contains 2 real roots.

The number of real roots is 4.

(iii) when f x is divided by 2 1x ,

21 1 1 1 1

f 2 3 1 1 22 2 2 2 2

1 9 1

f 12 8 8

[Analysis]

Need to construct f x . The equation, f 0x has two roots, 1 and 2 , so 1x and 2x

are factors of f x . Together with 2 3 1x x , that doesn’t contain the above two roots, then

a possible polynomial is 2 3 1 1 2x x x x . But the highest degree term is 42x , so a

factor 2 is needed to complete f x .

In Summary:

The difficulty in this question is in the construction of the correct polynomial.

Part (ii) needs not find the actual roots. Part(iii) needs to apply remainder theorem.



6.

The diagram shows a circle, centre O , with diameter AB . The pointC lies on the circle.

The tangent to the circle at A meets BC extended at D . The tangent to the circle at C

meets the line AD at E .

(i) Prove that triangles AEO and CEO are congruent.

(ii) Prove that E is the mid-point of AD .

Solution:

(i) Consider AEO and CEO

90OAE OCE (tangent radius)

OA OC (radii of the circle)

OE OE (common hypotenuse)

By RHS rule, AEO is congruent to CEO .

Therefore,

AOE COE

Hence, line OE is the angle bisector of AOC .

[Analysis]

In general, be ready to apply geometrical properties of circle; congruent triangles; as well as

similar triangles. Part (i) is about applying ASA, or SAS, or SSS, or RHS rules for

congruency. Part (ii) may apply mid-point or intercept theorem.

B

E

A

C

O

D



(ii)

2AOC AOE

2AOC ABC ( s at centre of circle = 2 s at circumference)

2 2AOE ABC

AOE ABC

Consider OAE and BAD ,

90OAE BAD (common angle)

AOE ABC (corresponding angle)

OAE and BAD are similar.

AO AE

AB AD

2

radius AE

radius AD

1

2

AE

AD

Therefore, E is the mid-point of AD .

Alternative solution for (ii):

90ACB (Right s in semi-circle)

90ACB ACD (Supplementary )

AEO CEO ,

AE CE

AEC is an isosceles triangle.

EAC ECA

90EAC ADC

90ECA DCE ACD

ADC DCE

EDC is an isosceles triangle.

EC ED

AE CE ,

AE ED , hence E is the mid-point of AD .

In Summary:

Before writing any proof, one shall think through the entire proving steps. This

thought process usually required looking at the result, and then work backward until it

meets the forward process, starting from the given conditions.



7. The function f is defined by f 4cos2 2x x .

(i) State the amplitude of f .

(ii) State the period of f .

The equation of a curve is 4cos2 2y x for 0 180x .

(iii) Find the coordinates of the minimum point of the curve.

(iv) Find the coordinates of the points where the curve meets the x -axis.

(v) Sketch the graph of 4cos2 2y x for 0 180x .

(vi) Sketch the graph of 4cos2 2y x for 0 180x .

Solution:

(i)

The amplitude of f is 4.

(ii)

The period of f is 360

1802

.

(iii)

4 4cos2 4x

6 4cos2 2 2x

The minimum value of y is 6 when cos2 1x

cos2 1x for 0 180x

12 cos 1x for 0 2 360x

Principal angle, 2 180x

90x

The coordinates of the minimum point of the curve is 90 , 6

[Analysis]

Part (i) and (ii) are straight forward about periodic trigo curve. The rest of the question, (iii),

(iv) and (v) will be better to start with part (v). Remember to mark clearly the intercepts and

turning points.



(iv)

The x -axis is at 0y .

0 4cos2 2x for 0 180x

0 2cos2 1x

1

cos 22

x

1 12 cos

2x

for 0 2 360x

Principal angle, 2 60x

2 60 ,300x

30 ,150x

The coordinates of the points where the curve meets the x -axis are 30 ,0 , 150 ,0

(v)

The graph of 4cos2 2y x for 0 180x .

x 0 45 90 135 180

y 2 2 6 2 2

0 x

y 3

2

90

4

30 60

2

6

120 150 180



(vi)

The graph of 4cos2 2y x for 0 180x .

x 0 45 90 135 180

y 2 2 6 2 2

In Summary:

It is best to have made a simple sketch of the equation first to get a “visual”

picture of the problem. Then transfers the result of part (iii) and (iv) onto this sketch.

Finally, construct a ‘proper’ sketch with the 5-point values table.

x 0

y

90 30 60

2

120 150 180

6

4



0

3y x ax b

2,0 x

y 8.

The diagram shows part of the curve 3y x ax b , where a and b are positive constants.

The curve has a minimum point at 2,0 . Find

(i) the value of a and of b ,

(ii) the coordinates of the maximum point of the curve,

(iii) the area of the shaded region.

Solution:

(i) At 2,0 ,

3

0 2 2a b

2 8a b ---------- (1)

23dy

x adx

2

0 3 2 a

12a

Substitute 12a into (1), we get 16b .

[Analysis]

The curve passes through point 2,0 and its gradient is zero at this point too. With two

unknown a and b , these will be enough to solve part (i). The turning points occur at zero

gradient, 0dy

dx . Of course, we already know the minimum point. The area is simply the

integration of the curve from 0 to 2 .



(ii) The equation of the curve,

3 12 16y x x

23 12dy

xdx

20 3 4x

0 3 2 2x x

2x or 2x

At 2x , 3

2 12 2 16 32y

The coordinates of the maximum point of the curve is at 2,32

(iii)

The area of the shaded region 2

3

0

12 16x x dx

24

2

0

6 164

xx x

4

226 2 16 2 0

4

12 square units

In Summary:

You should read through the entire question, make a plan on how to approach

them before beginning to answer them.



9.

The diagram shows a straight road OP . A runner leaves the road at O and runs 4km in a

straight line to a point A . She then turns through 90 and runs 2km in a straight line to a

point B . The angle POA is , where 0 90 , and the perpendicular distance of B

from the road OP is kmL .

(i) Show that 4sin 2cosL .

(ii) Express L in the form sinR , where 0R and 0 90 .

(iii) Find the value of for which 3L .

Solution:

(i) From the graph,

4sin 2cosL

4sin 2cosL

[Analysis]

For part (i), the challenge is to locate the value of 2cos . Study the diagram to find angle

around the length of 2 km. Part (ii) is about R-formulae. Part (iii) will need to solve a trigo

equation.

2km

P

A

B

O

4km

kmL

2km

P

A

B

O

4km

kmL



(ii)

sin sin cos cos sinR R

4sin 2cos sin cos cos sinL R R

cos 4R --------- (1)

sin 2R --------- (2)

2 2

1 2 , 2 2 2 2

sin cos 2 4R R

2 2 2sin cos 20R

2 20R

20 2 5R

2

1’

sin 2 1tan

cos 4 2

1 1tan 26.6

2

1 12 5 sin tan

2L

(iii)

When 3L ,

1 13 2 5 sin tan

2

1 1 3sin tan

2 2 5

1 11 3tan sin

2 2 5

Principal angle, 1 1tan 42.1

2

1 1tan 42.1 ,137.9

2

42.1 26.6 or 137.9 26.6

68.7 164.5 (inadmissible as 90 )

The value of for which 3L is 68.7 .




(ii)

2 24 2 20 2 5R

4cos

2 5

2sin

2 5

2 1

tan4 2

1 1tan 26.6

2

2 5 sin 26.6L

(iii)

When 3L ,

3 2 5 sin 26.6

3

sin 26.62 5

Let be the basic angle,

3sin

2 5

42.1

26.6 42.1 ,137.9

42.1 26.6 or 137.9 26.6

68.7 164.5 (inadmissible as 90 )

The value of for which 3L is 68.7 .

In Summary:

The method employed in the part (ii) of the alternative solution is more time

efficient. The part (iii) of the alternative solution uses the basic angle approach.

4

2 2 5



10. A curve is such that

2

6

2 1

dy

dx x

and 2,9P is a point on the curve. The normal to the

curve at P meets the y -axis at Q and the x -axis R .

(i) Find the coordinates of the mid-point of QR .

(ii) Find the equation of the curve.

A point ,x y moves along the curve in such a way that the x -coordinate increases at a

constant rate of 0.03 units per second.

(iii) Find the rate of change of the y -coordinate as the point passes through P .

Solution:

(i)

Let the equation of the normal be y mx q , the gradient of the curve at 2,9P ,

2

6 2

32 2 1

dy

dx

1 3

2m

dy

dx

3

9 22

q

12q

equation of the normal,

3

122

y x

When 0y , 8x , the coordinates of R is 8,0 ; the coordinates of Q is 0,12 .

The coordinates of the mid-point of QR is 8 12

, 4,62 2

.

[Analysis]

Gradient of the curve is dy

dx . The equation of the curve will be

dyy dx

dx

with a given

point. Gradient of normal, at a point is 1

gradient at the point



(ii)

The equation of the curve will be dy

y dxdx

2

6 3

2 12 1y dx c

xx

, where c is a constant

The curve passes through 2,9P ,

3

92 2 1

c

8c

The equation of the curve is 3

82 1

yx

.

(iii)

At 2,9P , 2

3

dy

dx , 0.03

dx

dt ,

By chain rule, dy dy dx

dt dx dt

2 30.02

3 100

dy

dt units per second

In Summary:

This is a very traditional question on finding the equation of a curve, given its

derivatives.



11.

The diagram shows two circles 1C and 2C . Circle 1C has its centre at the origin O .

Circle 2C passes through O and has its centre at Q . The point 8, 6P lies on both

circles and OP is a diameter of 2C .

(i) Find the equation of circle 1C .

(ii) Find the equation of circle 2C .

The line through Q perpendicular to OP meets the circle 1C at the points A and B .

(iii) Show that the x -coordinates of A and B are 3a b and 3a b respectively,

where a and b are integers to be found.

Solution:

(i) For 1C ,

228 6 10r OP

The equation of circle 1C ,

2 2 100x y --------- (1)

[Analysis]

The equation of a circle, centre at ,a b is 2 2 2x a y b r , where r is the radius of the

circle. Part (iii) is about solving simultaneous equations, a circle and a line.

8, 6P

x

y

A

B

O

1C

Q

2C



(ii)

The centre of circle 2C is the mid-point of OP , Q

8 6

, 4, 32 2

The radius of circle 2C , is half of OP , is 5

2 2

4 3 25x y

(iii)

Gradient of OP6 3

8 4

Gradient of AB1 4

3 3

4

The equation of line AB ,

4

3 43

y x

4 25

3 3y x --------- (2)

substitute (2) into (1), 2

2 4 25100

3 3x x

225 200 275 0x x 2 8 11 0x x

By quadratic formula,

28 8 4 1 11

2 1x

4 27x

4 3 3x

Therefore, 4a and 3b .




10OA OB

5OQ

QA QB (perpendicular radius bisects chord)

By Pythagoras Theorem,

5 3QA

Gradient of OP6 3

8 4

Gradient of AB1 4

3 3

4

By similar triangles with the gradient of AB and AQx A ,

3 3AQx

Therefore, 4 3 3, 3 4 3A , 4 3 3, 3 4 3B

4a and 3b

8, 6P

x

y

A

B

O

1C

4, 3Q

2C

5 3

4

3

A

B

4, 3 Q Ax

5



1. The expression 3 22 3x ax bx , where a and b are constants, has a factor of 1x and

leaves a remainder of 15 when divided by 2x . Find the value of a and b .

Solution (1):

Let 3 22 3f x x ax bx

By Factor Theorem, when 1 0x , 1 0f

3 2

1 2 1 1 1 3f a b

0 5a b

5a b ------------- (1)

By Remainder Theorem, when 2 0x , 2 15f

3 2

2 2 2 2 2 3f a b

15 4 2 13a b

4 2 28a b

2 14a b ----------- (2)

(1) + (2), we get

5a b ------------- (1)

(+) 2 14a b ------------- (2)

3 9

3

a

a

Substitute 3a into (1) or (2), we get 8b

[Analysis]

First, recognize that the expression 3 22 3x ax bx is a polynomial, and there are factor

1x and remainder 15 when this polynomial is divided by 2x . So naturally, we can

connect these to the Factor and Remainder Theorems of polynomial.



Solution (2):

When 3 22 3x ax bx is divided by 1x , the remainder is 0.

Therefore,

5 0a b

5a b --------------- (1)

Similarly, when 3 22 3x ax bx is divided by 2x , the remainder is 15 .

Therefore, from the long division below, we get

4 2 13 15a b

2 14a b -------------- (2)

So, (1) + (2), 3a , substitute 3a into (1) or (2), we get 8b

3 2

3 2

2

2

2 3

2 2

2 3

2 2

2 3

2 2

5

x ax bx

x x

a x bx

a x a x

a b x

a b x a b

a b

1x

22 2 2x a x a b

[Analysis]

Since the expression 3 22 3x ax bx is a polynomial, and when divided by 1x and 2x

leave a remainder 0 and 15 respectively. Then we may proceed with long division of this

polynomial. This long division can alternatively be carried out by synthetic division.



Solution (3):

Let 3 22 3 1 2x ax bx x x mx n kx p

By the inspection of the highest degree term, we get 2m .

when 1x , 1 0k p

k p --------- (3)

When 2x , 2 0k p

2 15k p -------- (4)

Substitute (3) into (4), we get 5k , 5p .

3 22 3 1 2 2 5 5x ax bx x x x n x

3 22 5 2 1 2 2x ax b x x x x n

3 2

3 2

2

2

2 3

2 4

4 3

4 2 4

2 8 3

2 8 2 2 8

4 2 13

x ax bx

x x

a x bx

a x a x

a b x

a b x a b

a b

2x

22 4 2 8x a x a b

[Analysis]

Since the expression 3 22 3x ax bx is a polynomial, and when divided by 1x and 2x

leave a remainder 0 and 15 respectively. We may propose rewriting this polynomial as

1 2x x mx n kx p . When 1x , 1 0k p . When 2x , 2 15k p .



By the inspection of the constant term, we get 1n .

3 22 3 1 2 2 1 5 5x ax bx x x x x

2

2 2

3 2 2

3 2

2 2 1 5 5

2 2 2 5 5

2 2 4 4 3

2 3 8 3

x x x x

x x x x x x

x x x x x

x x x

By equating the coefficients, we get 3a , 8b .

In Summary:

Solution (1) is the most direct and efficient way to solve this question. Factor

and Remainder Theorems are frequently tested. Selecting a time efficient method is

key factor in examination performance.



2. Given that ln x

yx

for 0x , find the set of values of x for which y is an increasing

function of x .

Solution :

Given that ln x

yx

for 0x , we can obtain the gradient by differentiation with quotient rule.

2

11 lnx x

dy x

dx x

2

11 ln

dyx

dx x

For the function to be increasing, the gradient 0dy

dx .

2

11 ln 0x

x

Given that 0x , 2 0x , then 2

10

x for 0x .

Therefore , 1 ln 0x

ln 1x

1x e

So the range of values of x is 0 x e .

[Analysis]

This question is most difficult as student may not understand what an increasing function is.

Let’s informally say that an increasing function is one that the value of the dependent

variable(y or f(x)) increases as the value of x increases. On a graph, this can be seen as an

upward sloping curve. If a curve is continuously upward sloping, then the gradient must be

non-negative , meaning, positive or zero.

In Summary:

This question is particularly challenging as the demand on understanding is

beyond the routine ones.



3. Without using a calculator, find the values of the integers a and b for which the solution

of the equation

24 3 6x x

is 7

a b.

Solution :

24 3 6x x

8 3 3 3 2x x

8 2x x

22

2

8 1 2

8 1 8 1 2 8 1

8 1 16 2

8 1 4 2

x

x

x

x

or

7 4 2x

Therefore, 4a , 2b .

[Analysis]

Observe that there is a common factor 3 in the equation, so we shall simplify it. Next, the

answer is to be expressed as 7

a bx

. This is the same as 7x a b . Collect the

common x terms together.

2

2

2

8 1 2

2

8 1

2 8 1

8 1 8 1

16 2

8 1

4 2

7

4 2

7

x

x

x

x

x

x

In Summary:

This question tests on simplifying surds, surd conjugates and rationalizing

surds. This topic is a regular visitor in O level.




(i) lg 14 lg 2 2lg5x x ,

(ii) 2 4log log 6y y .

Solution :

(i) lg 14 lg 2 2lg5x x OR (i) lg 14 lg 2 2lg5x x

214lg lg5

2

x

x

2lg 14 lg 2 lg5x x

14

252

x

x

lg 14 lg 25 2x x

14 25 2x x

64 24x

8 3x

8

3x

2

23

x

[Analysis]

Part (i) can be done with product rule or quotient rules of logarithm. A very straight forward

question. For lg 14x , 14 0x , so 14x . Similarly, for lg 2x , 2 0x , so

2x . Therefore the equation can only admit the values of 2x .

In Summary:

As in solving any equation, students are advised to check for correctness and

admissibility of the answers to the given equation. A logarithmic function of the type

loga x requires that 0x and, 0a but also 1a . This topic is regularly tested in O

level.



Solution(1) :

(ii) 2 4log log 6y y OR (ii) 2 4log log 6y y

22

2

loglog 6

log 4

yy 4

4

4

loglog 6

log 2

yy

22

2

loglog 6

2log 2

yy Since

1

24 4 4

1 1log 2 log 4 log 4

2 2 ,

Since 2log 2 1 , 44

loglog 6

1

2

yy

22

loglog 6

2

yy 4 42log log 6y y

2

3log 6

2y 43log 6y

2log 4y 4log 2y

42y 24y

16y 16y

[Analysis]

Part (ii) can be done with base change formula of logarithm. For lg y , 0y . Therefore the

equation can only admit the values of 0y .



Solution(2) :

Let 2log y n , then 2ny and 4log y m , then 24 2m my . We then get

22 2n m

2n m ---------- (1)

Given that 2 4log log 6y y , we get

6n m ------- (2)

Substitute (1) into (2), we get

3 6m

So, 2m

Hence, 2 22 16y

[Analysis]

Part (ii) can be done by exchanging the logarithm with index number where loga x n can be

written as nx a .

In Summary:

As in solving any equation, students are advised to check for correctness and

admissibility of the answers to the given equation. Remember that logarithm and

indices are the two sides of a coin. Solution(2) is a lot quicker.



5. The normal to the curve 1 3tany x , at the point where the curve crosses the y-axis,

passes through the point ,3k . Find the value of k .

Solution:

Given that 1 3tany x ,

23secdy

xdx

2

3

cos

dy

dx x

At 0x , 1y ,

2

3

cos 0

dy

dx

For cos 0 1 ,

3dy

dx

The gradient of the normal will be

1 1

3 3

.

The normal also passes through the point 0,1 , so the equation of the normal line is

1

1 03

y x

13

xy

The normal passes through the point ,3k , so

3 13

k

6k

[Analysis]

It is advisable to make a sketch of the curve 1 3tany x to better visualize the problem.

“the curve crosses the y-axis” at 0x , 1y , where tan 0 0 . Here x is in radian. The

normal is perpendicular to the tangent at point 0,1 where the gradient at this point is the

gradient of the tangent.

0

1

3tany x

1 3tany x

In Summary:

In this case, the correct sketch of the curve serves as a guide to check our work.

The equation of the normal matches that in the sketch.



6. A curve has the equation 22 6y x x c , where c is a constant.

(i) In the case where 20c , find the set of values of x for which 0y .

(ii) Find the value of c for which the line 2 8y x is a tangent to the curve.

Solution:

(i) In the case where 20c , find the set of values of x for which 0y .

Given that 22 6 20y x x

22 6 20 0x x

2 3 10 0x x

5 2 0x x OR

2 2

2 3 33 10 0

2 2x x

With the aid of the quadratic curve,

23 9 40

02 4 4

x

23 49

2 4x

49 3 49

4 2 4x

7 3 7

2 2 2x

7 3 3 3 7 3

2 2 2 2 2 2x

2 5x

the solution is 2 5x

(i)

[Analysis]

Part (i) is about solving quadratic inequality.

In Summary:

It is usually quicker to solve the inequality with graphical aided approach. The

completing the square method remain useful which must be master by all students. For 2x a , we then have x a or x a .



(ii) Find the value of c for which the line 2 8y x is a tangent to the curve.

Solution(1):

Given that 22 6y x x c , Given that 2 8y x ,

4 6dy

xdx

2 8y x

the gradient of the line is 2 .

Therefore, at the point of tangent,

2 4 6x

1x

When 1x , 2 1 8 6y

The point of tangent must also be on the curve.

2

6 2 1 6 1 c

10c

Solution(2):

Given that 22 6y x x c and 2 8y x ,

28 2 2 6x x x c

22 4 8 0x x c

The discriminant of this quadratic equation is to be zero.

2

4 4 2 8 0c

10c

[Analysis]

Part (ii) requires the point of tangent on the curve must have the same gradient as the line. the

coordinate of the point of tangent is to be on the line as well as on the curve.

[Analysis]

Part (ii) requires that the curve and the line touches at one point. This is equivalent to a

repeated root case with discriminant being zero.

In Summary:

By comparison, the nature of root method is much shorter than that of the

coordinate geometry approach.



7. Find the coordinates of the mid-point of the straight line joining the points of intersection

of the curve 2 22 5 68x y x and the line 2 3 9y x .

Solution(1):

Given 2 3 9y x ,

2 9 3y x ------------ (1)

Given 2 22 5 68x y x ,

222 2 10 136x y x ----------- (2)


222 9 3 10 136x x x

211 44 55 0x x 2 4 5 0x x

5 1 0x x

1x or 5x

When 1x , 9 3 1

62

y

.

When 5x , 9 3 5

32

y

.

The x coordinate of midpoint 1 5

22

The y coordinate of midpoint 3 6 3 1

12 2 2

Solution(2):

Given 2 3 9y x ,

2 9 3y x ------------ (1)

[Analysis]

Similar to Question 6, we can find the points of intersection by solving a linear and an non-

linear equation. Then determines the coordinates of the intersections, before finding the mid-

point of the two coordinates.

[Analysis]

We can make use of Sum of Roots to give an alternative solution.



Given 2 22 5 68x y x ,

222 2 10 136x y x ----------- (2)


222 9 3 10 136x x x

211 44 55 0x x 2 4 5 0x x

Let the two roots be 1x and 2x , then by sum of roots, we get

1 2 4x x

The x coordinate of midpoint 1 2 42

2 2

x x

The y coordinate of midpoint 1 2

2

y y

1 2

1 2

9 3 9 3

4

18 3

4

18 3 4

4

11

2

x x

x x

In Summary:

Notice that, it is easier to replace 2y into the curve rather than that of y . This

little trick makes the equation much easier to simplify.



8. (i) Show that 2cos3 cos 4sin cosx x x x .

(ii) Hence, or otherwise, solve, for 0 x radians, the equation

cos3 2cos 0x x .

(i) Show that 2cos3 cos 4sin cosx x x x .

Solution(1):

From L.H.S cos3 cosx x

cos2 cos sin 2 sin cosx x x x x

2 2 2cos sin cos 2cos sin cosx x x x x x

2 2 2cos 1 cos cos 2cos sin cosx x x x x x

2 22cos 1 cos 2cos sin cosx x x x x

2 2cos 2cos 1 2sin 1x x x

2 22cos cos 1 sinx x x

22cos 2sinx x

24sin cosx x

=R.H.S

Solution(2):

From L.H.S cos3 cosx x

cos 2 cos 2x x x x

2sin 2 sinx x

2 2cos sin sinx x x

24sin cosx x

=R.H.S

[Analysis]


formulae maybe productive. eg cos cos cos sin sinA B A B A B ;

cos cos 2sin sinA B A B A B .

[Analysis]


formulae maybe productive. eg cos cos cos sin sinA B A B A B ;

cos cos 2sin sinA B A B A B .



Solution(3):

From R.H.S 24sin cosx x

22 2sin cosx x

2 cos2 1 cosx x

2cos2 cos 2cosx x x

cos2 cos cos2 cos cos cosx x x x x x

cos2 cos 1 cos2 cos cosx x x x x

2cos2 cos 2sin cos cosx x x x x

cos2 cos 2sin cos sin cosx x x x x x

cos2 cos sin 2 sin cosx x x x x

cos3 cosx x

=L.H.S

(ii) Hence, or otherwise, solve, for 0 x radians, the equation

cos3 2cos 0x x .

Solution(1):

Given that cos3 2cos 0x x , for 0 x ,

cos3 cos 3cos 0x x x

24sin cos 3cos 0x x x

2cos 4sin 3 0x x

cos 0x or 24sin 3 0x

In Summary:

Take note that applying the factor formulae is most time efficient (solution(2)).

There are numerous ways to answer this question, of varying length.

[Analysis]

In this part, “Hence” is referring to using the result shown in part (i) to answer this question;

“or otherwise” means that students can answer this part as if it is a fresh one taking no

reference from the earlier part.



2

x

2 3sin

4x

3

sin2

x

3

sin2

x or 3

sin2

x

2

,3 3

x

Inadmissible


, ,3 2 3

x

Solution(2):

cos3 2cos 0x x

cos2 cos sin 2 sin 2cos 0x x x x x

2 2 2cos sin cos 2cos sin 2cos 0x x x x x x

2 22cos 1 cos 2cos sin 2cos 0x x x x x

2 2cos 2cos 1 2sin 2 0x x x

2 2cos 2cos 2sin 1 0x x x

2 2cos 2cos 2 1 cos 1 0x x x

2cos 4cos 1 0x x

cos 0x or 24cos 1 0x

2

x

2 1cos

4x

1

cos2

x

1

cos2

x or 1

cos2

x

3

x

2

3x


, ,3 2 3

x

In Summary:

Trigo questions are typically multiple approaches. Take note that solution(1) is

most efficient. The knowledge of trigo ratios of special angles and triangles are needed.



9. The function f is defined, for 0x , by f 3sin 13

xx

.

(i) State the maximum and minimum values of f x .

(ii) State the amplitude of f.

(iii) State the period of f.

(iv) Find the smallest value of x such that f 0x .

(v) Sketch the graph of 3sin 13

xy

for 0 540x .

Solution:

(i) State the maximum and minimum values of f x .

1 sin 13

x

3 3sin 33

x

3 1 3sin 1 3 13

x

4 f 2x

Therefore the minimum value is 4 , and the maximum value is 2 .

(ii) State the amplitude of f.

The amplitude is 3 .

(iii) State the period of f.

The period of the function f is 360

10801

3

[Analysis]

This question tests specifically all about trigo function and its related characteristics.



(vi) Find the smallest value of x such that f 0x .

3sin 1 03

x

1sin

3 3

x

19.473

x

58.4x (to the nearest .1 degree)

(vii) Sketch the graph of 3sin 13

xy

for 0 540x .

1

2

270 270

540 x

y

58.4 481.6

In Summary:

Trigo curve sketching is an important part in this paper. This is a rather easy

question.



10. The mass, m mg, of a radioactive substance decreases with time, t hours. Measured

values of m and t are given in the following table.

t (hours) 2 4 6 8 10

m (mg) 48.2 41.5 35.7 30.7 26.5

It is known that m and t are related by the equation 0

ktm m e , where 0m and k are

constants.

(i) Plot ln m against t for the given data and draw a straight line graph.

(ii) Use your graph to estimate the value of 0m and of k .

(iii) Estimate the number of hours for the mass of the substance to be halved.

Solution:

t (hours) 2 4 6 8 10

m (mg) 48.2 41.5 35.7 30.7 26.5

ln m 3.88 3.73 3.58 3.42 3.28

(i) Plot ln m against t for the given data and draw a straight line graph.

See the graph on the next page. 0ln lnm m kt

(ii) Use your graph to estimate the value of 0m and of k .

From the graph, 0ln 4.05m

4.05

0 57.4m e mg

3.95 3.300.077

1.2 9.6k

0.077k

[Analysis]

This question tests specifically on “straight line graph” or linear law. The word “Estimate” is

referred to reading from the graph.



(iii) Estimate the number of hours for the mass of the substance to be halved.

Given that 0

ktm m e , when 0

2

mm ,

00

2

ktmm e OR 0

0ln ln ln ln 2 4.05 0.69 3.362

mm m

1

2

kte From the graph, 9t hours

ln 2 kt

ln 2

tk

ln 2

90.077

t hours

In Summary:

Part (iii) the reading from the graph solution is preferred. In the exam, student

should use the full length of the graph paper for the vertical axis. The quality of your

solution may vary, largely dependent on the scales selected. It is advisable to make your

line as large as possible. This question is also very time-consuming, students beware.



0 2 4 6 8 10 t

3

ln m

4



11. Solutions to this question by accurate drawing will not be accepted.

The diagram shows a trapezium ABCD in which AB is parallel to DC and angle 90BAD .

The point A is 0,6 and the point D is 2, 2 .

(i) Find the equation of AB .

Given that B lies on the line y x , find

(ii) the coordinates of B .

Given that the length of DC is twice the length of AB , find

(iii) the coordinates of C ,

(iv) the area of the trapezium ABCD .

Solution:

(i) Find the equation of AB .

Given that A 0,6 , AB AD , D 2, 2 ,

Gradient of AD , 2 6

42 0

ADm

Gradient of AB , 1 1 1

4 4AB

AD

mm

Equation of the line AB ,

1

6 04

y x

1

64

y x

0 x

y

0,6 A

B

C

2, 2D



(ii) the coordinates of B .

Given that B lies on the line y x , B ,B Bx x ; Point B is on the line AB .

16

4B Bx x

36

4Bx

8Bx

the coordinates of B 8,8 .

(iii) the coordinates of C ,

Given that the length of DC is twice the length of AB ,

by similar triangle, we can deduce that 2C D B Ax x x x

2 2 8 0 18Cx

2C D B Ay y y y

2 2 8 6 2Cy

the coordinates of C 18,2 .

(iv) the area of the trapezium ABCD .

the length AD , 2 2

2 0 2 6 68AD

the length AB , 2 2

8 0 8 6 68AB

the area of the trapezium ABCD 1

2AB DC AD

1

22

AB AB AD

1

32

AB AD

13 68 68

2

102 square units

In Summary:

This is a fully coordinate geometry question. There are a number of ways to

find the area of the trapezium. Pay particular attention to part(iii).



12. A curve has the equation 2 1 4 1y x x .

(i) Express dy

dx in the form

4 1

kx

x , where k is a constant.

Hence

(ii) find the rate of change of x when 2x , given that y is changing at a constant rate

of 2 units per second,

(iii) evaluate 2

0

3

4 1

xdx

x .

Solution:

(i) Express dy

dx in the form

4 1

kx

x , where k is a constant.

Given that 2 1 4 1y x x ,

1 4

2 4 1 2 12 4 1

dyx x

dx x

2 4 1 2 2 1

4 1 4 1

x xdy

dx x x

OR

1 42 4 1 2 1

24 1 4 1

kxx x

x x

12

4 1

dy x

dx x

Let 2x ,

2 1 4

2 3 33 2 3

k

2 18 6k

12k

12

4 1

dy x

dx x

[Analysis]

Part (i) is about Product rule; part (ii) is a simple rate of change. Take note that “Hence”

requires part (i) to be used in these two parts.



(ii) find the rate of change of x when 2x , given that y is changing at a constant rate

of 2 units per second,

Given that 2dy

dt , we apply chain rule

dy dy dx

dt dx dt OR

1dy dy dt dy dy

dxdx dt dx dx dt

dt

OR dx dx dy

dt dy dt

1dx dy

dydt dt

dx

At 2x ,

12 28

4 2 1

dy

dx

Therefore , 2 8dx

dt

1

4

dx

dt

(iii) evaluate 2

0

3

4 1

xdx

x .

2

0

3

4 1

xdx

x

2

0

1 12

4 4 1

xdx

x

2

0

1 12

4 4 1

xdx

x

2

0

1

4

dydx

dx

2

0

1

4y

12 2 1 4 2 1 2 0 1 4 0 1

4

5 1

22 2

In Summary:

This is the only question on integration in this paper. Do pay particular

attention on part (ii). This is a rather easy question.



1. A and B are acute angles such that 3

sin8

A B and 5

sin cos8

A B . Without using a

calculator, find the value of

(i) cos sinA B ,

(ii) sin A B ,

(iii) tan

tan

A

B.

Solution :

Given that 0 , 90A B , 3

sin8

A B , 5

sin cos8

A B ,

(i) cos sinA B ,


cos sin sin cos sinA B A B A B

5 3 1

cos sin8 8 4

A B

(ii) sin A B ,


5 1 7

sin8 4 8

A B

(iii) tan

tan

A

B.

sin 5

tan sin cos 5cos 8sin 1tan cos sin 2

cos 4

A

A A BABB A B

B

[Analysis]

This question is solely based on the multiple angle formulae of sine ratio. Since 0 , 90A B ,

the ratios of sine and cosine of angles of A and, of B are positive(in quadrant I).

In Summary:

Please make sure that you are comfortable with operations with fractions. Be

careful with the sign of each ratios.



2. (i) Express 2

7

2 6x x in partial fractions.

(ii) Hence evaluate 9

23

7

2 6dx

x x .

Solution :

(i) Express 2

7

2 6x x in partial fractions.

22 6 2 3 2x x x x

2

7 7

2 6 2 3 2x x x x

Let

7

2 3 2 2 3 2

A B

x x x x

Therefore,

By covering up method, OR By equating coefficients, OR By assigning values

when 2x , 7 2 2 3A x B x 7 2 2 3A x B x

7

12 2 3

B

, 7 2 2 3A B x A B when 2x , 1B

when 3

2x ,

2 0

2 3 7

A B

A B

when 3

2x , 2A

72

32

2

A

1B , 2A

7 1 2

2 3 2 2 2 3x x x x

[Analysis]

Part (i) is a straight forward partial fractions. Part (ii) MUST use part (i) to complete the

question. The rule of integration is

'fln f

f

xdx x c

x .

[Analysis]

Part (i) is a straight forward partial fractions. Part (ii) MUST use part (i) to complete the

question. The rule of integration is

'fln f

f

xdx x c

x .



(ii) Hence evaluate 9

23

7

2 6dx

x x .

9

23

9

3

9 9

3 3

7

2 6

1 2

2 2 3

1 2

2 2 3

dxx x

dxx x

dx dxx x

9

23

9

3

9

3

7

2 6

1 2

2 2 3

ln 2 ln 2 3

dxx x

dxx x

x x

9 9

3 3ln 2 ln 2 3

ln 7 ln1 ln 21 ln 9

7 9ln

21

ln 3

1.10

x x

9

3

2ln

2 3

7 1ln ln

21 9

7 1ln

21 9

ln 3 1.10

x

x

Additional note:

If f Fx dx x , then f F Fb

ax dx b a .

So,

2

7 1 2 1 2

2 6 2 2 3 2 2 3dx dx dx dx

x x x x x x

2

ln 2 ln 2 3 ln2 3

xx x c c

x

9

23

7 9 2 3 2 7 1ln ln ln ln ln3

2 6 2 9 3 2 3 3 21 9dx c c

x x

In Summary:

This is a typical use of partial fractions with integration. More of this kind can

be expected. There are more ways to decompose a fraction than have been shown.

OR



3.

(i) Use the substitution 2xu to express the equation 28 2 15x x as an cubic

equation in u .

(ii) Show that 3u is the only real solution of this equation.

(iii) Hence solve the equation 28 2 15x x .

Solution :

(i) Use the substitution 2xu to express the equation 28 2 15x x as an cubic

equation in u .

Let 2 0xu ,

28 2 15x x

3

2 4 2 15 0x x

3 4 15 0u u

(ii) Show that 3u is the only real solution of this equation.

Let 3f 4 15u u u

3f 3 3 4 3 15 0 therefore 3u is a factor of the function f u .

2f 3 3 5u u u u by long division

Given that f 0u ,

23 3 5 0u u u

3 0u or 2 3 5 0u u

3u Discriminant 23 4 1 5 11 0 , no real root.

OR

2 23 3

5 02 2

u

23 11

02 4

u

, no real root

[Analysis]

Part (i) needs to apply indices rules for 3

3 3 38 2 2 2x

x x x u and

2 22 2 2 4 2 4x x x u . Part (ii) is to show that the quadratic factor is non-zero, or has no

real number root.



(iii) Hence solve the equation 28 2 15x x .

2 3x

lg 2 lg3x

lg31.58

lg 2x

Additional Note on factorization:

(1) by synthetic div

3 24 15 3 3 5u u u u u

(2) by equating coefficients

Let 3 24 15 3u u u au bu c

3 23 3 3au b a u c b u c

therefore, 1a , 5c , 3b

3 24 15 3 3 5u u u u u

(3) by inspection and assigning value

Let 3 24 15 3 5u u u u bu

When 1u , 1 4 15 2 1 5b

3b , 3 24 15 3 3 5u u u u u

In Summary:

This question comprises of topics from indices, exponential equation, solving

polynomial equation, nature of roots in quadratics and logarithm. There are

numerous ways to factorize a polynomial. In part (iii) ln 3

1.58ln 2

x also gives the

same result.

1

3

3

9

5

3 1 0 4 15

15

0



4.

The diagram shows an isosceles triangle ABC in which AC BC . Lines are drawn from

A and B to meet BC and AC at P and Q respectively. The lines AP and BQ intersect

at X . Given that PC QC , show that

(i) AXB is an isosceles triangle,

(ii) PX QX .

Solution :

(i)

Given that PC QC , AC BC

PC QC

BC BP AC AQ

Since BC AC , we get

BP AQ

For AQB and BPA ,

AB BA , common side

BAQ ABP , ABC is isosceles

AQ BP , as shown above

AQB and BPA are congruent(SAS).

Therefore, ABQ BAP

ABX BAX

Therefore, AXB is an isosceles triangle.

[Analysis]

Essentially, this is the only pure geometrical proof question in Add Math 2009. In part (i), to

show that AXB is an isosceles, we just need to show either AX BX or XAB XBA . In

part (ii), PX QX is the same as AP AX BQ BX , with AP BQ and AX BX from

part (i).

A B

C

P Q

X



(ii)

Since AQB and BPA are congruent,

BQ AP

BX XQ AX XP

Since AXB is an isosceles,

BX AX

XQ XP

PX QX


(i)

For ACP and BCQ ,

AC BC , ABC is an isosceles

ACP BCQ , common angle

PC QC , as given

ACP and BCQ are congruent (SAS)

therefore, CAP CBQ

also AP BQ

Since CAB CBA , then

CAP PAB CBQ QBA

But CAP CBQ , therefore

PAB QBA

so XAB XBA

Therefore, AXB is an isosceles.

(ii)

Since AXB is an isosceles, then AX BX .

AP BQ ,from above

AX XP BX XQ

Therefore, PX QX

In Summary:

Before making attempt to answer a geometrical proof, students should look at

what the given conditions and what these mean to the diagram. More importantly,

look at what is required to prove and work backward to the point where it meets the

given conditions. Typically, there are multiple ways to prove a geometrical question.



5. (i) Write down the first three terms in the expansion, in ascending powers of x , of

24

nx

, where n is a positive integer greater than 2 .

The first two terms in the expansion, in ascending powers of x , of 1 24

nx

x

are

2a bx , where a and b are constants.

(ii) Find the value of n .

(iii) Hence find the value of a and of b .

Solution:

(i) Given that 24

nx

,

0 1 2

1 2 0

0 1 22 2 2 2 24 4 4 4 4

n n

n n n n n n n

n

x x x x xC C C C

2 21 1 22

24 32

nnn

n n xn x

(ii) Given that 21 24

nx

x a bx

,

2 21 1 22

1 2 1 24 4 32

n nnn

n n xx n xx x

2 2 2 31 1 21 2 1 22 2

2 24 32 4 32

n nn nn n

n n x n n xn x n xx

1 2 1 24 2 2 1 2 8 2

24 32

n n n n

nn x n n n x

1 2 28 2 17 2

24 32

n n

nn x n n x

[Analysis]

This is regarding binomial theorem, 0 1 1 0

0 1

n n n n n n n

na b C a b C a b C a b .



1 2 2

28 2 17 2

24 32

n n

nn x n n x

a bx

18 2

04

nn xx

Since 12 0n , therefore

8 0n

8n

(iii) 82 256a

68 9 2144

32b


(a)

1 21

2 2 1 2 14 8 8 2! 8 8

nn n

n nn nx x x x x

n

21

2 18 128

nn n xnx

21 22

28 128

nnn

n n xn x

2 21 1 22

24 32

nnn

n n xn x

(b)

Let 2

0 1 2f 24

n

n

n

xx c c x c x c x

When 0x ,

02n c

[Analysis]

Rewrite binomial theorem, 1 2

11 1

2!

n nn n n

n nb b b ba b a a n

a a a a

.



1

' 1 1

1 2

1f 2 2

4 4

n

n

n

xx n c c x nc x

When 0x ,

1

1

2

4

nnc

2 2

'' 2

2

1f 1 2 2 1

4 4

n

n

n

xx n n c n n c x

When 0x ,

2

2

1 2

32

nn nc

Therefore,

2 21 1 22

2 24 4 32

n nnn

n n xx n x

In Summary:

The difficulty in this question is in the care of indices. Alternative solution (a) is

a little trick to make the expansion much easier to handle. Alternative solution (b) uses

differentiation to expand the series.



6.

The diagram shows part of the curve 1 2cosy x , meeting the x -axis at the points A

and B .

(i) Show that the x -coordinate of A is 2

3

and find the x -coordinate of B .

(ii) Find the total area of the shaded regions.

Solution:

(i)

1 2cos 0x

1cos

2x

x is in 2nd

and 3rd

quadrants.

,3 3

x

2 4

,3 3

x

Therefore, 2

3A

,

4

3B

[Analysis]

Part (i) is about solving simple trigo equation; part (ii) is integrating area under the curve.

Take note that the area below the x-axis is negative, so the two shaded regions are to be

integrated separately.

0

1 2cosy x

A B x

y



(ii) the total area of the shaded regions.

42

332

03

1 2cos 1 2cosx dx x dx

42

33203

2sin 2sinx x x x

2 2 4 4 2 2

2sin 0 2sin 0 2sin 2sin3 3 3 3 3 3

2 3 2 3 3

2 2 23 2 3 2 2

2 2

3 2 33 3

2 2

3 3 33 3

since 3 3 0

2 23 3 3

3 3

3 3 5.20 square units

In Summary:

Be mindful of the trigo ratio of special angles and triangles.



7. (i) Find the coordinates of all the points at which the graph of 3 5 2y x meet the

coordinate axes.

(ii) Sketch the graph of 3 5 2y x .

(iii) Solve the equation 3 5 2x x .

Solution:

(i)

When 0y , the x-axis intercepts

3 5 2 0x

3 5 2x

When 2

13

x , When 2

13

x ,

3 5 2x 3 5 2x

7 12

3 3x 1x

When 0x , the y-axis intercept

5 2 3y

(ii)

[Analysis]

Part (i) is about y- and x- axes intercepts. Part (ii) should show the turning point and symmetry.

Part (iii) is the solution of the line y x and 3 5 2y x .

0 x

y 3

2

3 1

1 2

1

2

21 , 2

3

12

3



(iii)

3 5 2x x ,

2 3 5x x since 3 5 0x , then 2 0x . Therefore, 2x

When 2

13

x , When 2

13

x ,

2 3 5x x 2 3 5x x

2 7x 4 3x

7 13

2 2x

3

4x


Part (iii)

3 5 2x x

2 3 5x x since 3 5 0x , then 2 0x . Therefore, 2x

2

2 3 5x x

2 2

2 3 5x x

2 2

2 3 5 0x x

2 3 5 2 3 5 0x x x x

4 3 2 7 0x x

4 3 0x or 2 7 0x

3

4x

7 13

2 2x

In Summary:

Be mindful that when simplifying modulus, there is a particular range of values

of x for each of them. Any solution that does not fall in this range will need to be

rejected. All solutions MUST be tested with the given equation.



8. A motorcycle is driven along a straight horizontal road. As it passes a point A the brakes

are applied and the motorcycle slows down, coming to rest at a point B . For the journey

from A to B , the distance, s metres, of the motorcycle from A , t seconds after passing

A , is given by

10400 1 16t

s e t

(i) Find an expression, in terms of t , for the velocity of the motorcycle during the

journey from A to B .

(ii) Find an expression, in terms of t , for the acceleration of the motorcycle during the

journey from A to B .

(iii) Find the velocity of the motorcycle at A .

(iv) Show that the time taken for the journey from A to B is approximately 9.163

seconds.

(v) Find the average speed of the motorcycle for the journey from A to B .

Solution:

(i) Let the velocity be V ,

101

400 1610

tds

V edt

10 m40 16s

t

V e

[Analysis]

It will be best to organize all the pieces of information into a diagram:

A B s

10s 400 1 16t

e t

0t Bt t

V 0V



(ii) Let the acceleration be a ,

101

4010

tdV

a edt

102

m4s

t

a e

(iii) At A when 0t , at the start of the journey,

0 m40 16 24s

V e

(iv) Let the time taken for the journey from A to B be Bt , where the velocity is zero.

100 40 16Bt

e

102

5

Bt

e

2ln

10 5

Bt

510ln 9.163 s

2Bt

(v) Let the distance from A to B be Bs

10400 1 16Bt

B Bs e t

2

400 1 16 9.163 93.392 93.3m5

Bs

Average speed93.393 m10.19 10.2

s9.163

In Summary:

If you are to introduce some symbols, should state them clearly before use.



9.

The diagram shows a circle with centre 2, 1C and radius 5 .

(i) Given that the equation of the circle is 2 2 2 2 0x y gx fy c , find the value of

each of the constants g , f and c .

The points A and B lie on the circle such that the line AC is parallel to the x -axis and

the line AB passes through the origin 0 .

(ii) Write down the coordinates of A .

(iii) Find the equation AB .

(iv) Find the coordinates of B .

Solution:

(i) Centre 2, 1C and radius 5 ,

2g

1 1f

2 2 22 1 5 20c

2, 1C

x

y

A

B

0

[Analysis]

If you can remember that 2 22 2 2 2 2 22 2 0x y ax by a b r x a y b r where

,a b being the centre of the circle, you may quickly pick up a few points.



(ii) Point A is on 1y and 5 units to the left of 2x ,

the coordinates of 3, 1A .

(iii) The line AB passes through the origin, 0,0O and 3, 1A .

Let the equation AB be y kx ,

1 3k

1

3k

equation AB : 1

3y x

(v) To find the coordinates of B .

2 2 4 2 20 0

3

x y x y

y x

2 23 4 3 2 20 0y y y y

210 10 20 0y y 2 2 0y y

2 1 0y y

1y or 2y

When 2y , 1

23

x

6x

the coordinates of 6,2B .

In Summary:

This is the only coordinate geometrical question on circle. Questions relating to

circle can be expected in the future installations in this exam. It can be set with

geometrical properties of circles.




Part (i)

Equation of the circle,

2 2 22 1 5x y

2 24 4 2 1 25x x y y 2 2 4 2 20 0x y x y

Therefore,

2g

1f

20c



10. A curve is such that 2

26 6

d yx

dx . The curve passes through the point 3,10 and at this

point the gradient of the curve is 12 . Find the coordinates of the stationary point of the curve

and determine the nature of this stationary point.

Solution:

Given that 2

26 6

d yx

dx , the gradient of the curve

2

2

dy d ydx

dx dx

6 6dy

x dxdx

23 6x x c

At 3,10 , the gradient of the curve is 12 .

2

12 3 3 6 3 c

3c

Therefore, the equation of the gradient

23 6 3dy

x xdx

The stationary points are where the gradient is zero.

20 3 6 3x x 20 2 1x x

2

0 1x

0 1x

1x

There is only one stationary point, with 2

26 1 6 0

d y

dx , it has not determined the nature of

this stationary point.

[Analysis]

You remember that gradient of the curve, 2

2

dy d ydx

dx dx

.



x 1 1 1

dy

dx 0 0 0

From the table, we can deduce that the stationary point is a point of inflexion.

In Summary:

This is a very traditional question on derivatives of a curve and the determination

of a stationary point.



11.

The diagram shows three fixed points 0 , A and D such that 17cmOA , 31cmOD and

angle 90AOD . The lines AB and DC are perpendicular to the line OC which makes an

angle with the line OD . The angle can vary in such a way that the point B lies between

the point O and C .

(i) Show that 48cos 14sin cmAB BC CD .

(ii) Find the values of for which 49cmAB BC CD .

(iii) State the maximum value of AB BC CD and the corresponding value of .

Solution:

(i)

Given that 90AOD , COD , therefore

90AOB

As 90ABO , then

OAB

17cosAB

As 17sinOB , and 31cosOC ,

BC OC OB

31cos 17sin

O 31cm

17cm

A

B

C

D

[Analysis]

Remember point B is to be in between OC .



31sinCD

17cos 31cos 17sin 31sin

AB BC CD

48cos 14sin cm

(ii)

2 214 48 50

48cos 14sin 49

48 14 49cos sin

50 50 50

48 24tan

14 7

49sin cos cos sin

50 73.74

49

sin50

78.52 , 101.15

78.52 or 101.15

78.52 73.74 101.15 73.74

4.78 4.8 27.41 27.4

(iii)

48cos 14sin

50sin

AB BC CD

50 sin 50

For maximum value of 50AB BC CD ,

sin 1

90

90 73.74 16.26 16.3

In Summary:

One must make sure that OB OC for all solutions.



Additional information:

For OB OC , we can determine the largest .

From the diagram above, we get

31tan

17

61.26

Therefore 61.26 .

17

31



1. The function f is defined by 4 3f 4x x x kx , where k is a constant.

(i) Given that 2x is a factor of f x , find the value of k .

(ii) Using the value of k found in part (i), find the remainder when f x is divided by

2x .

Solution :

(i) Given that 2x is a factor of f x , then f 2 0

4 3f 2 2 2 2 4k

0 4 2k

2k

(ii) When 2x is to divide the polynomial f x , then the remainder is f 2 .

4 3

f 2 2 2 2 2 4

f 2 24


(i) By long division,

4 3 3 24 2 4 2 2 4 4x x kx x x x k x k

Since 2x is a factor of f x , the remainder 2 4 4 0k .

2 4 4 0k

2k

[Analysis]

A very routine application of Factor & Remainder Theorems of polynomial, which states that

“If x a is a factor of a polynomial f x , then f 0a .” Likewise, “If x a is to divide a

polynomial f x , then the remainder is f a .”



(ii) When 2x is to divide the polynomial f x , we just need to apply the similar

method as in part (i) .

By Long Division,

4 3 3 22 4 3 6 14 2 24x x x x x x x

the remainder is 24 .

In Summary:

This is a rather routine question. Factors(Multiplication) and Division are two

sides of the same coin. Students MUST be able to reason this well. The question can

be quickly done with Synthetic divisions.



2. (i) Show that 2

sin cos 1 sin 2x x x .

(ii) Hence find, in terms of , the value of 2

2

0

sin cosx x dx

.

Solution :

(i)

L.H.S. 2

sin cosx x

2 2sin cos 2sin cosx x x x

Since 2 2sin cos 1x x and 2sin cos sin 2x x x ,

1 sin 2x

= R.H.S.

(ii)

2

2

0

sin cosx x dx

2

0

1 sin 2x dx

2

0

cos 2

2

xx

cos cos0

2 2 2

1 1

2 2 2

12

[Analysis]

Observe that there is a double angle in the equation in part(i). So, the use of double angle

maybe productive.




(i)

R.H.S. 1 sin 2x

Since 2 21 sin cosx x and sin 2 2sin cosx x x ,

2 2sin cos 2sin cosx x x x

2

sin cosx x

= L.H.S.

(ii)

Consider

2

sin cosx x dx

1 sin 2x dx

cos 2

2

xx c where c is the integration constant

Therefore,

2

2

0

sin cosx x dx

cos 2cos 2 02

02 2 2

c c

1 1

2 2 2

12

In Summary:

Part (ii) MUST be solved with the result from part (i).



3. Using a separate diagram for each part, represent on the number line the solution set of

(i) 3 2 18x x ,

(ii) 23 5 1x x .

State the set of values of x which satisfy both of these inequalities.

Solution :

(i)

3 2 18x x ,

3 x

The solution set is : 3x x

(ii)

23 5 1x x

23 14 0x x

3 7 2 0x x

2x or 1

23

x

The solution set is 1

: 2 or 23

x x x

[Analysis]

This is an interesting and unexpected question on inequalities. Part (i) is on linear inequality,

part (ii) is of quadratic inequality. Last part requires answer to be in Set notation.

In Summary:

Take note of the set notation! Other form of set notation may also be

acceptable.

3

3 x

x

2

2x

x

12

3x

12

3



4. A curve has the equation sin 2 3cosy x x .

(i) Find the gradient of the curve when 6

x

.

(ii) Given that x is increasing at a constant rate of 0.06 units per second, find the rate of

change of y when 6

x

.

Solution :

(i)

Given that sin 2 3cosy x x ,

2cos 2 3sindy

x xdx

when 6

x

,

2cos 2 3sin6 6

dy

dx

2cos 3sin3 6

.

1 12 3

2 2

12

2

(ii)

When6

x

, given that 0.06dx

dt

dy dy dx

dt dx dt

5 6

0.152 100

dy

dt units per second

[Analysis]

Gradient of a curve is dy

dx. Rate of change

dy dy dx

dt dx dt . This kind of question has been

repeated in recent years.



5. (i) Sketch the curve 29y x for 5 5x .

(ii) Find the x -coordinates of the points of intersection of the curve 29y x and

the line 27y .

Solution:

(i) Given that 29y x for 5 5x ,

When 0x , 9y .

When 0y , 20 9 x ,

3x

When 5x , 16y .

(ii)

Given that 29y x , when 27y ,

227 9 x

Since the horizontal line can only be intersecting on the reflected part of the quadratic curve,

227 9 x

236 x

6x

the x -coordinates of the points of intersection are 6 and 6 .

[Analysis]

Part (i) is a quadratic modulus. Let’s begin by sketching the quadratic before applying the

modulus. Part (ii) is to solve the simultaneous equations of a line and this modulus.

In Summary:

To change from y = f(x) to y = |f(x)|, is to reflect the part of curve that are below

the x-axis to that of above the axis. The reflected parts have the equation y = - f(x).

x

y

29y x

3 3 O

9

5 5

16



6. (i) Differentiate 2xxe with respect to x .

(ii) Use your answer to part (i) to show that

1 22

0

1

4

x exe dx

.

Solution:

(i) By product rule, let 2xy xe

2 21 2x xdye x e

dx

2 22 x xxe e

(ii) From part (i),

2 22 x xdyxe e

dx

2 22 x xdydx xe dx e dx

dx

2 22 x xdyxe dx dx e dx

dx

2

222

xx e

xe dx y c where c is a integration constant

2 2

2

12 4

x xx xe e

xe dx c where 12

cc is a integration constant

1 2 2 0 02

1 1

0

0

2 4 2 4

x e e e exe dx c c

2 1

4 4

e

2 1

4

e

[Analysis]

Use product rule for part (i). Then a little trick to use part (i) for integration in part (ii).




(i) Let 2xy xe ,

2ln ln xy xe

ln ln 2y x x

1 1

2dy

y dx x

1

2dy

ydx x

2 12xdy

xedx x

2 22x xdye xe

dx

(ii) Given from part (i), 2 22x xdye xe

dx ,

2

21

2 2

xxdy e

xedx

1 1 22

0 0

1

2 2

xx dy e

xe dx dxdx

1 1

2

0 0

1 1

2 2

xdydx e dx

dx

11 2

0 0

1 1

2 4

xy e

1

2 2

0

1 11

2 4

xxe e

2 21 11

2 4e e

2 1

4

e

In Summary:

Do pay particular attention to how the integration is carried out. This

approach has been required frequently in recent exam.



7. The table shows experimental values of two variables x and y , which are connected by an

equation of the form nyx k , where n and k are constants

(i) Using a scale of 1cm to 0.1unit on each axis, plot lg y against lg x and draw a

straight line graph.

(ii) Use your graph to estimate the value of k and of n .

Solution:

(i) Given that nyx k ,

lg lg lgy k n x

(ii)

From the graph, the vertical intercept

lg 1.94k

87.1k

From the graph, the gradient

0.8

0.6n

4 1

13 3

n

x 2 8 14 20

y 33.00 5.07 2.38 1.47

x 2 8 14 20

y 33.00 5.07 2.38 1.47

lg x 0.30 0.90 1.15 1.30

lg y 1.52 0.71 0.38 0.18

[Analysis]

This is a question on linear graph. The given equation, nyx k , can be expressed as

lg lg lgy k n x , where n and lg k are gradient and vertical axis intercept respectively.



O lg x

lg y

0.5 1.0 1.5

0.5

1.0

1.5

In Summary:

This question is very straight forward and very time consuming. Students are

mindful of the time spent on the drawing of the graph.



8. The equation of a curve is 3 23 9y x x x k , where k is a constant.

(i) Find the set of values of x for which y is decreasing.

(ii) Find the possible values of k for which the x -axis is a tangent to the curve.

Solution:

(i) Given that 3 23 9y x x x k ,

23 6 9dy

x xdx

For decreasing range of the curve, 0dy

dx ,

23 6 9 0x x

2 2 3 0x x

3 1 0x x

3 1x

The solution set is : 3 1x x .

(ii) From part (i), 23 6 9dy

x xdx

, let 0dy

dx ,

23 6 9 0x x

3 1 0x x

3x or 1x

When 3x , 0y ,

3 2

0 3 3 3 9 3 k

27k

When 1x , 0y ,

3 20 1 3 1 9 1 k

5k

[Analysis]

Part (i) tests on decreasing function where 0dy

dx . Part (ii) test on gradient being zero, 0

dy

dx

and 0y .




(ii)

From part (i), 23 6 9dy

x xdx

, let 0dy

dx ,

23 6 9 0x x

3 1 0x x

3x or 1x

When 3x , 0y ,

23 23 9 3x x x k x x p

Let 3x , 27 0k

27k

Similarly,

When 1x , 0y ,

23 23 9 1x x x k x x q

Let 1x , 5 0k --------- (2)

5k

In Summary:

This is a good question with a bit of non-routine.



9. Given that the roots of 23 2 1 0x x are and , find the quadratic equation whose

roots are 2 and 2 .

Solution:

Given that 23 2 1 0x x ,

2

3

2

3

1

3

the quadratic equation whose roots are 2 and 2 ,

Sum of roots, 2 2 3 2

Product of roots, 22 22 2 2 5 2

22 1

23 3

11

9

2 112 0

9x x

29 18 11 0x x

In Summary:

Modulus is a key topic to understand as it will be use in ‘A’ level. It is best by

going through this solution very carefully so as to enhance your understanding.

[Analysis]

This question is about Sum of roots, b

a and Product of roots,

c

a , where

2 0ax bx c .



10. Without using a calculator, show that

(i) tan 75 2 3 ,

(ii) 2sec 75 4tan75 .

Solution:

(i) tan 75 tan 45 30 where tan 45 1 , 3

tan 303

tan 45 tan 30

1 tan 45 tan 30

31

3

31

3

3 3

3

3 3

3

3 3

3 3

2

3 3

3 3 3 3

12 6 3

6

2 3

[Analysis]

This is an un-usual question. For part (i), we can make use of sum of angles for tangent and the

ratios of special triangles. As for part (ii), we may try the Pythagoras identity, and then apply

the result from part (i) to finish the question.



(ii) 2 2sec 75 1 tan 75

2

1 tan 75

2

1 2 3

1 7 4 3

8 4 3

4 2 3

4tan75


(i) Construct a right angled triangle of 75 , ABC .

75BAC , 15ACB , 60BAD , 15CAD

therefore, ACD is an isosceles triangle.

AD DC

Let 1AB , 2AD , then 3BD

3 2BC BD DC BD AD

2 3

tan 75 2 31

BC

AB

(ii) To show 2sec 75 4tan75 , is to show

2sec 75 4tan75 0

2sec 75 4tan75 21 tan 75 4tan75

2

1 2 3 4 2 3

1 7 4 3 8 4 3

0

In Summary:

Knowledge on surds, rationalization of denominators, ratios of special triangles,

and Pythagoras identities are a must.

75

60

A B

C

D



11. A curve is such that 2

d 82

d

y

x x .

(i) Given that the curve passes through the point 1,5 , find the equation of the curve.

(ii) Find the x -coordinates of the stationary points of the curve.

(iii) Obtain an expression for 2

2

d

d

y

x and hence, or otherwise, determine the nature of each

stationary point.

Solution:

(i) Given that 2

d 82

d

y

x x ,

2

d 8d 2 d

d

yx x

x x

let

1u

x ,

2

1du

dx x

8

2y x cx

At 1,5 , 8

5 2 11

c

15c

the equation of the curve , 8

15 2y xx

.

(ii) Let d

0d

y

x ,

2

80 2

x

2 4x

2x

[Analysis]

Using integration of 2

d 82

d

y

x x in Part (i) to get the equation of the curve that also passes

through the point 1,5 . Let d

0d

y

x , the gradient to be zero to find the stationary points.

Lastly, to determine the nature of each stationary point.



(iii) Given that 2

d 82

d

y

x x ,

2

2 3

d 16

d

y

x x

When 2x ,

2

2

d2 0

d

y

x

At 2x , this is a locally maximum point

When 2x ,

2

2

d2 0

d

y

x

At 2x , this is a locally minimum point


(iii) Given that 2

d 82

d

y

x x ,

When 2x ,

At 2x , the turning point is a locally maximum.

Similarly,

When 2x ,

At 2x , the turning point is a locally minimum.

x 2 ( 1.9) 2 2 2.1

d

d

y

x 0 0 0

x 2 ( 2.1) 2 2 1.9

d

d

y

x 0 0 0

In Summary:

A routine question on first, identifying the curve; secondly, locating the turning

points; lastly, the type of turning points.



12. (i) Write down the equation of the circle with centre 3,2A and radius 5 .

This circle intersects the y -axis at points P and Q .

(ii) Find the length of PQ .

A second circle, centre B , also passes through P and Q .

(iii) State the y -coordinate of B .

Given that the x -coordinate of B is positive and that the radius of the second circle is 80 ,

find

(iv) the x -coordinate of B .

The equation of the circle, centre B , which passes through P and Q , may be written in the

form 2 2 2 2 0x y gx fy c .

(v) State the value of g and of f , and find the value of c .

Solution:

(i) Given that the circle with centre 3,2A and radius 5 , the equation of this circle is

2 2 23 2 5x y

(ii) Points P and Q are on the y -axis 0x .

2 2 20 3 2 5y

2

2 16y

2 4y

6y or 2y

the length of PQ , 6 2 8PQ units

[Analysis]

This question asks very routine questions about circle and co-ordinate geometry. Remember,

the y -axis is on the line 0x . The equation of circle is 2 2 2x a y b r where point

,a b is the centre of the circle with radius .r



(iii) The y -coordinate of B is the mid-point of PQ , 6 2

22

y

(iv) Let the centre of , 2B a , the equation of the circle is

22 2

2 80x a y

2 2

2 80x a y

At point 0,6P ,

2 2

0 6 2 80a

2 64a

8a or 8a (rejected, since the x -coordinate of B is positive.)

the x -coordinate of B is 8x

(v) The equation of circle, centre B ,

2 2

8 2 80x y

2 22 8 64 2 2 4 80 0x x y y

2 2 2 8 2 2 12 0x y x y

Equate 2 2 2 2 0x y gx fy c with the above equation, we get

8g

2f

12c

In Summary:

Remember that 2 2 2 2 0x y gx fy c is the same as 2 2 2x a y b r .

Part (iii) makes use of the geometrical property of circle “ Perpendicular of radius

bisects chord” . It is advisable to make a sketch of the two circles to “visualize” the

question.



1. Solve the equation 23cot 10cosec 5 for 0 360 .

Solution :

Given that 23cot 10cosec 5 ,

23 cosec 1 10cosec 5

23cosec 10cosec 8 0

cosec 4 3cosec 2 0

cosec 4 or 2

cosec3

(Rejected, since sin 1 , then cosec 1 )

1

sin4

1 1sin

4

where is in 3

rd and 4

th quadrants.

Principal angle, 14.47

180 14.47 ,360 14.47

194.5 ,345.5

Alternative solution :

For solving 1

sin4

, is in 3rd

and 4th

quadrants.

Let basic angle be ,

1sin

4

14.47

180 14.47 ,360 14.47

194.5 ,345.5

[Analysis]

The given trigo equation contains the un-common ratios of cot and cosec . It has been a

common practice to change these ratios into cos

cotsin

and

1cosec

sin

. If this

approach is taken, then the question actually would have become more difficult. Rather, it is

more productive to apply Pythagoras identity, 2 21 cot cosec .




Given that 23cot 10cosec 5 ,

2cos 1

3 10 5sin sin

where sin 0

2 23cos 10sin 5sin

2 23 1 sin 10sin 5sin 0

23 10sin 8sin 0

28sin 10sin 3 0

4sin 1 2sin 3 0

1

sin4

or 3

sin2

(Rejected, as sin 1 )

(The rest of the solutions are as shown before.)

In Summary:

Be well verse in the 3 lesser use ratios. Do not simply convert them to sine,

cosine or tangent ratios. Students MUST be very competent in solving simple trigo

equations either by principal angle or basic angle.



2.

The diagram shows a triangular piece of land PQR in which angle 90PQR , 8mPQ

and 12mQR . A rectangle QUVW is to be used as the base of a greenhouse, where U ,

V and W lie on QR , RP and PQ respectively, mQU x and mQW y .

(i) Show that 2

83

xy .

(ii) Express the area, 2mA , of the base of the greenhouse in terms of x .

(iii) Given that x can vary, find the maximum value of A .

Solution :

(i)

Given that 0 8y and 0 12x ,

,V x y , 0,0Q , 0,8P , 12,0R .

Let equation of the line PR be

y mx c

When 0, 8x y ,

8c

When 12, 0x y ,

12 8 0m

2

3m

Therefore, 2

83

xy ,

[Analysis]

We will first cast this question into coordinate geometry.

12m

W

P

R

V

Q

8m

U

my

mx

W

0,8P

12,0

R

,V x y

Q U

y

x



(ii)

The area of the greenhouse,

A xy

2

83

xx

(iii)

2

83

xA x

22

83

xA x

2212

3A x x

2 2 2212 6 6

3A x x

226 36

3A x

22

6 243

A x

When 6x , the maximum value of 224mA .

Alternative solutions:

(i)


,V x y , 0,0Q , 0,8P , 12,0R .

The gradient of PR ,

8 0 2

0 12 3m

equation of the line PR ,

2

8 03

y x

2

83

xy

(i)


,V x y , 0,0Q , 0,8P , 12,0R .

By intercepts formula of straight line, equation

of the line PR ,

112 8

x y

2 3 24x y

3 24 2y x

2

83

xy



(i)


,V x y , 0,0Q , 0,8P , 12,0R .

Knowing two points on the line, we have

8 0 8

0 12 0

y

x

After rearrangement,

2

83

xy

(i)

By similar triangles of RVU and RPQ , we

have

RU UV

RQ QP

12

12 8

x y

After rearrangement,

2

83

xy

(iii)

2

83

xA x

22

83

xA x

d 4

8d 3

A x

x

Let d

0d

A

x ,

40 8

3

x

6x

2

2

d 4

d 3

A

x

When 6x , 2

2

d 40

d 3

A

x , the turning point is a maximum point.

2 6

6 83

A

24A

When 6x , the maximum value of 224mA .

In Summary:

The question can be tackled by different means, of which, the coordinate

geometrical approach is most preferred. In solving for min/max value, students should

be well aware of the conditions that 0 8y and 0 12x .



3. Without using a calculator, solve, for x and y , the simultaneous equations

32 2 1x y ,

1

123 27 81x y x .

Solution :


32 2 1x y ,

52 2 1x y

5 02 2x y

5 0x y

5y x --------- (1)

Given that 1

123 27 81x y x

1123

8127

x

xy

412

3

33

3

x

xy

4

12 33 3x y x

412 3x y

x --------- (2)


4

12 15x xx

4

16 12xx

0x , 216 12 4 0x x

24 3 1 0x x

4 1 1 0x x



2 2log 32 2 log 1x y

2 2log 32 log 2 0x y

5

2 2log 2 log 2 0x y

2 25 log 2 log 2 0x y

25 log 2 0x y

5 0x y --------- (3)

Given that 1

123 27 81x y x

112

3 3

3log log 81

27

x

xy

1

12

3 3 3log 3 log 27 log 81x y x

4

12 3

3 3 3log 3 log 3 log 3 0x y x

3 3 3

412 log 3 3 log 3 log 3 0x y

x

3

412 3 log 3 0x y

x

4

12 3 0x yx

--------- (4)

3 3 4 ,

216 12 4 0x x (the rest of the

solution follows as shown on the left.)

[Analysis]

Notice that 532 2 in the first equation; 3 427 3 ,81 3 in the second equation. Naturally, we

can start by applying indices laws to simplify the equations. Alternatively, we may try

logarithmic laws.



1

4x or 1x

When 1

4x ,

5

4y .

When 1x , 5y .

In Summary:

Students should check the answers against the given equations to guard against

mistakes.



4. (i) Given that the constant term in the binomial expansion of

8

3

kx

x

is 7 , find the

value of the positive constant k .

(ii) Using the value of k found in part (i), show that there is no constant term in the

expansion of 8

4

31

kx x

x

.

Solution :

(i) Given that 0k ,

8

3

kx

x

8

8

41

kx

x

Each term of this expansion, 8 8 4

4

8 81

rn r r r

r

kT x k x

r rx

For the constant term,

8 4 0r

2r

Therefore, 2 0

2

8

2T k x

, since 2 7T , we have

28!7

6!2!k

2 1

4k

1

2k

Since 0k , 1

2k

[Analysis]

A very routine question that simply applies binomial theorem. A constant term is one that is

free of the factor x .



(ii)

Given that 8

4

31

kx x

x

8 8

4

3 3

k kx x x

x x

8 8

8 4 12 4

0 0

8 8r rr rk x k xr r

For the constant term, 2r , for the first expansion; 3r , for the second expansion.

The sum of the constant terms,

38 7 6 1

73! 2

8 7 1

71 8

7 7

0

Hence, there is NO constant term.


(i) Given that 0k ,

8

3

kx

x

Each term of this expansion, 8 8 4

3

8 8r

r r r

r

kT x k x

r rx

The rest will be the same as before.

In Summary:

A routine question, Be competent in handling the term expression, Tr .



5. The curve 25 xy e intersects the coordinates axes at the points A and B .

(i) Given that the line AB passes through the point with coordinates ln 5,k , find the

value of k .

(ii) In order to solve the equation ln 9x x , a graph of a suitable straight line is

drawn on the same set of axes as the graph of 25 xy e . Find the equation of this

straight line.

Solution:

(i)

When 0x , 05 4y e .

When 0y , 25 0xe ,

2 5xe

ln 5

2x

The coordinates of the points A and B are either 0,4 or ln 5

,02

.

Let the equation of the line AB be y mx c .

therefore , 4c .

At ln 5

,02

, ln 5

0 42

m

8

ln 5m

Equation of the line AB , 8

4ln 5

y x

At ln 5,k , 8

ln 5 4ln 5

k

4k

[Analysis]

It will be beneficial to sketch the curve of 25 xy e to as to visualize the problem. Part (i) is

then quite straight forward. Part (ii) may not be clear at first. The trick is to create 25 xe out

of ln 9x x . We need the e term which can be obtained by taking anit-log on the given

equation.



(ii) Given that ln 9x x ,

9xe x

22

9xe x

2 9xe x

2 9xe x

25 4xe x

Therefore, the line to add in the graph of 25 xy e in order to solve the equation ln 9x x ,

is 4y x


(i)

When 0x , 05 4y e .

When 0y , 25 0xe ,

2 5xe

ln 5

2x

The coordinates of the points A and B be 0,4 and ln 5

,02

respectively.

Let ln5,C k , ln5,0D .

90AOB CDB

ln 5

2OB DB

OBA DBC (vertically opposite angle)

AOB CDB , the two triangles are congruent.

4OA DC ,

therefore 4k

In Summary:

It is troublesome to form the equation of the line, before finding k. Part (ii) is

usually flexible. Just ask how to get 25 xe from ln 9x x .

x

y

25 xy e

O

4

ln 5

2

B ln 5

k

A

C

5

D



6.

The diagram shows a point X on a circle and XY is a tangent to the circle. Points A , B

and C lie on the circle such that XA bisects angle YXB and YAC is a straight line. The

line YC and XB intersect at D .

(i) Prove that AX AB .

(ii) Prove that CD bisects angle XCB .

(iii) Prove that triangles CDX and CBA are similar.

Solution:

(i) Given that XA bisects angle YXB ,

YXA BXA

YXA XBA (alternate segment theorem)

Therefore, XBA BXA

Therefore, AXB is an isosceles triangle.

Hence, AX AB

(ii)

XCA XBA ( s in same segment)

BCA BXA ( s in same segment)

[Analysis]

Part (i) is to show that ABX is an isosceles triangle. This is to say that we need to show

ABX BXA . Part (ii) is to show that XCD BCD .

We already know by then from part (ii) that XCD BCD , we will just need to show that

CXD CAB in Part (iii).

A

Y

C

X

D

B



From part (i),

XBA BXA

Therefore, XCA BCA

XCD BCD ( D is on the line YAC )

Therefore, CD bisects angle XCB .

(iii)

Consider CDX and CBA ,

XCD BCD (from part (ii))

CXD CXB CAB ( s in same segment)

therefore, XDC ABC

So, CDX CBA

In Summary:

Before writing any proof, one shall think through the entire proving steps. This

thought process usually required looking at the result, and then work backward until it

meets the forward process, starting from the given conditions.



7.

The diagram shows part of the curve 2 5y x passing through the point P and meeting the

x -axis at the point Q . The line 2x passes through P and intersects the x -axis at the point S .

Lines from Q meet 2x at the points R and T such that QR is parallel to the tangent to the

curve at P , and RS ST . Find

(i) the equation of QR ,

(ii) the area of the shaded region.

Solution:

(i) At Q , 0y .

0 2 5x

2 5 0x

5

2x

Coordinates of point 5

,02

Q

O

2 5y x

x

y

Q

P

R

T

S

2x

[Analysis]

Since the line QR is parallel to the tangent of the curve that passes through point

where , 2x its gradient can be found with dy

dx. Knowing the point Q , we can then find the

equation of QR .

Recognise that QRS is congruent to QTS , the area of the shaded region will be the area

under the curve and the area under the line QR .



Gradient of the curve, 2 5y x , by chain rule,

1

2d 1

2 5 2d 2

yx

x

d 1

d 2 5

y

x x

At P , 2x .

d 1 1

d 32 2 5

y

x

As QR is parallel to the tangent of the curve at P , the gradient of QR is 1

3.

The equation of QR ,

1 5

03 2

y x

1 5

3 6y x

(ii)

As RS ST , QS is common, 90QSR QST , by SAS, QRS is congruent to QTS .

Area of QTS area of QRS .

Area of the shaded region,

2 2

5 5

2 2

52 5d d

3 6

xx x x

2 2

5 5

2 2

1 2 53 2 5d d

3 6 6

xx x x

22 23

2

5 52 2

1 52 5

3 6 6

xx x

3

21 4 10 25 25

9 03 6 6 24 12

14 25

96 24

27

98

3

128

square units




(ii)

Gradient of 1

3

TS RSQT

QS QS

The equation of QT , 5

,02

Q

.

1 5

03 2

y x

1 5

3 6y x

Area of the shaded region,

2 2

5 5

2 2

52 5d d

3 6

xx x x

2 2

5 5

2 2

1 2 53 2 5d d

3 6 6

xx x x

2 2

5 5

2 2

1 2 53 2 5d d

3 6 6

xx x x

22 23

2

5 52 2

1 52 5

3 6 6

xx x

3

21 4 10 25 25

9 03 6 6 24 12

14 25

96 24

27

98

3

128

square units

In Summary:

The alternative solution on part (ii) uses the idea that the area enclosed by the

curve and a line. The same method can also be applied for the area enclosed by two

curves.



8. Two particles, P and Q leave a point O at the same time and travel in the same direction

along the same straight line. Particle P starts with a velocity of 9m/s and moves with a

constant acceleration of 21.5m/s . Particle Q starts from rest and moves with an

acceleration of 2m/sa , where 12

ta and t seconds is the time since leaving O . Find

(i) the velocity of each particle in terms of t ,

(ii) the distance travelled by each particle in terms of t .

Hence find

(iii) the distance from O at which Q collides with P ,

(iv) the speed of each particle at the point of collision.

Solution:

(i)

For particle P , Velocity of P , V . Acceleration of P ,

d 3

d 2

V

t

1

d 3 3d d

d 2 2

VV t t t c

t where 1c is the integration constant.

When 0t , 9V .

1

39 0

2c

1 9c

3

92

V t

[Analysis]

The curve passes through point 2,0 and its gradient is zero at this point too. With two

unknown a and b , these will be enough to solve part (i). The turning points occur at zero

gradient, 0dy

dx . Of course, we already know the minimum point. The area is simply the

integration of the curve from 0 to 2 .



For particle Q , Velocity of Q , V . Acceleration of Q ,

d

1d 2

V t

t

2

2

dd 1 d

d 2 4

V t tV t t t c

t where 2c is the integration constant.

When 0t , 0V .

2

2

00 0

4c

2 0c

2

4

tV t

(ii) When 0t , 3

9 02

V t ; 2

04

tV t .

For particle P , distance travelled of P , S .

dS V t

3

9d2

t t

2

3

39

4

tt c

When 0t , 0S .

3 0c

23

94

tS t

For particle Q , distance travelled of Q , S .

dS V t

2

d4

tt t

2 3

42 12

t tc

When 0t , 0S .

4 0c

2 3

2 12

t tS



(iii)

When Q and P meet, the distance travelled must be the same.

2 3 23

92 12 4

t t tt

2 3 26 9 108t t t t

3 23 108 0t t t

2 3 108 0t t t

9 12 0t t t

0t or 9t or 12t

N.A. N.A.

When 12t , 2 312 12

2162 12

S m

(iv)

When 12t ,

for particle P ,

3

12 9 272

V m/s

for particle Q ,

212

12 484

V m/s

In Summary:

You should read through the entire question, make a plan on how to approach

them before beginning to answer them.



9.

The diagram shows a triangle ABC with vertices at 0,5A , 8,14B and runs ,15C k .

Given that AB BC ,

(i) find the value of k .

A line is drawn from B to meet the x -axis at D such that AD CD .

(ii) Find the equation of BD and the coordinates of D .

(iii) Show that the area of the triangle ABC is 2

7 of the area of the quadrilateral ABCD .

Solution:

(i) Given that AB BC ,

2 2 2 2

0 8 5 14 8 15 14k

2

64 81 8 1k

2

8 144k

8 12k

[Analysis]

For part (i), the AB BC . Part (ii) is to sketch out D and BD to visualize the problem.

Recognise that the two diagonals BD and AC are perpendicular to one another. Part (iii) will

need to find the length ratio.

x

,15C k

0,5 A

8,14B

O

y



20k or 4k , Rejected since 0k

(ii) Let the diagonals BD and AC intersect at M . Quadrilateral ABCD is a kite . Therefore,

the two diagonals are perpendicular to each other.

Gradient AC , 15 5 1

20 0 2

Gradient BD , 1

21

2

The equation of BD ,

14 2 8y x

2 30y x

When 0y ,

0 2 30x

15x


(iii) Quadrilateral ABCD is a kite where AM CM , or M is the mid-point on AC

the coordinates of 0 20 5 15

,2 2

M

.

10,10M

area of the triangle ABC 1

2AC BM

area of the quadrilateral ABCD1

2AC BD

1area of triangle 2

1area of quadrilateral

2

AC BMABC

ABCDAC BD

BM

BD

2 2

2 2

10 8 10 14

15 8 0 14

20

245

x

20,15C

0,5 A

8,14B

O

y

D

M



4

49

2

7


(iii)

area of the triangle ABC

0 20 8 01

5 15 14 52

1

0 280 40 100 1202

50 square units

area of the quadrilateral ABCD

0 15 20 8 01

5 0 15 14 52

1

225 280 40 75 1202

1

545 1952

175 square units

area of triangle 50

area of quadrilateral 175

ABC

ABCD

2

7

In Summary:

The method employed in the part (iii) of the alternative solution is an easier way

to solve this question.



10. (i) Given that

2

22

3 4 20

2 1 42 1 4

x x A Bx C

x xx x

, where A , B and C are constants,

find the value of A and of B and show that 0C .

(ii) Differentiate 2ln 4x with respect to x .

(iii) Using the results from part (i) and (ii), find

2

2

3 4 20d

2 1 4

x xx

x x

.

Solution:

(i)

Given that

2

22

3 4 20

2 1 42 1 4

x x A Bx C

x xx x

,

22

2 2

4 2 13 4 20

2 1 4 2 1 4

A x Bx C xx x

x x x x

2 23 4 20 4 2 1x x A x Bx C x

2 23 4 20 2 2 4x x A B x B C x A C

Equating the coefficients,

2 3A B --------- (1)

2 4B C --------- (2)

4 20A C --------- (3)

2 2 1 , 4 5C A --------- (4)

4 3 4 , 17 85A

5A

Substitute 5A into 1 , 4B

Substitute 5A into 3 , 0C

[Analysis]

Part (i) is a partial fractions with an irreducible quadratic, 2 4x .

Apply chain rule for part (ii). Apply

'fln f

f

xdx x c

x

.



(ii)

Let 2ln 4y x and 2 4u x ,

thus lny u . d 1

du

y

u .

d2

d

ux

x

By chain rule,

d d d

d d d

y y u

x u x .

d 1

2d

yx

x u

2

d 2

d 4

y x

x x

(iii)

2

2

3 4 20d

2 1 4

x xx

x x

2

5 4d

2 1 4

xx

x x

2

5 4d d

2 1 4

xx x

x x

2

5 2 2d 2 d

2 2 1 4

xx x

x x

25ln 2 1 2ln 4

2x x c where c is an integration constant


(i) Given that

2

22

3 4 20

2 1 42 1 4

x x A Bx C

x xx x

,

2 23 4 20 4 2 1x x A x Bx C x

When 1

2x , 5A . 2 23 4 20 5 4 2 1x x x Bx C x

28 4 2 1x x Bx C x

4 2 1 2 1x x Bx C x , Hence 4B , 0C .

In Summary:

A very-much guided question. There are many ways to solve the partial

fractions.



11.

The diagram shows the curves 4cosy x and 2 3siny x for 0 2x radians. The

points A and B are turning points on the curve 2 3siny x and the point C is a turning

point on the curve 4cosy x . The curves intersect at points D and E .

(i) Write down the coordinates of A , B and C .

(ii) Express the equation 4cos 2 3sinx x in the form cos x a k , where and k

are constants to be found.

(iii) Hence find, in radians, the x -coordinate of D and of E .

Solution:

(i) For points A and B ,

2 3siny x

d

3cosd

yx

x

[Analysis]

Recognise that points A , B and C are the min/max locations in each curve. We can apply

differentiation to find the turning points on each curve.

Next, is a R-formula application, where cos cos cos sin sinx a x a x a .

Lastly, the points D and E are the intersections of the two curves.

2 x

y

A

B O

C

D

E

4cosy x

2 3siny x



Let d

0d

y

x ,

0 3cos x

cos 0x

3

,2 2

x

When 2

x

, 2 3sin 2 3 52

y

When 3

2x

,

32 3sin 2 3 1

2y

coordinates of ,52

A

, 3

, 12

B

Similarly,

For point C ,

4cosy x

d

4sind

yx

x

Let d

0d

y

x ,

0 4sin x

sin 0x

0, , 2x

When 0x , 4cos 0 4y , Rejected, not point C

When x , 4cos 4y

When 2x , 4cos 0 4y , Rejected, not point C

coordinates of , 4C

(ii)

Given that 4cos 2 3sinx x ,

4cos 3sin 2x x

4 3 2

cos sin5 5 5

x x

2

cos cos sin sin5

x x

4

3 5



2

cos5

x where 1 3tan

4

(iii) To find D and E , is to solve 2 3sin

4cos

y x

y x

.

4cos 2 3sinx x

From part (ii), we get

2

cos5

x where 1 3tan 0.6435

4

1 2cos

5x

angle x is in Quadrant 1st and 4

th.

Principal angle, 1.1592x

1.1592x or 2 1.1592x

1.1592x 2 1.1592x

0.516x radians 4.48x radians

the x -coordinate of D is 0.516x radians ; and of E is 4.48x radians


(i) For points A and B , given that 2 3siny x ,

1 sin 1x

1 2 3sin 5x

When sin 1x , 2

x

, 5y , ,52

A

.

When sin 1x , 3

or2 2

x

, 1y , 3

, 12

B

.

For point C ,

4cosy x

1 cos 1x

4 cos 4x

When cos 1x , x , 4y , , 4C .



1. Given thatx

yxe

for 0x , find the range of values of x for which y is a decreasing

function of x. [4]

Solution :

Given thatx

yxe

for 0x ,

2

ee

d

d

x

x

x

y xx

2

1e

d

d

x

x

x

y x

Let 0d

d

x

y,

01e

2

x

xx

Because 0e,2 xx for all value of x, therefore

01x

1x

Combine with the valid range of x, we have

10 x

Since 1x , is the turning point of the function, it is at the lowest value. So, this point is to be

included. The final range of the decreasing function is

10 x

[Analysis]

A function is decreasing only if its first derivative is negative, i.e. 0d

d

x

y. To differentiate a

function of the form v

yu

, we apply quotient rule, that is 2

d

d

d

d

d

d

v

x

vu

x

uv

x

y

. If the end

points are turning points, then it must be included in the range. Take note that the function

begins with 0x .




Given thatx

yxe

for 0x ,

2

ee

d

d

x

x

x

y xx

2

1e

d

d

x

x

x

y x

Let 0d

d

x

y,

0

1e2

x

xx

Because 0e,2 xx for all value of x, therefore

01x

1x ,is the turning point of the function.

When 1x , 01x

01e

2

x

xx

When 1x , 01x

01e

2

x

xx

1x is a minimum point.

For 0d

d

x

y, 10 x

Since 1x , is the turning point of the function, it is at the lowest value. So, this point is to be

included. The final range of the decreasing function is

10 x

In Summary:

Students are to be familiar with the monotonicity of basic functions such as

,,log,1

,1

,,,2

32 xaxxx

xxx . Each year, there will be one question on this.



2.

The diagram shows part of a straight line graph drawn to represent the equation bx

ay

,

where a and b are constants. Given that the line passes through (2, 6) and has gradient 3 ,

find the value of a and of b. [4]

Solution :

Given that bx

ay

,

abyyx

abyxy , the line equation

Since the gradient is 3 ,

b3

3b

Sub. (2, 6) into the line equation,

a 236

12a

O

6,2

y

xy

[Analysis]

Observe that the coordinates (2, 6) is for (y, xy). The equation bx

ay

does not match the

line equation in the graph, so it needs to re-write first, xy in terms of y.




Given that the line passes through (2, 6) and has gradient 3 ,

32

6

y

xy

636 yxy

123 yxy

123 xy

3

12

xy

Compare with bx

ay

, we get

12a and 3b

In Summary:

Be very careful with the handling of the coordinate (y, xy). There is little

difficulty with the coordinate geometry and re-writing equation.



3. Without using a calculator, find the value of x6 , given that xx 22 123 . [4]

Solution :

xx 22 123

x

x

12

1239

2

16123 xx

1636 x

1662 x

166 x

46 x (Rejected 46 x as 06 x )


xx 22 123

12lg23lg2 xx

12lg12lg23lg23lg xx

3lg12lg212lg3lg x

4lg236lg x

36lg

4lg2x

6lg

4lgx

4log6x

46 x

[Analysis]

This question is asking for the value of x6 , not the value of x. The first attempt is to transform

the equation into x6 term, by applying laws of indices.

In Summary:

Log and Indices are two of a kind. These topics are a regular set in this exam.

Be very competent with Log and Indices laws.



4. Solve the equation cos542cos2 for 20 , giving your answers, in radians,

correct to 2 decimal places. [5]

Solution :

Given that cos542cos2 ,

cos5422cos4 2

06cos52cos4 2

02cos3cos4 .

03cos4 or 02cos

4

3cos 2cos , Rejected as 1cos

Principal value, 418.2

418.2 or 418.2

418.22

864.3

42.2 (2 d.p.) 86.3 (2 d.p.)


Given that cos542cos2 ,

06cos52cos4 2

02cos3cos4 .

4

3cos 2cos , Rejected as 1cos

Basic angle = 722.0

722.0 or 722.0

418.2 863.3

42.2 (2 d.p.) 86.3 (2 d.p.)

[Analysis]

Changing the double angle with 1cos22cos 2 . Take note that the answer is rounded to

2 decimal places.



5. The variable x and y increase in such a way that, when 2x , the rate of increase of x with

respect to time is twice the rate of increase of y with respect to time. Given that

14 xky , where k is a constant, find the value of k. [5]

Solution:

Given that 14 xky ,

4142

1

d

d2

1

xk

x

y

14

2

d

d

x

k

x

y

When 2x , t

y

t

x

d

d2

d

d

124

2

d

d

k

x

y

3

2

d

d k

x

y

t

x

x

y

t

y

d

d

d

d

d

d

t

xk

t

y

d

d

3

2

d

d

t

yk

t

y

d

d2

3

2

d

d

4

3k

[Analysis]

Rate of change t

x

x

y

t

y

d

d

d

d

d

d . Given that

t

y

t

x

d

d2

d

d , need to find the value of

x

y

d

d when

2x .




When 2x , t

y

t

x

d

d2

d

d

t

x

t

y

d

d

2

1

d

d

Therefore,

2

1

d

d

x

y

Given that 14 xky ,

4142

1

d

d2

1

xk

x

y

14

2

d

d

x

k

x

y

When 2x ,

124

2

d

d

k

x

y

3

2

d

d k

x

y

3

2

2

1 k

4

3k

In Summary:

Rate of change has been featured regular since year 2009. It is usually simple

and direct.



x

x

y

6. A solid rectangular block has a square base of side x cm and a height of y cm. The total

surface area of the rectangular block is 120 cm2 and the total length of the 12 edges is 54 cm.

Show that 02092 xx and find the possible values of x and of y. [5]

Solution:

25448

112042 2

yx

xyx

From (2), we get

34272 xy

From (1), we get

120222 2 yxx

Subst (3) into the above equation,

12042722 2 xxx

0120546 2 xx

02092 xx

054 xx

4x or 5x

When 4x , 5.5y .

When 5x , 5.3y .

In Summary:

This type of question was usually set in E-math paper in the past.

[Analysis]

Sketch a cuboid to assist in formulating the two equations: surface areas and edges.



7. The coefficient of x in the expansion of 6132 axx is 3.

(i) Find the value of the constant a. [4]

(ii) Using your value of a, find the coefficient of x2 in the expansion of 6132 axx .

[2]

Solution:

(i)

26

!2

56611 axaxax

221561 xaax

226156132132 xaaxxaxx

2222 1561315612 xaaxxxaax

22 18303122 xaaxa

For the x term,

3123 a

2

1a

(ii)

For the x2 term, the coefficient is

aa 1830 2

2

18

4

30

2

3

[Analysis]

This is a question can be solved with expansion of the first three terms of 61 ax or by the

general term, r + 1 term expression.




(i) For 61 ax , the general term expression is rr axC6 , where 6,,2,1,0 r

rr

r

r axCxaxCaxx 66

632132 , where 6,,2,1,0 r

For the x term, the coefficient is

3320

06

1

16 aCaC

3312 a

2

1a

(ii)

For the x2 term, the coefficient is

116

2

26 32 aCaC

1263152 aa

aa 1830 2

2

18

4

30

2

3

In Summary:

This question is very straight forward. This topic is a regular visitor in this

exam.



8.

The diagram shows part of the graph of 123 xy .

(i) Find the coordinates of the points A, B and C. [3]

(ii) Solve the equation xx 123 . [3]

Solution:

(i) Given that 123 xy , when 0y

1230 x

123 x

When 012 x , When 012 x ,

123 x 123 x

1x 2x

0,1C 0,2A

When 012 x , 2

1x , 3y

3,

2

1B

A

O

B

x

y

C

[Analysis]

(i) A and C are the x-axis intercepts where 0y . B is the turning point of the linear

modulus where 012 x .

(ii) Need to check for the validity of the solutions.



(ii)

xx 123

123 xx

When 012 x , When 012 x ,

123 xx 123 xx

3

2x 4x


(i) Given that 123 xy , when 0y

1230 x

123 x

2123 x

22 123 x

223120 x

22420 xx

2x or 1x

0,2A 0,1C

When 012 x , 2

1x , 3y

3,

2

1B

(ii)

xx 123

123 xx

22123 xx

012322 xx

0324 xx

4x 3

2x

In Summary:

It is good to know solve linear modulus equation and the related properties in

graph.



9. The equation of a curve is baxy 3 where a and b are constants. The equation of the

normal to the curve at the point where 1x is 1225 xy . Find the value of a and of b.

[6]

Solution:

Given that 1225 xy , when 1x , 2y

12 ba

5

12

5

2 xy

The tangent to the curve at 1x , 2y , is 2

5

d

d

x

y

the curve is baxy 3

23d

dax

x

y

When 1x ,

ax

y3

d

d

a32

5

6

5a

Subst, 6

5a into (1),

b6

52

6

7b

In Summary:

Very simple application of differentiation on normal and tangent.

[Analysis]

This equation 1225 xy is normal to the curve. The tangent on the curve passes through

the same point on the normal and perpendicular to this normal.



10.

The diagram shows a rectangle ABCD in which 322 AB cm. Given that the area of

the rectangle is 39 cm2, find, without using a calculator,

(i) the exact value of BC in the form ba cm, where a and b are integers, [4]

(ii) the exact value of (AC)2 in the form dc cm

2, where c and d are integers. [3]

Solution:

(i) 69322 BC

69322322322 BC

3521538 BC

32355 BC

323 BC

318 BC

(ii) By Pythagoras theorem,

222323322 AC

62322

AC

24322

AC

A B

C D

322 cm

[Analysis]

This is about surd conjugates, simplifying surd. As for part (ii), we may try the Pythagoras

theorem for the right triangle ABC.

In Summary:

Simplifying surds and solving surd equation are a must.



11.

The diagram shows a rhombus ABCD in which the coordinates of the points A and C are

2,2 and 2,2 respectively. Given that the point B lies on the y-axis, find

(i) the coordinates of B and of D. [6]

(ii) the area of the rhombus. [2]

Solution:

(i) Let the mid-point of the two diagonals be M,

2

82,

2

102M

5,4M

The gradient of AC

210

28

2

1

The gradient of BD 2

A 2,2

O

B

x

y

C 8,10

D

[Analysis]

The property of a rhombus is such that the two diagonals are perpendicular to each other. The

two diagonals bisect each other.

A 2,2

O

B

x

y

C 8,10

D

M



Let ByB ,0 ,

240

5

By 13By

13,0B

Let DD yxD , , 5,4M

42

0

Dx 5

2

13

Dy

8Dx 3Dy

3,8 D

(ii) the area of the rhombus,

BDAC2

1

2222 1681262

1

22 2182162

1

5862

1

120 cm2


(ii) the area of the rhombus,

238132

281002

2

1

6641301630262

1

200402

1

120

the area of the rhombus is 120 cm2

In Summary:

A simple routine question on coordinate geometry. Familiarity with

parallelogram is most useful.



12. A particle P leaves a fixed point O and moves in a straight line so that, t s after leaving O, its

displacement, s m, from O is given by

ttts 1ln

Find, when 20t ,

(i) the displacement of P from O, [1]

(ii) the velocity of P, [4]

(iii) the acceleration of P. [4]

Solution:

(i) Given that the displacement ttts 1ln ,

When 20t ,

20120ln20 s

2021ln20 s

890.40s

9.40s m (3 s.f.)

(ii) As velocity, t

sv

d

d ,

11

11ln

d

d

t

tt

t

s

1

11ln

ttv

When 20t ,

21

121ln v

9969.2v

00.3v m-1

s (3 s.f.)

[Analysis]

This question asks very routine questions about derivation from displacement to velocity and

then acceleration.



(iii) the acceleration t

va

d

d

21

1

1

1

d

d

ttt

v

21

2

t

ta

When 20t ,

221

22a

221

22a

04988.0a

0499.0a m-2

s (3 s.f.)

In Summary:

Be familiar with the chain rule, product rule and quotient rule of

differentiation.



13.

The diagram shows a rod OA, which is hinged at O, and a rod AB, which is hinged at A.

The rods can move in the xy-plane with origin O where the x and y axes are horizontal and

vertical respectively. The rod OA can turn about O and is inclined at an angle to the

y-axis, where 900 . The rod AB can turn about A in such a way that its inclination to

the horizontal is also . The lengths of OA and AB are 5 m and 3 m respectively.

Given that B is h m from the x-axis,

(i) find the values of the integers a and b for which

sincos bah . [2]

Using the values of a and b found in part (i),

(ii) express h in the form sinR , where 0R and 900 . [4]

Hence

(iii) state the maximum value of h and find the corresponding value of , [3]

(iv) find the value of when 17h . [2]

Solution:

A

O

B

x

y

3 m

5 m

h m

[Analysis]

Reading the statements in this question maybe a little confusing at first. Even with the aide of

the diagram, it is possible to answer part (i).



(i) Refer to the diagram,

sin3cos5 h

Therefore,

5a , 3b

(ii) cossinsincossin RRR

3cos

5sin

R

R

3

5tan

036.59

3

5tan 1

222 35 R

34R

0.59sin34 h

(iii) h is maximum when 10.59sin .

10.59sin

1sin0.59 1

900.59

0.31

Therefore 34h when 0.31 .

(iv) When 17h ,

0.59sin3417

2

10.59sin

2

1sin0.59 1

Principal value, 450.59

450.59 or 451800.59

14 rejected 76

A

O

B

x

y

3 m

5 m

h m

sin3

cos5



1. The equation of a curve is kkxxy 62 2 , where k is a constant.

(i) Find the range of values of k for which the curve lies completely above the x-axis.

[4]

(ii) In the case where 2k , find the values of m for which the line 4mxy is a

tangent to the curve. [4]

Solution :

(i) The curve is to hang completely above x-axis.

That is kkxx 620 2 has no real root.

It is necessary and sufficient that the discriminant

of the equation kkxx 620 2 is negative.

06242

kk

04882 kk

0412 kk

From the sketch,

412 k

(ii) When 2k , kkxxy 62 2 has no real root.

4

62 2

mxy

kkxxy has repeat roots.

4224 2 xxmx

8220 2 xmx

The discriminant,

824202 m

22820 m

6100 mm

10m or 6m

[Analysis]

This question is on nature of roots, in relation to quadratic equation. A sketch of its graph

would have made the problem a lot clearer. Need to know how to answer quadratic

inequality. Part (ii) is about a curve tangent to a line, or a repeated root where discriminant is

zero.

x

y

O

y

k O

x

y

O 4

4 mxy




(i) Given that the quadratic curve is above x-axis, the minimum value of y must be positive.

ymin is located on the line of symmetry.

Line of symmetry 422

kkx

kk

kk

y

6

442

2

min

68

2

min kk

y

Since ymin > 0,

068

2

kk

04882 kk

0412 kk

From the sketch,

412 k

(ii) Given that the line 4 mxy is tangent to the curve kkxxy 62 2 , when 2k ,

the gradient at the point of tangent must be the same.

24d

d x

x

y , must be the same as the gradient of the line, m.

24 xm

422424 2 xxxx

042 x

2x or 2x

When 2x , 6224 m

When 2x , 10224 m

In Summary:

The topic on Quadratics remains the most important content to be mastered.

k O

x

y

O

ymin



2. The function f is given by bxax tan:f , where a and b are positive integers and

22

x .

(i) Given that 0f x when 2

x , find the smallest possible value of b. [1]

(ii) Using the value of b found in part (i) and given that the gradient of the graph of xy f

is 12 at the point where 8

x , find the value of a. [3]

(iii) Sketch the graph of xy f . [3]

Solution :

(i) Refer to the sketch, the period of bxay tan is 2

.

2

b

2b

(ii) When 8

x , 12

d

d

x

y

xax

y2sec2

d

d 2

82sec212 2

a

4sec6 2

a

4cos6 2

a 2

1

4cos

3a

(iii) as shown above.

[Analysis]

Tangent curve might not be familiar to many students. A sketch of the curve would be useful.

Part (i), the period of tan x is 180 . Part (ii) is about differentiation.

4

O

y

x

4

2

2



3. The expression baxx 32 , where a and b are constants, has a factor of 2x and leaves a

remainder of 35 when divided by 3x .

(i) Find the value of a and of b. [4]

(ii) Using the values of a and b found in part (i), solve the equation 02 3 baxx ,

expressing non-integer roots in the form 2

dc , where c and d are integers. [4]

Solution :

(i) Let baxxx 32p

By Factor Theorem, 02p ,

ba 2222p3

12160 ba

By Remainder Theorem, 353p ,

ba 3323p3

ba 35435

23190 ba

12 ,

7a

Sub. 7a into (1),

2b

272p 3 xxx

(ii) 0272 3 xx

By long division,

1422272 23 xxxxx

01422 2 xxx

2x or

2

22

22

124164

x

[Analysis]

This is a typical Factor and Remainder Theorems question. Part (ii) simply asks for

application of quadratic formula or completing square after factorised x+2.



4. The roots of the quadratic equation 0542 2 xx are and .

(i) Show that 5

222

. [4]

(ii) Given that the roots of 02 baxx are 2

and 2

, find the value of a and

of b, where a and b are constants. [4]

Solution :

(i) Given that 0542 2 xx ,

22

4

2

5

12

5242

222

5

2

2

5

122

(ii)

5

184

5

2422

22

5

212524122

22

05

21

5

182 xx

[Analysis]

Recognize that this is a Sum-of-roots, Product-of-roots question, a

b and

a

c of

a quadratic equation 02 cbxax .

In Summary:

A routine question, Be competent in handling the algebraic indentity.



5. (a) Solve the equation xx 273 log4log .

(b) The curve naxy , where a and n are constants, passes through the points (2, 40),

(3, 135) and (4, k). Find the values of n, a and k.

Solution:

(a) Given

xx 273 log4log

27log

log4log

3

33

xx

3

3

333

3log

log3log4log

xx

3log3

log3loglog

3

34

33

xx

3

log3loglog 34

33

xx

4

33

3 3log3

loglog

xx

4

33 3loglog3

2x

4

33 3log2

3log x

2

34

33 3loglog

x

6

33 3loglog x

63x

729x

[Analysis]

Part (a) requires base change formula. Part (b) simply find the values of the unknowns.



(b)

When 2x , 40y , na240

When 3x , 135y , na3135

n

n

3

2

135

40

n

3

2

27

8

n

3

2

3

23

3

n

3

2

3

23

3n

therefore 3240 a

5a

When 4x , ky , 345k

320k

In Summary:

Application of Log and Indices laws are key elements in this paper.



6.

A garden is being designed to include a semicircular pond and a lawn. The radius of the pond

is r m and the length of the lawn, which is in the shape of a rectangle with the semicircle

removed, is l m as shown in the diagram above.

(i) Given that the area of the lawn is 400 m2, express l in terms of r. [2]

(ii) Given that the perimeter of the lawn is P m, show that r

rP400

22

3

. [2]

(iii) Given that r and l can vary, find the value of r for which P has a stationary value and

determine whether this value of P is a maximum or a minimum. [5]

Solution:

(i) Area of the lawn = 400 m2,

4002

12 2 rrl

rr

l 4

1

2

400

2r m

l m

r m

POND

LAWN

[Analysis]

Part (i) is to form an equation with the area. Part (ii) is to form an equation with the perimeter,

and then replace l from part (i) to obtain the desired expression. Part (iii) is to differentiate P,

then set it to zero to locate the turning points.



(ii)

rlrP 22

rrr

rP

42

40022

r

rP400

22

3

(iii)

2

4002

2

3

d

d

rr

P

When 0d

d

r

P,

2

4002

2

30

r

2

434002

r

43

8002

r

43

800

r take 0r only.

72.7r m (3 s.f.)

0800

d

d32

2

rr

P the turning point is a minimum.



7.

The diagram shows two intersecting circles C1 and C2, with centres P and Q respectively.

The point R lies on both circles and the line PR is a tangent to C2. A line L passes through Q.

The point E lies on C2. The line PE is perpendicular to L and intersects C2 at A, C1 at B and

L at D.

(i) By considering triangles QAD and QED show that AD = ED. [4]

(ii) Show that PEPAADPD 22 . [3]

(iii) Hence show that 222 PBADPD . [2]

Solution:

(i) Consider QDA and QDE

EQAQ (radius of circle C2)

QDQD (common side)

90QDEQDA

Therefore, QDA QDE (RHS)

Therefore, AD = ED

A

D

L E

B

C1

R

P

Q

C2

[Analysis]

Part (i) requires proof of congruent triangles. Part (ii) applies algebraic identity and finally part

(iii) uses tangent-secant theorem.



(ii)

22 ADPD

ADPDADPD

PADEPD DEAD , from part (i)

PAPE

(iii)

By Tangent-Secant Theorem,

2PRPAPE

222 PRADPD PEPAADPD 22

222 PBADPD PBPR , radius of circle C1

In Summary:

This geometrical proof is the easiest since 2009.



8. (i) Express 22

12

138

xx

xx in partial fractions. [5]

(ii) Hence find

x

xx

xxd

12

1382

2

. [5]

Solution:

(i) Let

22

2

121212

138

x

C

x

B

x

A

xx

xx

CxxxBxAxx 121213822

When 0x ,

A1

CxxxBxxx 121213822

CxxxBxxx 121213822

CxxxBxx 124 2

When 1x ,

)1(33 CB

When 1x ,

)2(5 CB

(1)+(2),

B28

B4

9C

22

2

12

9

12

41

12

138

xxxxx

xx

[Analysis]

First recongise that the numerator is of degree 2 and the denominator is of degree 3. Therefore

this is a proper fraction. Again, observe that 2x + 1 is a repeated factor. Apply

'fln f

f

xdx x c

x

in part (ii)



(ii)

x

xx

xxd

12

1382

2

xxxx

d12

9

12

412

cx

xx

12

1

2

912ln2ln

In Summary:

This s a typical question that combining partial fraction with integration.



9.

The diagram shows part of the curve 2

sin3x

y . The line 3

4x meets the curve at P and

the x-axis at Q. The tangent to the curve at P meets the x-axis at R. Find

(i) the length of QR.

(ii) the area of the shaded region.

Solution:

(i)

3

2sin3,

3

4 P for

2

3

3sin

3

2sin

2

33,

3

4P

x

P

Q R O

y 3

4x

2sin3

xy

[Analysis]

For part (i), PQ can be found by simply substituting 3

4x into

2sin3

xy . Then by

finding the gradient with 2

cos2

3

d

d x

x

y , then QR is found with similar triangles. Part (ii) is to

find the area under the curve from 0x to 3

4x by integration, then combines with the

area of triangle PQR.



2cos

2

3

d

d x

x

y

When 3

4x ,

3

2cos

2

3

d

d

x

y

4

3

d

d

x

y for

2

1

3

2cos

Therefore,

4

3

QR

PQ

PQQR3

4

32QR units for 2

33PQ

(ii) the area of the shaded region = area under the curve from 0x to 3

4x + area of triangle

PQR.

the area of the shaded region

QRPQxx

2

1d

2sin3

3

4

0

323

4

2

1d

2sin

2

16

3

4

0

xx

3

34

2cos6

3

4

0

x

3

340cos

3

2cos6

3

341

2

16

3.163

349

unit

2 (3 s.f.)



10. Without using a calculator

(i) find the exact value of 75cos15cos , [3]

(ii) find the exact value of 75cos15cos , [2]

(iii) show that 2

375cos15cos 22 , [2]

(iv) state the acute angle such that 75cossin , [1]

(v) use the results of parts (iii) and (iv) to find the exact value of 15cos2 . [3]

Solution:

(i)

Given that 75cos15cos

30sin45sin2

30sin45sin2 for xx sinsin

2

1

2

12

2

1

2

175cos15cos

(ii)

Given that 75cos15cos

30cos45cos2

30cos45cos2 for xx coscos

2

3

2

12

[Analysis]

This question examines trigo identities and ratios of special angles.



2

3

2

3

(iii)

Given that 75cos15cos 22

75cos15cos75cos15cos

2

1

2

3

2

3

(iv)

Given that 75cossin

15sinsin

15 or 165 (rejected)

(v)

75cos2

315cos 22

275cos2

3

215sin2

3

2

315sin15cos 22

2

3115cos2 2 for xx 22 cos1sin

2

1

4

315cos2

4

2315cos2



11. A circle, centre C, has a diameter AB where A is the point (1, 1) and B is the point (7, 9).

(i) Find the coordinates of C and the radius of the circle. [4]

(ii) Find the equation of the circle. [2]

(iii) Show that the equation of the tangent to the circle at B is 5734 xy . [3]

The lowest point on the circle is D.

(iv) Explain why D lies on the x-axis. [1]

(v) Find the coordinates of the point at which the tangents to the circle at B and D intersect.

[1]

Solution:

(i)

5,42

91,

2

71CC

coordinates of 5,4C

(ii)

106436191722

AB

255422 yx

(iii)

Gradient of AB, 3

4

17

19

ABm

Gradient perpendicular of AB, 4

3ABm

[Analysis]

Part (i) makes use of midpoint formula; part (ii) first find the length AB; part (iii) is to first find

the gradient AB then forming the equation of tangent; Part (iv) is to related the centre of the

circle, its radius and x-axis; Part (v) is to set 0y .



The equation of the tangent at B,

74

39 xy

5734 xy

(iv)

For point 5,4C , and radius of 5 units, the circle touches x-axis. So, the lowest point, D on

the circle is also tangent to the x-axis.

(v)

The tangent of the circle at D is the x-axis ,i.e. 0y .

When 0y ,

57304 x

19x

coordinates of the point of intersection 0,19



1. The equation of a curve is 3

12

x

xy . Given that x is changing at a constant rate of

0.4 units per second, find the rate of change of y when 2x . [4]

Solution :

Given that3

12

x

xy for 3x ,

23

1223

d

d

x

xx

x

y

23

5

d

d

xx

y

Let 2x , 2.0d

d

x

y

08.04.02.0d

d

t

y units per second


Given that3

12

x

xy for 3x ,

3

532

3

562

x

x

x

xy

3

52

xy

[Analysis]

Recognise that the question is about rate of change, i.e. t

x

x

y

t

y

d

d

d

d

d

d . Given that 4.0

d

d

t

x, to find

t

y

d

d at 2x . To differentiate a function of the form

vy

u , we apply quotient rule, that is

2

d

d

d

d

d

d

v

x

vu

x

uv

x

y

. Take note that the function implies that 3x .



23

5

d

d

xx

y

Let 2x , 2.0d

d

x

y

08.04.02.0d

d

t

y units per second

In Summary:

Simple rate of change question. The alternative solution by-passes the quotient

rule, after reducing the improper fraction to proper.



2.

The diagram shows a triangle ABC in which angle A is 6

radians, angle B is a right angle,

M is the mid-point of AB and the length of AC is 8 m. Without using a calculator, find the

value of the integer k such that angle

14sin 1 k

ACM . [5]

Solution :

Since ABC is a 30°-60°-90° triangle, by similar triangles, we have

ABCABC

321 .

Given that 8CA m, then, 4BC m and 34AB m.

Since M is midpoint on AB, then 32 MBAM m.

By Pythagoras theorem on triangle MBC, we get

2843222

MC m.

[Analysis]

Observe that the triangle ABC is a 30°-60°-90°. 82

130sin

AB . Rewrite

angle

14sin 1 k

ACM as 14

sink

ACM . With right-angled triangle, sine ratio or sine rule are

likely to be productive. 0k , since k . Take note that “Without using a calculator” in the

question which implies approximated values from calculator are inadmissible.

M A B

C

8 m

6

rad



Applying Sine rule on triangle ACM, we get

AM

ACM

CM

sin6sin

AMCM

ACM

6

sin

sin

28

332

28

2

1

sin ACM

142

3

14

k

21k

21k


Since ABC is a 30°-60°-90° triangle, , we have

2

1

86sin

BC and

2

3

86cos

AB

Therefore, 4BC m and 34AB m.

Since M is midpoint on AB, then 32 MBAM m.

By Pythagoras theorem on triangle MBC, we get

2843222

MC m.

MCBACM 3

MCBACM

3sinsin

MCBMCBk

sin

3coscos

3sin

14



Since 2

3

3sin

,

2

1

3cos

,

28

32sin

MC

MBMCB ,

28

4cos

MC

BCMCB , we get

28

32

2

1

28

4

2

3

14

k

28

3

28

32

14

k

28

3

14

k

142

3

14

k

21k

21k

In Summary:

Once a student is able to recognize the special 30°-60°-90° triangle and able to

recall the trigo ratios of this triangle, the question becomes very easy.

In the alternate solution, we can make use of the trigo identity of sin (A – B) to find the

answer.



3. (i) Find xxx

lnd

d 2 . [2]

(ii) Hence find xxx dln [3]

Solution :

(i) By product rule,

xxxxxx

x

1ln2ln

d

d 22

xxx ln2

(ii) From part (i), we have

xxxxxx

ln2lnd

d 2

xxxxxxxx

dln2dlnd

d 2

xxxxxxx ddln2ln2

cx

xxxxx 2dln2ln

22

cx

xxxxx 2lndln2

22

Cxxx

xxx 42

lndln

22

, where C is an arbitrary constant.

[Analysis]

This question is essentially the same as question 4 in Paper 1, 2008, (4018). Part (i) requires

product rule to differentiate. Part (ii) is to make use of the result of part (i) to integrate.

In Summary:

Very simple, direct question. Log and Exponential operations are very

important topics in this paper. These topics are a regular set in this exam. Be very

competent with Log and Indices laws.



4. Find the inverse of the matrix

1753 and hence solve the simultaneous equations

.0235,0117

baba [5]

Solution :

Given that

1753M , the determinant, 38353

1753

3751

38

11M

Given that

02350117

baba ,

117253

abab

112

1753

ab

112

3751

38

1ab

1957

38

1ab

2

12

3

ab

5.1b or 5.0a

[Analysis]

There are two parts to this question, firstly is to find the inverse matrix of M, M–1

. M × M–1

= I,

where I is the identity matrix. Need to find determinant first before formulating the M–1

. Secondly

is to apply the inverse matrix to solve for the value of a and of b.




Given that

1753M , let

srqp1M

M × M–1

= I

1001

1753

srqp

153 rp , 053 sq ,

07 rp , 17 sq

0535

153rp

rp

5535053sq

sq

138 p 38

5q

38

1p

38

3r

38

7r

38

3

38

738

5

38

1

M 1-

In Summary:

The first time a Matrix question is set on a 4038 paper. Students may not be

familiar with the way to find the inverse Matrix. The alternative should be a useful

one.



5. In a chemical reaction, a solid substance slowly dissolves in a liquid. At the start of the

reaction, there are M grams of the substance. After t minutes, there are x grams of the

substance still to be dissolved. It is known that x and t are related by the equation ktMx e ,

where k is a constant. The table below shows measured values of x and t.

t (minutes) 1 2 3 4 5

x (grams) 6.55 5.36 4.40 3.60 2.94

(i) Plot xln against t and draw a straight line graph. [2]

(ii) Us your graph to estimate the value of M. [2]

(iii) Estimate the time taken for half of the substance to be dissolved. [2]

Solution:

(i) Given that ktMx e ,

ktMx elnln

ktMx elnlnln

ktMx lnln

[Analysis]

Wow, it is just like in an exercise. Put you practice in full benefit to pick up the 6 marks. Part (i)

hints that taking xln will be the place to start.

1.2

1.4

1.6

1.8

2.0

t

1

3 4 5

•

•

•

•

•

O 1 2

lnx



t (minutes) 1 2 3 4 5

x (grams) 6.55 5.36 4.40 3.60 2.94

lnx 1.879 1.679 1.482 1.281 1.078

(ii) From the graph, the lnx-axis intercept is 2.08.

08.2ln M

08.2eM

00.8M grams

(iii) 15

88.108.1

k

4

8.0k

2.0k

tx 2.008.2ln

2

Mx

ktMM

ln2

ln

ktMM ln2lnln

47.32.0

2lnt min

In Summary:

Direct copy from textbook.



6. (i) Express tansec2sin2 as a quadratic expression in sin . [3]

(ii) Use your answer to part (i) to find, for 20 , the exact solutions of the equation

03tansec2sin2 [3]

Solution:

(i) Given that,

tansec2sin2

cos

sin

cos

1cossin22

x

cos

sin1cossin22

2sin4sin4

(ii) Given that 03tansec2sin2

03sin4sin4 2

03sin4sin4 2

01sin23sin2

2

3sin or

2

1sin

(rejected) 6

11

62

,

6

7

6

In Summary:

Be careful that the question asks for EXACT solutions. Another very direct

question.

[Analysis]

Part (i) requires a quadratic expression in sin , so apply double angle to 2sin and rewriting

sec and tan in sin and cos .

Part (ii) is to use part (i) to solve the given equation. 1sin1 and 20 .



7. In the expansion of 62 pxqx , the coefficient of 6x is 7 and there is no term in 5x .

Given that 0p , find the values of the constants p and q. [6]

Solution:

2415066

26

16

06 pxpxpxpx

4256 156 xppxx

4256615622 xppxxqxpxqx

42564256 1561562 xppxxqxppxxx

5267 630122 xpqpxqpx

For the x6 term,

qp 127 …….. (1)

For the x5 term,

pqp 6300 2

qpp 560

0p , qp 50 …….. (2)

(1) – (2), p77

1p

Subst 1p into (1),

5q


For 6px , the general term expression is rr pxr

66

, where 6,,2,1,0 r

rrrr pxr

qpxr

xpxqx

666 6622 , where 6,,2,1,0 r

[Analysis]

For all binomial expansion, extreme care is usually in order, as an expansion can be messy.



rrrr pxr

qpxr

pxqx

676 6622 , where 6,,2,1,0 r

For the 6x term, the coefficient is

706

162 01

pqp

712 qp …….. (1)

For the 5x term, the coefficient is

016

262 12

pqp

06152 2 qpp

056 qpp

0p , qp 50 …….. (2)

(1) – (2), p77

1p

Subst 1p into (1),

5q

In Summary:

An usual suspect in this paper every year. Be careful in handling the

expansion.



8. The function f is defined, for all values of x, by

cx

x

2cos2f ,

where c is a constant. The graph of xy f passes through the point

2,

3

2.

(i) Find the value of c. [2]

(ii) State the amplitude and period of f. [2]

(iii) Sketch the graph of xy f for 20 x . [2]

Solution:

(i) Given that cx

y

2cos2 , when

3

2x , 2y

c

3

2

2

1cos22

c

3cos22

Since 2

1

3cos

,

c

2

122

1c

[Analysis]

(i) Substitute the coordinate

2,

3

2 into the function to determine c.

(ii) Need to state Amplitude, a, which is the coefficient of the cosine and period as 2π/b

where bxacos .

(iii) Remember the domain, 20 x , end points, intercepts and shape.



(ii)

Amplitude = 2 units

Period = 422

(iii)

In Summary:

A direct curve sketching question.

O

3

x

y

1.16π

–1

2π π



9. A particle moves in a straight line, so that, t seconds after leaving a fixed point O, its

velocity v ms –1

, is given by

31

122

tv . Find

(i) an expression for the acceleration of the particle in terms of t, [2]

(ii) the distance travelled by the particle before it comes to instantaneous rest. [5]

Solution:

(i)

Given that

31

122

tv ,

31122

d

d t

t

v

21

24

d

d

tt

v

(ii) When v = 0 ,

31

120

2

t

21

41

t

412t

21 t

1t or 3t (rejected)

Distance = 1

0

dtv

[Analysis]

(i) Acceleration = t

v

d

d.

(ii) Distance = tv d .



dtt

1

0

23

1

12

1

0

1

0

2d3d

1

12tt

t

1

0

1

0

2d3d

1

112 tt

t

013

0

1

1

112 t

t

01312

112

36

3 m

In Summary:

Very simple application of differentiation and integration.



10. A curve has the equation cxxxy 243 23 , where c is a constant.

(i) Find the x-coordinates of the two stationary points on the curve. [4]

(ii) Hence find the value of c for which the minimum point of the curve lies on the x-axis.

[3]

Solution:

(i) Given that cxxxy 243 23 ,

2463d

d 2 xxx

y

0d

d

x

y, 24630 2 xx

820 2 xx

024 xx

4x or 2x

(ii) To determine the nature of stationary points,

66d

d2

2

xx

y

when 4x , 0d

d2

2

x

y, minimum point.

when 2x , 0d

d2

2

x

y, maximum point.

So, when 4x , 0y

c 424434023

80c

[Analysis]

(i) stationary points are points with zero gradient, i.e. 0d

d

x

y.

(ii) use the result in part (i) to find c when the minimum point touches the x-axis.




(ii) A sketch of the curve cxxxy 243 23 ,

So, when 4x , 0y

c 424434023

80c


(ii) A sketch of the curve cxxxy 243 23 ,

4x is a repeated root. Let the other root be k.

kxxcxxx 223 4243

kxkxkxcxxx 168168243 2323

5k

80165 c

In Summary:

A straight forward question on stationary points.

O

c

x

y

–2 4

O

c

x

y

–2 4 k



11. Given that the roots of 052162 xx are 3 and 3 , find a quadratic equation

whose roots are and . [7]

Solution:

Let and being the roots of quadratic equation 02 cbxax , then a

b and

a

c where 0a .

1633 5233

164 523103 22

4 52103 22

4a

b 521023

2

ab 4 5210243 2

44

1

1a

c

ac

therefore, 02 cbxax

042 aaxax

0142 xx

In Summary:

Although the question is set in reverse order to the exercises, the approach in

solving it is no different from these exercises.

[Analysis]

Usually this type of question is that given and being the roots of quadratic equation

02 cbxax , then a

b and

a

c where 0a .



12. Without using a calculator, find

(i) the value of x such that x

x

27

139

3 , [4]

(ii) the values of the integers a and b such that 32

325

325

3

ba . [4]

Solution:

(i) Given that x

x

27

139

3 ,

xx 2733232

xx 3223 33

xx 346 33

xx 346

9

4x

(ii) 32

325

325

3

ba,

32

3332

325

3

ba

3232

32331

325

3

ba

34

3332361

325

3

ba

331325

3

ba

325343 ba

35386203 ba

33143 ba

14a and 3b

[Analysis]

This question asks indices laws and surds.



13. A circle, whose equation is 03758622 yxyx , has centre C and radius r.

(i) Find the coordinates of C and the value of r. [3]

The point 12,9P lies on the circle.

(ii) Show that CP passes through the origin O. [2]

(iii) Find the equation of the circle for which OP is a diameter. [3]

Solution:

(i) Given the equation of circle, 03758622 yxyx

The centre of this circle, 4,3 C

The radius of this circle, 2043375 22 r units

(ii) 12,9P , 4,3 C

39

412

3

4

x

y

12

16

3

4

x

y

3

4

3

4

x

y

124123 xy

xy3

4 , equation of CP

The line CP intercept the y-axis at the origin.

(iii) The mid-point of OP,

6,

2

9

2

12,

2

9. length OP

2

156

2

9 2

2

2

2

2

2

156

2

9

yx

[Analysis]

This question asks very routine questions about circle.

In Summary:

No surprises. Routine question.



1. The equation of a curve is xxy cos3sin2 for x0 .

(i) Write down expression for x

y

d

d and

2

2

d

d

x

y. [3]

(ii) Find the value of x for which the curve has a stationary point. [2]

(iii) Determine the nature of this stationary point. [2]

Solution :

(i) Given that xxy cos3sin2 ,

xxx

ysin3cos2

d

d

xxx

ycos3sin2

d

d2

2

(ii) When 0d

d

x

y,

xx sin3cos20

3

2tan x

55.2x rad or 70.5x rad (rejected)

(iii) When 55.2x rad,

55.2cos355.2sin2d

d2

2

x

y < 0, hence a maximum point.

[Analysis]

This question is on differentiation and nature of turning points.




(i) Given that xxy cos3sin2 ,

xxy cos

13

3sin

13

213

sincoscossin13 xxy where 2

3tan

xy sin13 983.0 rad

xx

ycos13

d

d

xx

ysin13

d

d2

2

(ii) When 0d

d

x

y,

xcos130

2

x or

2

3 x

55.2x rad 70.5x rad (rejected)

(iii) When 55.2x rad,

2sin13

d

d2

2

x

y < 0, hence a maximum point.

In Summary:

Be very careful in handling the radian. The actual value of 2nd

derivative is not

as important, as we just need to know the sign of it only.



2. (i) Given that xBxAxx coscoscos3cos , state the value of A and of B. [1]

(ii) Hence, or otherwise, solve the equation 0cos23cos xx for 1800 x . [5]

Solution :

(i)

2

3cos

2

3cos2cos3cos

xxxxxx

xxxx cos2cos2cos3cos

Hence, 2A , 2B

(ii) Given that 0cos23cos xx for 1800 x , so 36020 x

0coscos2cos2 xxx

012cos2cos xx

12cos2 x or 0cos x

2

12cos x 90x

principle angle, 1202x

240,1202x

120,60x


(i) xxxxxxx cossin2sincos2coscos3cos

xxxxx coscossin2cos2cos 2

xxx 2sin212coscos

xxcos2cos2

[Analysis]

Part (i) apply the sum of angle identity. Part (ii) is about solving a trigo equation.

In Summary:

Be very careful in handling equation in part (ii). Take note of the double angle

solution, because the range of 2x is larger.



3. (i) Express 383

202 xx

in partial fractions. [4]

(ii) Find x

xxd

383

202

and hence evaluate xxx

d383

207

22

. [4]

Solution :

(i) Since 313383 2 xxxx ,

let 313383

12

x

B

x

A

xx

when 3x , 10

1

10

1

B ,by cover-up method

when 3

1x ,

10

3

33

1

1

A

3

10

1

13

10

3

383

12

xxxx

3

2

13

6

383

202

xxxx

(ii) x

xxd

383

202

x

xxd

3

2

13

6

x

xx

xd

3

12d

13

32

cxx 3ln213ln2 where c is an integrating constant

[Analysis]

Part (i) is a simple 2 parts decomposition. Part (ii) will need integration into ln function.



xxx

d383

207

22

273ln213ln2 cxx

cc 32ln216ln237ln2121ln2

cc 5ln25ln210ln220ln2

2ln2

39.1 (3 s.f.)

In Summary:

Nothing complicated in this question, except the integration into ln function.

Students are reminded that exponential and log functions are VERY important.



4.

Two circles C1 and C2, intersect at P and Q as shown in the diagram. The tangent to C1 at P

meets C2 at R and the tangent to C2 at P meets C1 at S. Prove that

(i) triangle PQR and SQP are similar, [3]

(ii) Show that 2QPQRQS . [2]

Points A and B lie on the circumferences of C1 and C2 respectively,

(iii) Prove that angle SAP = angle PBR. [3]

Solution :

(i) Consider PQR and SQP

PSQRPQ (Alternate Segment Theorem)

QPSQRP (Alternate Segment Theorem)

Therefore, PQR SQP (AA)

(ii) Since PQR SQP ,

PS

RP

QP

QR

SQ

PQ .

QP

QR

SQ

PQ

QRSQQPPQ 2QPQRQS

[Analysis]

Part (i) needs to find 2 congruent angles in both triangles. Part (ii) can be arranged into length ratio.

Part (iii) may need properties of cyclic quadrilaterals.

S

A

B

C1

R

P

Q

C2



(iii)

SAPQ and PBRQ are cyclic quadrilaterals.

180SQPSAP 180PQRPBR

PQRPBRSQPSAP

Since PQRSQP , from part (i) SQP PQR , therefore

PBRSAP

In Summary:

Need a keen sense of observation of the given diagram. When part (i) is solved,

the rest of the question becomes straight forward.



5. The equation of a curve is xx BAy ee2 , where A and B are constants. The point P(0, 4)

lies on the curve and the gradient of the tangent to the curve at P is 1 .

(i) Find the value of A and of B. [4]

(ii) Using your values of A and B from part (i), obtain an expression for xy d and hence

evaluate xy d1

0 corrected to one decimal place. [4]

Solution:

(i) Given xx BAy ee2 ,

At P(0, 4),

BA4 --------- (1)

xx BAx

y ee2d

d 2

At P(0, 4), 1d

d

x

y

BA 21 --------- (2)

(1) + (2),

1A

3B

(ii)

xxy e3e2

xy d

xxx de3e2

xx xx de3de2

[Analysis]

Part (i) requires differentiation. Part (ii) simply find the integration and the value of the definite

integration.



xx xx de3d2e

2

1 2

Cxx e3e2

1 2 where C is an arbitrary integrating constant

therefore xy d1

0

CC 0012 e3e

2

1e3e

2

1

CC 0012 e3e

2

1e3e

2

1

32

1e3e

2

1 12

1.5 (1 d.p.)

In Summary:

Application of differentiation and integration of exponential functions.



6. (a) (i) Given that ux 4

3

8 log4log , express u in terms of x. [3]

(ii) Find the value of x for which 4log

1log5log

3

3

8

2

4 xxx . [3]

(b) Solve the equation 152ee yy . [3]

Solution:

(a)

(i) Given that ux 4

3

8 log4log ,

2

2

2

3

2

3

2

2log

log4

2log

log ux

2

log4log 2

2

ux

ux 222 log2log8log2

ux 2

8

2

2

2 log2loglog

ux

2

2

2 log256

log

256

2xu

(ii) Given that 4log

1log5log

3

3

8

2

4 xxx where 0x and ( 0x or 5x )

4log

3log

2log

log

2log

5log

2

2

3

2

3

2

2

2

2

2 xxx

2

3loglog

2

5log 22

2

2

xxx

3loglog5log 2

2

2

2

2 xxx

3log5

log 22

2

2

x

xx

35

2

2

x

xx

22 35 xxx

[Analysis]

(a) Part (i) requires base change formula. Part (ii) simply find the values of the log equation.

(b) to solve a routine exponential equation.



052 2 xx

0x or 5.2x

(rejected)

(b) Given that 152ee yy ,

015e2e2 yy

03e5e yy

3e y or 5e y

(rejected) 61.15ln y (3 s.f.)

In Summary:

Part (a) requires application of log rules, part (b) is just another exercise.



7. (i) Find the value of a and of b for which 232 2 xx is a factor of

bxxaxx 234 32 . [6]

(ii) Using the value of a and b found in part (i), solve the equation

032 234 bxxaxx . [3]

Solution:

(i) Given that 212232 2 xxxx , let bxxaxxx 234 32f .

When 2x , 02f .

ba 2223222f234

ba 280 --------- (1)

When 2

1x , 0

2

1f

.

ba

2

1

2

1

2

13

2

12

2

1f

234

ba 4

3

2

10 --------- (2)

(1) – (2),

a4

5

2

150

a 60

6a

b 6280

4b

[Analysis]

212232 2 xxxx , apply factor theorem to find a and b. Part (ii) factorizes completely to

find the roots.



(ii) Given that 046632 234 xxxx

223246632 22234 kxxxxxxxx

When 1x ,

2123246632 k

133 k

0k

223246632 22234 xxxxxxx

02232 22 xxx

2x or 2

1x or 22 x

or 2x or 2x

In Summary:

Without fail, factor theorem and remainder theorem are regulars in this paper.



8. (i) Find the set of values of m for which the curve 22 xxy and the line 1 mxy do

not intersect. [4]

(ii) Sketch the curve 22 xxy , giving the coordinates of the maximum point and of the

points where the curve meets the x-axis. [3]

(iii) Find the number of solutions of the equation 12 2 mxxx when

(a) 1m , (b) 2

1m [4]

Solution:

(i) Given 22 xxy and 1 mxy ,

221 xxmx

0122 xmx

Discriminant of the equation is to be negative,

0422

m

422m

222 m

40 m

The set of values 40: mm

(ii) Given 22 xxy , when 0y ,

220 xx

xx 20

0x or 2x

Line of symmetry, 12

02

x

The locally maximum value, 112 y

The coordinate of the maximum point is (1, 1)

[Analysis]

(i) Solving simultaneous equations with negative discriminant.

(ii) Sketch the graph by flipping up the negative part to the upper half.

(iii) Add lines to the sketch in part (ii) to determine the number of solutions.



(a) 1m , 1 xy (b) 2

1m , 1

2

1 xy

number of solutions = 2 number of solutions = 3

In Summary:

This is a typical curve sketching question. Remember to label your curve and

lines clearly.

x

(1, 1)

2 O

y 1x

22 xxy

–1

1 xy

12

1 xy



9.

In the diagram ABC is a structure consisting of a rod AB of length 10 cm attached at B to a

rod BC of length 24 cm so that angle ABC = 90°. Small rings at A and B enable A to move

along a vertical wire Oy and B to move along a horizontal wire Ox. Angle OAB = Ɵ and can

vary. The horizontal distance of C from the vertical wire Oy is L cm.

(i) Explain clearly why cos24sin10 L . [2]

(ii) Express L in the form cosR , where R > 0 and 900 . [4]

(iii) Find the greatest possible value of L and the value of at which this occurs. [3]

(iv) Find the value of for which L = 20. [2]

Solution:

(i) sin10OB cm CBx , cos24xBC

cos24sin10 xBCOBL

(ii) sinsincoscoscos RRR

sinsincoscoscos24sin10 RRL

24cos R and 10sin R

12

5

24

10tan

6.22

12

5tan 1

[Analysis]

Part (i) is to formulate the projected horizontal length. Part (ii) is to express it into cosR .

Part (iii) is to find the maximum value of L. Part (iv) is simply solving the trigo equation.

O x

y

B

A

C

10 cm

24 cm

O x

y

B

A

C

10 cm

24 cm

Cx



22222410sincos RR

26R

cos26L where

12

5tan 1

(iii) When 1cos , L is at its maximum.

The largest 26L when 1cos

0

6.22

(iv) When 20L ,

cos2620

13

10cos

7.39

13

10cos 1

3.626.227.39

In Summary:

Question on R formula is a regular in this paper.



10.

The diagram shows a trapezium ABCD in which AD is parallel to BC and angle BCD = 90°.

The vertices of the trapezium are at the points A(8, k), B(3, 6), C(1, 2) and D. The line with

equation 253 xy passes through M, the mid-point of AB, and through N, which lies on

AD.

(i) Show that k = 11. [3]

(ii) Find the coordinates of N and of D. [6]

(iii) Find the area of the triangle AMN. [2]

Solution:

(i)

2

6,

2

11

2

6,

2

38 kkM

Since M lies on the line 253 xy ,

252

113

2

6

k

63350 k

11k

A(8, k)

O

253 xy

y

x

C

(1, 2)

(3, 6) B

M

N

D

[Analysis]

Part (i), mid-point M, is on the line. Part (ii) needs to find the equation of AD using gradient BC.

And also need to find the equation of CD. Part (iii) begins with finding the area of BNA.



(ii)

11,8A AD // BC, gradient of BC = gradient of AD 213

26

Equation of AD ,

8211 xy

52 xy

At point N, 52 xy intersects 253 xy ,

25352 xx

305 x

6x

7512 y

7,6N

CD BC, gradient of CD 2

1

Equation of CD ,

12

12 xy

5.22

1 xy

At point D, 52 xy intersects 5.22

1 xy ,

5.22

152 xx

5.75.2 x

3x

156 y

1,3D

(iii)

length CD 5211322 units

length AN 201178622 units

Area of triangle BNA 52052

1 square units

Area of triangle AMN = 2

1 of Area of triangle BNA 5.25

2

1 square units



11.

The diagram shows part of the curve xy 45 , meeting the x-axis at the point A and the

line 1x at the point B. The normal to the curve at B meets the x-axis at the point C. Find

(i) the coordinates of C , [6]

(ii) the area of the shaded region. [6]

Solution:

(i) When 1x , 345 y , coordinates of 3,1B

xxx

y

45

2

45

4

2

1

d

d

When 1x ,

3

2

45

2

d

d

x

y

The gradient of the tangent at B is 3

2, therefore the normal at B is

2

3 .

Equation of BC ,

12

33 xy

5.42

3 xy

x

B

A C O

y 1x

xy 45

[Analysis]

Part (i) requires differentiation to find gradient then the normal equation. Part (ii) breaks the shaded

area into two regions.



When 0y ,

5.42

30 x

3x

coordinates of 0,3C

(ii) When 0y ,

x450

4

5x

coordinates of

0,

4

5A

area of the shaded region

3132

145

1

4

5

dxx

34542

3

6

11

4

5

dxx

3456

11

4

52

3

x

30456

12

3

2

3

3276

1

35.4

5.7 square units

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