H2 Promo Revision Paper 1 IJC Promo 2012(2) 2 H2 Promo … · 2019-09-07 · H2_Promo Revision...

H2_Promo Revision Paper 1_IJC Promo 2012(2) 2

H2_Promo Revision Paper 1_IJC Promo 2012_Solution(1) 9

H2_Promo Revision Paper 2_MI_1_Promo 2012(1) 21

H2_Promo Revision Paper 2_MI_1_Promo 2012_Solutions 26

H2_Promo Revision Paper 3_MI_2_Promo 2012 30

H2_Promo Revision Paper 3_MI_2_Promo 2012_Solutions 33

H2_Promo Revision Paper 4_NJC Promo 2012(1) 38

H2_Promo Revision Paper 4_NJC Promo 2012_Solution 44

H2_Promo Revision Paper 5_NYJC Promo 2012 53

H2_Promo Revision Paper 5_NYJC Promo 2012_Solution 60

H2_Promo Revision Paper 6_SAJC Promo 2012 71

H2_Promo Revision Paper 6_SAJC Promo 2012_Solution 78

H2_Promo Revision Paper 7_ACJC Promo 2012(1) 93

H2_Promo Revision Paper 7_ACJC Promo 2012_Solution 99

sgfreepapers.com 1

sgfreepapers.com

H2 Mathematics MJC 2013

Page 1 of 7

HH22 MMAATTHHEEMMAATTIICCSS ((99774400))

PPrroommoo RReevviissiioonn PPrraaccttiiccee PPaappeerr 11 Name: __________________ Total marks: ____/ 80 Class: ________ Duration: 2 hours 24 minutes

1 Show that, when x is sufficiently small for 3x and higher powers of x to be neglected,

2

sin 2 1 34

1 2cos 22

x

x xx

π +

≈ + −

. [5]

2 Referred to the origin O, the points P and Q have position vectors 2 2− +i j k and 2−i k

respectively. The point M lies on PQ produced such that PM : PQ = 3 : 2. Find the position

vector of M and hence find the size of angle POM, giving your answer to the nearest 0.1� . [5]

3 The equation of a curve is given by 3 2y ax bx cx d= + + + , where a, b, c and d are constants. It

is known that the curve has a minimum point at ( )3, 6− . When the curve is translated 2 units

in the direction of y-axis, the curve passes through the points ( )0,5 and

201,

3

− . Determine

the equation of the curve. [5]

4 A closed circular cylinder of radius x is inscribed in a right circular cone of radius 2h and

vertical height 3h, where h is a constant. One circular end of the cylinder lies on the base of

the cone and the circumference of the other circular end is in contact with the inner surface of

the cone.

2h

3h

x

sgfreepapers.com 2


Page 2 of 7

(i) Show that 2 13 ( )

2V x h xπ= − , where V is the volume of the cylinder. [2]

(ii) If x is made to vary, find the maximum volume of the cylinder in terms of h. [4]

5 A sequence is such that 13

2u = and

( )

( )

1

1

3 2 1

1

n

n n

nu u

n n

−

−

−= +

+ for 2n ≥ .

(i) Write down the values of 2u , 3u and 4u . [1]

(ii) Make a conjecture for nu , where 1n ≥ . [1]

(iii) Prove your conjecture in part (ii) using mathematical induction. [4]

(iv) Hence determine if the sequence is convergent. [1]

6 (a) The graph of f ( )y x= has a non-stationary point of inflexion at the origin and the

equations of the asymptotes are 2y = ± and 2x = ± .

On separate diagrams, sketch the graphs of

(i) 2 f ( )y x= , [2]

(ii) f ( )y x= . [2]

y

y = 2

x = 2

x

y = f(x)

x = –2

y = –2

O

sgfreepapers.com 3


Page 3 of 7

(b) The diagram below shows the graph of h( )y x= .

Given that h ( ) 0x′ < for , 0x x∈ ≠� , sketch the graph of h( )y x= . [2]

7 Given that ( )1tan ln 1y x− = − , show that

(i) 2d(1 ) 1 0

d

yx y

x− + + = , [2]

(ii) ( ) ( )23 2

3 2

d d d1 2 1 2 0

d d d

y y yx y

x x x

− + − + =

. [3]

Find the Maclaurin series for y, up to and including the term in 3x . [3]

Write down the equation of the tangent to the curve ( )tan ln 1y x= − at the point where

0x = . [1]

8 Relative to the origin O, the points A, B and C have position vectors 6 2+ −i j k , 3 2 5− + −i j k

and 3 2 2− +i j k respectively. The line L passes through the points A and C.

(i) Find a vector equation of L. [2]

(ii) Find the exact length of projection of AB��

onto L. [3]

(iii) Hence or otherwise, find the shortest distance from B to L, leaving your answer in exact

form. [2]

D is a point such that ABCD is a parallelogram. Use a vector product to find the exact area of

the parallelogram. [3]

y

x

O

y = a

h( )y x=

b

sgfreepapers.com 4


Page 4 of 7

9 A curve has parametric equations 32sinx θ= , 3cosy θ= , for 0 θ π≤ ≤ .

(a) Sketch the graph of the curve, showing clearly the points of intersection with the axes if

any. [2]

(b) Find the equations of the tangent and normal to the curve at the point P where 6

πθ = .

The tangent and normal at P meet the y-axis at A and B respectively. Hence find the

exact area of triangle ABP. [7]

(c) Given that x

wy

= and θ is decreasing at a constant rate of 0.1 rad s-1, find the value of

d

d

w

t when

6

πθ = . [5]

10 (a) A factory manufactures light bulbs. Using a newly bought machine, the number of light

bulbs manufactured on the first day is 4130. Due to wear and tear, the number of light

bulbs manufactured on each subsequent day is 11 less than that on the previous day. The

machine will be deemed uneconomical when the number of light bulbs manufactured is

less than 100. The machine will then be condemned on that day after use.

(i) How many light bulbs are manufactured on the 85th day? [2]

(ii) Find the total number of light bulbs manufactured by the machine when it is

condemned. [4]

(b) A bank offers a cash loan of $10,000. To make the loan attractive, the bank offers the

following repayment plan.

Repay a fixed amount of $x to the bank on the 15th of every month. At the end of each

month, the bank will add an interest at a fixed rate of 5% on the remaining amount owed.

When the amount owed is less than $x, only the balance will have to be paid on the 15th

of the following month.

John takes up the loan on 1st October 2012.

(i) How much will he owe the bank on 31st October 2012 after the interest has been

added? Leave your answer in terms of x. [1]

(ii) Show that the total amount of money John owes the bank at the end of n months is

given by ( ) ( )$ 10000 1.05 21 1.05 1n nx − − . [3]

(iii) If John repays $500 every month to the bank, find the total number of months for

the loan to be repaid fully. [3]

sgfreepapers.com 5


Page 5 of 7

Qn/N

o Topic Set Answers

1 Maclaurin’s Series -

2 Vectors 11

12

8

−

, 128.0°

3 System of Linear Equations

3 213 3

3y x x x= − − +

4 Applications of Differentiation

(ii) 316

9hπ

5 Mathematical Induction (i) 2 3u = , 3

27

4u = and 4

81

5u =

(ii) 3

1

n

nun

=+

(iv) not convergent

6 Graphs (a)(i)

(ii)

y = 2

x = 2

y

x

y2 = f(x)

x = –2

y = – 2

sgfreepapers.com 6


Page 6 of 7

(b)

7 Maclaurin’s Series 2 31 2

...2 3

Equation of tangent:

y x x x

y x

= − − − +

= −

8 Vectors

(i)

1 1

6 4

2 2

λ

= + − −

r

(ii) 6

21

(iii) 11

57

10 33

x = 2

y

x

y = f(x

)

x = –2

y = –2

sgfreepapers.com 7


Page 7 of 7

9 Applications of Differentiation (a)

(b) 2 5 3

33y x= − + ,

3 11 3

2 8y x= + ,

7 3

192

(c) 1

18

−

10 Arithmetic and Geometric Series (a) (i) 3206 (ii) 777,032

(b) (i) ( )( )$ 10000 1.05x−

(iii) 63

x

y

( )0,3

( )0, 3−

( )2,0

sgfreepapers.com 8

1

Promo Revision Practice Paper 1 (Solutions)

1 sin 2 sin cos 2 cos sin 2

4 4 4

cos cos

x x x

x x

π π π + +

=

1 1cos 2 sin 2

2 2

cos

x x

x

+

=

2

2

1 4 11 (2 )

22 2

12

xx

x

− +

≈

−

2

2

1 4 11 (2 )

22 2

12

xx

x

− +

≈

−

( )

12

211 2 2 1

22

xx x

−

= + − −

( )

221

1 2 2 1 ...22

xx x

= + − + +

2 21 11 2 2

22x x x

≈ + + −

21 31 2

22x x

= + −

sgfreepapers.com 9

2

2

( )

2

3

13

2

1 21

3 0 12

2 2

11

12

8

OM OPOQ

OM OQ OP

+=

= −

= − − −

= −

��

��

3or

2

13

12

4

1 23

1 12

4 2

11

12

8

PM PQ

PM

OM

=

−

= −

−

= + − −

= −

��

��

��

1

1 2

1 1

8 2ˆcos66 9

2 1 16

3 66

5ˆcos66

5ˆ cos66

127.985

128.0 (1d.p)

POM

POM

POM−

• − − =

− −=

−=

− =

= °

= °

sgfreepapers.com 10

3

3 ( ) ( )3 2

6 3 3 3a b c d− = + + +

( )27 9 3 6 ................... 1a b c d+ + + = −

Since ( )3, 6− is a minimum point, 0dy

dx= when 3x =

23 2

dyax bx c

dx= + +

( )27 6 0 ............... 2a b c+ + =

Equation of y after translation: 3 2 2y ax bx cx d= + + + +

At ( )0,5 : 5 2d= +

3d =

At 20

1,3

−

: ( ) ( ) ( )3 220

1 1 1 23

a b c d= − + − + − + +

( )14

................. 33

a b c d− + − + =

Substitute d = 3 into (1) and (3)

( )27 9 3 9 ................... 4a b c+ + = −

( )5

................. 3'3

a b c− + − =

From GC, 1

, 1, 33

a b c= = − = −

Equation of 3 213 3

3y x x x= − − +

Alternative Method

( )f 3 6 27 9 3 6a b c d= ⇒ + + + = − -----------(1)

( )f 3 0 27 6 0a b c′ = ⇒ + + = -----------(2)

( )f 0 3 3d= ⇒ = ------------(3)

( )14 14

f 13 3

a b c d− = ⇒ − + − + = ------------(4)

From GC, 1

, 1, 3, 33

a b c d= = − = − =

Equation of 3 213 3

3y x x x= − − +

sgfreepapers.com 11

4

4

(i)

Let y be the height of the cylinder.

By similar triangles

2 2

3 3

2

33

2

3 3

2

13 ( ) ( )

2

h y h

x h

h y x

y h x

V x y x h x shownπ π

−=

⇒ − =

⇒ = −

= = −

(ii) For maximum V,

2d 33 (2 ) (3 ) 0

d 2

9(6 ) 0

2

Vh x x

x

x h x

π π= − =

⇒ − =

40 (rejected) or

3x x h⇒ = =

2

2

3

4 3 43 ( )( )

3 2 3

16( )

9

16

9

V h h h

h h

h

π

π

π

= −

=

=

2

2

4When , 6 12 6 0

3

d Vx h h h h

dxπ π π= = − = − <

sgfreepapers.com 12

5

5(i) Using GC, 2 3u = , 3

27

4u = and 4

81

5u = .

(ii) 3

1

n

nun

=+

(iii) Let nP be the statement

3

1

n

nun

=+

, n+∈� .

Consider n = 1.

LHS = 1

3

2u = (given)

RHS = 13 3

1 1 2=

+

1P∴ is true.

Assume kP is true for some k

+∈� , i.e. 3

1

k

kuk

=+

.

Then for 1kP + ,

LHS

1ku +=

( )

( )( )

3 2 1

1 2

k

k

ku

k k

+= +

+ +

( )

( )( )

3 2 13

1 1 2

kk k

k k k

+= +

+ + +

( )( )

2 131

1 2

k k

k k

+= +

+ + ( )3 2 2 1

1 2

kk k

k k

+ + +=

+ +

( )3 3 3

1 2

kk

k k

+=

+ +

( )

13 1

1 2

kk

k k

+ +=

+ +

13

2

k

k

+

=+

= RHS

Since 1P is true, kP is true ⇒ 1kP + is true. Hence by mathematical induction, nP

is true for all n+∈� .

(iv) Since

3lim lim

1

n

nn n

un→∞ →∞

= = ∞

+ , the sequence is not convergent.

OR

Since 3

1

n

n→ ∞

+ as n → ∞ , the sequence is not convergent.

sgfreepapers.com 13

6

6(a)

(i)

(ii)

(b)(i)

y = – a

h( )y x=

x O

y

b

x = 2

y

x

y = f( x )

x = –2

y = –2

y = 2

x = 2

y

x

y2 = f(x)

x = –2

y = – 2

sgfreepapers.com 14

7

7(i) ( )1

2

2

2

tan ln 1

1 d 1

1 d 1

d(1 ) (1 )

d

d(1 ) 1 0 ( shown)

d

y x

y

y x x

yx y

x

yx y

x

−= −

= −+ −

− = − +

− + + = ∴

(ii) ( )

2

2

d d d1 2 0

d d d

y y yx y

x x x− − + = ---------- (*)

( ) ( )2

2

d d1 2 1 0

d d

y yx y

x x− + − =

( ) ( )3 2 2

3 2 2

d d d d d1 2 1 2 0

d d d d d

y y y y yx y

x x x x x

− − + − + =

------- (**)

( ) ( ) ( )23 2

3 2

d d d1 2 1 2 0 shown

d d d

y y yx y

x x x

− + − + = ∴

( ) ( )

1

2

2

2

When 0,

tan ln1 0

0

dSub into (1 ) 1 0

d

d1

d

d dSub into 1 2 1 0

d d

x

y

y

yx y

x

y

x

y yx y

x x

−

=

= =

⇒ =

− + + =

⇒ = −

− + − =

( ) ( )

2

2

23 2

3 2

3

3

d1

d

d d dSub into 1 2 1 2 0

d d d

d4

d

y

x

y y yx y

x x x

y

x

⇒ = −

− + − + =

⇒ = −

2 31 2...

2 3

Equation of tangent:

y x x x

y x

∴ = − − − +

= −

sgfreepapers.com 15

8

8

(i) 3 1 2 1

2 6 8 2 4

2 2 4 2

AC

= − − = − = − −

��

1 1

: 6 4

2 2

L λ

= + − −

r or equivalent

(ii) 3 1 4

2 6 4

5 2 3

AB

− −

= − = − − − −

��

4 1

4 4

3 2 4 16 6 6

21 21 21AN

− − − +

= = =

i

(iii) 16 16 9 41AB = + + =��

2 2 3641

21

275 115

7 7

BN AB AN= − = −

= =

Area of parallelogram BA CA= ×��

4 2

4 8

3 4

−

= × −

40 4

10 10 1 10 33

40 4

− −

= = =

units2

sgfreepapers.com 16

9

9

(a)

(b) 2d6sin cos

d

xθ θ

θ= ,

d3sin

d

yθ

θ= −

2

d 3sin 1

d 2sin cos6sin cos

y

x

θ

θ θθ θ

−= = −

When 6

πθ = ,

1

4x = ,

3 3

2y = ,

d 2

d 3

y

x= −

Tangent:

3 3 2 1

2 43y x

− = − −

⇒

2 5 3

33y x= − +

Normal:

3 3 3 1

2 2 4y x

− = −

⇒

3 11 3

2 8y x= +

Area of triangle

1 5 3 11 3 1

2 3 8 4

= −

7 3

192=

(c) d0.1

dt

θ= − ;

322sin 2

sin tan3cos 3

wθ

θ θθ

= =

2 2d 2 2(2)sin cos tan sin sec

d 3 3

wθ θ θ θ θ

θ= +

When 6

πθ = ,

d 2 1 3 1 2 1 4 5(2)( )( )( ) ( )( )

d 3 2 2 3 4 3 93

w

θ= + =

( )d d d 5 1

0.1 or 0.0556d d d 9 18

w w

t t

θ

θ= = × − = − −

(c) Alternative 1

x

y

( )0,3

( )0, 3−

( )2,0

sgfreepapers.com 17

10

d0.1

dt

θ= − ;

322sin 2

sin tan3cos 3

wθ

θ θθ

= =

( )

2d 2 d 2 dtan sin 2 tan

d 3 d 3 d

2 dtan tan sin 2

3 d

w

t t t

t

θ θθ θ θ

θθ θ θ

= +

= +

When 6

πθ = ,

( )

( )

d 2 1 1 30.1

d 3 23 3

2 3 5 30.1

3 3 6

1

18

w

t

= − +

= −

= −

(c) Alternative 2

2 2

3 3 3 3 1 3d d2 4 4 2d 5d d

d 93 3

2

x yy x

w

y

θ θθ

− − − = = =

( )d d d 5 1

0.1 or 0.0556d d d 9 18

w w

t t

θ

θ= = × − = − −

sgfreepapers.com 18

11

10(a)

(i)

Let nu be the number of light bulbs manufactured by the machine on the nth

day.

Given 1 4130u = and 11d = − ,

85 1 84 4130 84( 11) 3206u u d= + = + − =

(ii) ( )1 1 100nu u n d= + − <

( )( )4130 1 11 100n+ − − <

367.3636n >

∴ the machine will be condemned on the 368th day.

Total number of light bulbs that has been manufactured

( ) ( )

368

3682 4130 367 11

2

777,032

S=

= + −

=

(b)

(i)

On the 31st Oct 2012 (at the end of 1st month), the amount John owes the bank

( )( )$ 10000 1.05x= −

or ( ) ( )$ 10000 1.05 1.05x= −

(ii) At the end of 2nd month, the amount John owes the bank

( ) ( ) ( )$ 10000 1.05 1.05 1.05x x= − −

( ) ( ) ( )2 2

$ 10000 1.05 1.05 1.05x x = − −

At the end of 3rd month, the amount John owes the bank

( ) ( ) ( ) ( )2 2

$ 10000 1.05 1.05 1.05 1.05x x x = − − −

( ) ( ) ( ) ( )3 3 2

$ 10000 1.05 1.05 1.05 1.05x x x = − − −

……

At the end of nth month, the amount John owes the bank

( ) ( ) ( ) ( )2

$ 10000 1.05 1.05 1.05 1.05n n

x x x = − − − −

�

sgfreepapers.com 19

12

( )( )1.05 1.05 1

$ 10000 1.051.05 1

n

n x − = −

−

( ) ( )$ 10000 1.05 21 1.05 1n n

x = − −

(iii) Let x = 500

For the loan to be repaid fully,

( ) ( )( )10000 1.05 21 500 1.05 1 0n n− − ≤

Using GC,

when n = 62, amount John owed the bank = 203.10

when n = 63, amount John owed the bank = 311.70−

∴ the number of complete months required is 63.

Let x = 500

For the loan to be repaid fully,

( ) ( )( )10000 1.05 21 500 1.05 1 0n n− − ≤

Using GC, 63n ≥

∴ the number of complete months required is 63.

sgfreepapers.com 20


Page 1 of 5


PPrroommoo RReevviissiioonn PPrraaccttiiccee PPaappeerr 22

Name: __________________ Total marks: ____/ 69

Class: ________ Duration: 2 hours 5 minutes

1

Kelly, Ginny, Joseph and Jaron are teachers who share the marking load of a particular

examination in a school with three pre-university levels. The number of scripts to be

marked by each teacher and the total time taken by Kelly, Ginny and Joseph is shown in

the following table.

Kelly Ginny Joseph Jaron

Pre-University 1 58 77 65 81



Total Time 2000 2726 2474

Assuming that the marking speed for marking the same level is the same for all teachers,

calculate the total time taken by Jaron to mark his load of scipts. [5]

2 The annual population of village at the end of years, , can be modelled by

, where represents the initial population.

(i) Express , and in terms of . [3]

(ii) Show that . [2]

(iii) Explain what will eventually happen to the population. [1]

3 (i) Find the expansion of

in ascending powers of up to and including the term in [3]

(ii) State the range of values of for which this expansion is valid. [1]

(iii) By substituting a suitable value for estimate to 3 significant figures. [2]

sgfreepapers.com 21


Page 2 of 5

4 Without using a calculator, find the range of values of for which

[4]

Hence solve the inequality

[3]

5 The curve has equation ( )

22 21.

9 4

xy ++ = The curve has equation

(i) Sketch and on the same diagram, stating the exact coordinates of any points of

intersection with the axes and the equations of any asymptotes. [5]

(ii) Hence, or otherwise, solve [2]

6 Given the recurrence relation , and ,

(i) find , , and , [2]

(ii) show that a conjecture for is , [1]

(iii) prove the conjecture in (ii) by mathematical induction. [4]

7 (i) Show that ( ) ( )

4 2 2.

1 1 1 1r r r r= −

− + − + [1]

(ii) Hence, find

22

1

1

n

r r= −∑ , leaving your answer in the form of , where

is a constant and is a rational expression to be determined. [3]

(iii) Use your answer to part (ii) to find

( )3

1.

2= −∑

n

r r r [2]

(iv) Give a reason why

23

1

1r r

∞

= −∑

converges and find its value.

[3]

sgfreepapers.com 22


Page 3 of 5

8 (a) The sum of the first terms of a series is given by .

(i) Find an expression for the term and show that the series is an

arithmetic series. [4]

(ii) Find the value of if the sum of the first terms of the series is 1220. [2]

(b) The sum to infinity of a geometric series is 6. By squaring all terms of the series, a

new geometric series is formed and its sum to infinity is 18. Find the first term and

the common ratio of the first series. [4]

9 State clearly the coordinates of point(s) of intersections with the axes, stationary

points, and equation(s) of asymptotes in the following sketches.

(a) The curves and for a certain function , are given below.

Sketch the graph of [3]

(b) Given that

2,

1

xy

x=

− sketch on separate diagrams the graphs of

(i) [3]

(ii)

1,

( )=y

f x [3]

(iii)

[3]

sgfreepapers.com 23


Page 4 of 5

Qn/No Topic Set Answers 1 System of linear equations 10000, 7500, 5000

2372

A B C= = =

2 Recurrence relation

The population stabilises to 500.

3 Binomial Expansion

4 Inequalities or

5 Graphing Techniques

6 Mathematical Induction

7 Summation

8 Arithmetic and

Geometric Progressions

y

x

(0,3)

(0,0) (-4,0)

1

2x = −

( )ln 2 1 3y x= + +

( )22 2

19 4

xy ++ =

3 1,0

2

e− −

sgfreepapers.com 24


Page 5 of 5

9 Graphing Techniques

y=f(x)

(0,0)

(1,-2)

x

x=-1 x=1 y

x

y2=f(x)

O

y

x

x=-0.5 x=0.5

(0,0)

x=-1

sgfreepapers.com 25

Promo Revision Practice Paper 2 – Solutions

1i Let the time spent marking each script for pre-university 1, 2 and 3 be �, �, and � respectively.

58� + 60� + 50� = 2000 77� + 72� + 78� = 2726 65� + 68� + 72� = 2474

�58 60 5077 72 7865 68 72� �� = �

200027262474�

Using GC, �� = �

101214�

1ii Time taken by Jaron = 81 × 10 + 52 × 12 + 67 × 14 = 2372

2i �� = 0.9�� + 50 �� = 0.9�0.9�� + 50� + 50 = 0.9�� + 50�0.9� + 50 �� = 0.9�0.9�� + 50�0.9� + 50� + 50 = 0.9�� + 50�0.9�� + 50�0.9� + 50

2ii �� = 0.9�� + 50�0.9�� +⋯+ 50

= 0.9�� + 50�1 − 0.9��1 − 0.9

= 0.9�� − 500� + 500

2iii As # → ∞,�� → 0�� − 500� + 500 = 500. The population stabilises to 500.

3i � + 1√1 − 2� = �� + 1��1 − 2��

≈ �� + 1� (1 + )−12* �−2�� + +−12, +− 32,2 �−2�� + +−12, +−32, +− 52,6 �−2��-

≈ � + �� + 32 �� + 1 + � + 32 �� + 52 ��

= 1 + 2� + 52 �� + 4��

3ii Valid for |−2�| < 1 ⇒ |�| < �� . 3iii when � = �1 , 234�5� �+23, ≈ 1 + 2 +�1, + 6� +�1,� + 4+�1,�

94√3 ≈ 8364

√3 ≈ 1.73

4 �� − 3�4�� − 9 ≥ 14

4�� − 9 − 4�� + 12�4�� − 9 ≤ 0

4� − 3�2� − 3��2� + 3� ≤ 0

� < − �� 9 ≤ � < �� Replace � with :; . :; < − �� (NA) or

�9 ≤ :; < ��

sgfreepapers.com 26

<# +�9, ≤ � < <# +��,

5i

5ii Using GC, � = −0.361, � = −0.497

6i �� = 99, �� = 95, �� = 91, �9 = 87

6ii �� = 103 − 4�1�, �� = 103 − 4�2�, �� = 103 − 4�3�, �9 = 103 − 4�4� ∴ �� = 103 − 4#

6iii Let Pn be the statement “�� = 103 − 4#” for all # ∈ ℤ, # ≥ 0.

when # = 0, LHS= �� = 103.

RHS= 103 − 4�0� = 103 =LHS

P0 is true.

Assume Pk true for some @ ∈ ℤ, @ ≥ 0,�A = 103 − 4@. To prove Pk+1 is true, �A4� = 103 − 4�@ + 1�.

when # = @ + 1, LHS= �A4�

= �A − 4

= 103 − 4@ − 4

= 103 − 4�@ + 1� =RHS

Since P0 is true and Pk true implies Pk+1 is true, Pn true for all # ∈ ℤ, # ≥ 0.

7i 2B − 1 − 2B + 1 = 2B + 2 − 2B + 2�B − 1��B + 1� = 4�B − 1��B + 1�

7ii C 1B� − 1�DE� = 14CF 2B − 1 − 2B + 1G

�DE�

= 14 F21 − 23 + 22 − 24 + 23 − 25 +⋯+ 2# − 2 − 2# + 2# − 1 − 2# + 1G = 34 − 2# + 12#�# + 1� 7iii C 1B�B − 2� =

�DE� C 1B� − 1

� �DE�

= 34 − 2# − 12#�# − 1� 7iv As # → ∞, ��4��4�� → 0. C 1B� − 1

HDE� =C 1B� − 1

HDE� − 13

� = ln�2� + 1� + 3

��9 + �� + 2��4 = 1

�: � − 12 , 0�

(0,0) (-4,0)

� = −12

(0,3)

y

x

sgfreepapers.com 27

C 1B� − 1HDE� → 34 − 0 − 13 = 512

8ai K� = L� − L� � = 3#� + # − 3�# − 1�� − �# − 1� = 6# − 2 K� − K� � = 6# − 2 − 6�# − 1� + 2 = 6, which is a constant, ∴AP

8aii 1220 = 3#� + # Using GC, # = 20 or # = − M�� (rej)

8b 6 = N1 − B ⇒ N = 6 − 6B

18 = N�1 − B� 3 = 186 = N1 + B ⇒ N = 3 + 3B

6 − 6B = 3 + 3B ⇒ B = 13

N = 4

9a

9bi

9bii

y

x

y=f(x)

(1,-2)

(0,0)

x= -1

x

y x=1 x=-1

O

�� = O��

y

x

y=x

x=0

(1,0) (-1,0) O

1

( )y

f x=

sgfreepapers.com 28

9biii

sgfreepapers.com 29


Page 1 of 3


PPrroommoo RReevviissiioonn PPrraaccttiiccee PPaappeerr 33 Name: __________________ Total marks: ____/ 61 Class: ________ Duration: 1 hour 50 minutes

1 Arthur put $22500 into three different accounts of varying risk levels at the start of a

particular year. The accounts are denoted by Account A, Account B and Account C. The yearly interest rate for each of the three accounts is as shown below.

Account Yearly Interest Rate A 1% B 3% C 6%

At the start of the year, the total amount of money he put into Account B and Account C exceeded the amount of money he put into Account A by $2500. At the end of the year, he had made $625 in interest. Calculate the amount of money

Arthur put into each of the accounts at the start of the year. [4]

2 By using the standard series expansion for xe and sin ,x find the Maclaurin’s

series for sin ,xy e= up to and including the term in 3.x [3]

By using a suitable value for ,x find an approximate value for 0.5.e [2]

3 Without using a calculator, solve the inequality 1

4 5 .xx

− ≤ − [3]

Hence solve the inequality 1

5 4 .x

xe

e− ≥ [3]

4 (a) The th8 term of an arithmetic progression is 38 and the sum of the first 17 terms is

850. It is given that the sum of the first n terms is greater than 700. Find the least

possible value of .n [4] (b) The sum, ,

nS of the first n terms of a sequence 1 2 3, , , ...T T T is given by

( )1,

1

n

n

k vS

v

−−=

−

where k and v are constants.

sgfreepapers.com 30


Page 2 of 3

(i) Find nT in terms of ,k v and ,n and show that the terms form a geometric

progression. [3]

(ii) Find the range of values of v for which nS converges and find, in

terms of k and ,n the sum to infinity of the sequence. [2]

5 A curve C has parametric equations

tx e= and 3 .y t= −

(i) The point Q on the curve has parameter .q Show that the equation of tangent at Q is

( )4 .q qe y x e q+ = −

Find also the equation of normal at .Q [5]

(ii) Sketch ,C giving the coordinates of any axial intercept and the equation of any asymptote. [3]

(iii) The tangent and normal at point Q meet the y − axis at the points A and B

respectively. Show that the area of triangle ABQ is

( )2 1

2

q qe e +

units2. [3]

6 The line 1l passes through the point A, whose position vector is

7

1

1

−

, and is

parallel to the vector

0

2

1

. The line 2l also passes through the point A and is parallel to

the vector

1

1

1

−

. The plane p has equation 3x y zα β+ − = .

(i) Given that 1l is parallel to p but does not lie on ,p what can be said about

the values of α and ?β [4]

(ii) For the case where 2α = and 0,β = find the position vector of the point

of intersection N between 2l and .p [3]

sgfreepapers.com 31


Page 3 of 3

(iii) Find the perpendicular distance from N to 1.l Hence find the length of

projection of AN onto 1.l [5]

(iv) For the case where 2,α = find the acute angle between 2l and .p [2]

7 (i) Show that ( ) ( ) ( )2

1

3 1 2 1 ,n

r

r r n n an=

− + = + +∑ where a is a constant to be

determined. [4] (ii) Prove your answer in (i) by mathematical induction. [5]

(iii) Hence find ( )( )2

1

3 1 2n

r n

r r= +

− +∑ . [3]

Qn/No Topic Set Answers

1 System of linear equations

10000, 7500, 5000= = =A B C

2 Maclaurin series (standard series)

2 311 0 ;1.66

2≈ + + +y x x x

3 Inequality 10 or 1; ln 4 0

4< ≤ ≤ − ≤ ≤x x x

4 AP/GP 16; ; 1;

1

− >−

n kkv v

v

5 Tangent/ Normal

23− = − −q qy e x q e

6 Vectors (line and plane)

43 3

2, 18; 4 ; 70, 55 5

4

67.8

α β

= ≠ = −

°

��

ON

7 Summation and Mathematical Induction

( )24; 7 12 1= + +a n n n

sgfreepapers.com 32

Promo Revision Practice Paper 3 - Solutions

1 22500

2500 2500

0.01 0.03 0.06 625

+ + =

+ = + ⇒ − + + =

+ + =

A B C

B C A A B C

A B C

Using GC, 10000, 7500, 5000= = =A B C

2

3

sin

6

2 33 3

3

3 2 3

2 3

6 61

6 2 6

16 2 6

11 0

2

−

=

≈

− −

≈ + − + +

≈ + − + +

≈ + + +

x

xx

y e

e

x xx x

xx

x x xx

x x x

when ,6

xπ

=

2

0.5 11

6 2 6

1.66

eπ π

≈ + +

≈

3

( )( )

2

14 5

4 5 10

4 1 10

10 or 1

4

xx

x x

x

x x

x

x x

− ≤ −

− +≤

− −≤

< ≤ ≤

15 4

14 5

x

x

x

x

ee

ee

− ≥

− ≤ −

Replace x with .xe

10 (NA) or 1

4

ln 4 0

x xe e

x

< ≤ ≤

− ≤ ≤

4(a)

( )

( ) ( )

7 38

172 16 850

2

12, 46

2 46 12 1 7002

15.97 or 7.30(NA)

least 16

a d

a d

d a

nn

n n

n

+ =

+ =

= = −

− + − >

> < −

=

sgfreepapers.com 33

4(bi)

( ) ( )

( )

( )

1

1

1

1 1

1 1

1

1

1

n n n

n n

n n

n

n

T S S

k v k v

v v

k v v

v

kv v

v

kv

−

− − +

− + −

−

−

= −

− −= −

− −

−=

−

−=

−

=

Since 1

1

1n

n

n

n

T kv

T kv v

−

− +

−

= = which is a constant, the terms form a geometric sequence.

Alternatively,

( )1

1

11

1

11

11

n

n

n

n

k vS

v

k

v v

v

v

k

v v

v

−−=

−

− =

−

− =

−

Sequence is geometric with 1

1, .

kT r

v v= =

11

n

n

n

kT

v v

kv

−

−

=

=

4(bii) 1

1 1vv

< ⇒ >

1

11

1

kvS

v

k

v

−

∞ =

−

=−

5(i)

( ) ( )

1

1

13

= = −

= −

− − = − −

t

t

q

q

dx dye

dt dt

dy

dx e

y q x ee

Equation of tangent: ( )4q qe y x e q+ = −

( ) ( )3 q qy q e x e− − = −

Equation of normal: 23− = − −q qy e x q e

5(ii) y

sgfreepapers.com 34

5(iii) ( ) ( )20, 4 0,3 qA q B q e− − −

( )

( )

2

2

1Area 4 3

2

1

2

q q

q q

e q q e

e e

= − − + +

+=

6(i)

Since 1l is parallel to

0 3

, 2 1 0.

1

p

α

• = −

2α =

Since 1l does not lie on

7 3

, 1 1 .

1 2

p β

− • ≠ −

18β ≠

6(ii)

2

7 1

: 1 1 ,

1 1

3

: 1 0

2

7 3

1 1 0

1 2

3

4

4

4

l r

p r

ON

λ λ

λ

λ

λ

λ

= − + ∈ −

= −

+

− + = − −

= −

= −

�

i

i

��

6(iii) 3

3

3

AN

−

= −

��

x

C

( )3 , 0e

0x =

sgfreepapers.com 35

3 0

3 2

3 1perpendicular distance

5

126 370

5 5

−

− ×

=

= =

3

Alternatively, 3 2

3

0 3 0

2 3 2 2 0

1 3 1

3

5

33

perpendicular distance 1.8 705

3.6

λ

λ

λ

λ

λ

= + − +

⊥ ⇒ + ⋅ = − +

= −

= = −

��

��

NF

NF

27AN =��

126length of projection 27

5

9 35

5 5

= −

= =

6(iv) 1 3

1 1

1 2sin

3 14

67.8 (1dp)

θ

θ

− −

=

= °

i

7(i) ( )( ) ( )

( )( ) ( )

( )

2

1 1

2

3 1 2 3 5 2

3 1 2 1 5 1 26 2

4 1

4

= =

− + = + −

= + + + + −

= + +

=

∑ ∑n n

r r

r r r r

n nn n n n

n n n

a

7(ii) Let Pn be the statement “ ( )( ) ( )2

1

3 1 2 4 1n

r

r r n n n=

− + = + +∑ ” for .n+∈�

when 1,n =

( )( )

( )( )

1

1

2

LHS 3 1 2 6

RHS 1 1 4 1 1 6 LHS

r

r r=

= − + =

= + + = =

∑

∴ P1 is true.

sgfreepapers.com 36

Assume Pk true for some ,k+∈� ( )( ) ( )2

1

3 1 2 4 1k

r

r r k k k=

− + = + +∑

To show Pk+1 is true, ( )( ) ( )( )1

2

1

3 1 2 1 6 6k

r

r r k k k+

=

− + = + + +∑

when 1,n k= +

( )( )

( ) ( )( )

( )( )

1

1

2

3 2 2

3 2

2

LHS 3 1 2

4 1 3 2 3

4 3 11 6

7 12 6

1 6 6

RHS

k

r

r r

k k k k k

k k k k k

k k k

k k k

+

=

= − +

= + + + + +

= + + + + +

= + + +

= + + +

=

∑

Since P1 is true and Pk true ⇒ Pk+1 is true, Pn true for .n+∈�

7(iii) ( )( ) ( )( ) ( )( )

( ) ( )

( )

2 2

1 1 1

2 2

2

3 1 2 3 1 2 3 1 2

2 4 8 1 4 1

7 12 1

n n n

r n r r

r r r r r r

n n n n n n

n n n

= + = =

− + = − + − − +

= + + − + +

= + +

∑ ∑ ∑

sgfreepapers.com 37


Page 1 of 6



Name: __________________ Total marks: ____/ 64


1 (i) Show that ( )

2 1

1 !

r r

r

+ −

+ can be expressed in the form

( ) ( )1 ! 1 !

A B

r r−

− +, where A and B

are some constants to be determined. [1]

(ii) Find ( )

2

1

1

2 1 !

N

r

r r

r=

+ −

+∑ in terms of N. [2]

(iii) Deduce the exact value of ( )

2

6

( 2) 3

2 1 !r

r r

r

∞

=

− + −

−∑ . [3]

2 The diagram shows a right pyramid TABCD with height h m and a square base ABCD of

sides x m. TA = TB = TC = TD = 8 m. The volume of the pyramid, V, increases at a constant

rate of 10 3 1m s− as x and h vary.

(i) Show that 2

2164 .

3 2

xV x= − [1]

(ii) Calculate the rate of increase of the side of the square base at the instant when the

height of the pyramid is 7 m. [4]

[The volume of a pyramid is 1

base area height3

× × .]

x

T

A B

C D h

x

8

sgfreepapers.com 38


Page 2 of 6

3 An arithmetic series and a convergent geometric series both have first term a and second term

24. The third term of the geometric series is greater than the third term of the arithmetic series

by 4.

(i) Show that 2 52 576 0a a− + = . [3]

(ii) Hence, find the common ratio of the geometric series. [2]

(iii) Find the least value of M such that the sum of the first M terms of the geometric series

is at least 8

9 times the sum to infinity of the geometric series. [2]

4 (i) Show that 23 9 8x x− + is always positive for all real values of x . [1]

(ii) Without the use of graphic calculator, solve the inequality

( )( )3

14

20≥

+− xx. [3]

(iii) Hence, solve the inequality ( ) ( )

203

4 sin 1 sinx x≥

+ − , where 0 2πx≤ ≤ . [3]

5 A sequence 1 2 3, , ,..., ,...nu u u u , is defined by the recurrence relation

1

31

( 2)n nu u

n n−

= −

+

where 1 2u = .

(i) Prove by mathematical induction that 3

2n

nu

n

+= for any positive integer n. [4]

(ii) State the limit of nu as n → ∞ . [1]

(iii) Find 2

1

N

r

r

r u=

∑ , leaving your answer in the form ( )( 1)

6

N N aN b+ +, where a and b are

constants to be found. [3]

2

1

( 1)(2 1)You may use the result:

6

n

r

n n nr

=

+ +=

∑

sgfreepapers.com 39


Page 3 of 6

6 The curve C is defined by the parametric equations

2e , e for .t tx t y t

−= = ∈�

(i) Show that 2d e

d (2 )

ty

x t t=

−, and find the equations of the tangent to C that are parallel to

the y-axis. [3]

(ii) Find the exact equation of the normal to C at 1t = , and also the value of t where this

normal meets C again. [4]

7 (a) Given that π

tan cos2 10 ,4

x x x

− − = −

and x is sufficiently small for 3x and

higher powers of x to be neglected, show that a cubic equation for x is

3 26 5 1 0.x x x+ + − =

Hence, find an approximate value for x, leaving your answer to 3 significant figures.

[4]

(b) Given that 11 tan ,y x−= + show that

( )

22

22 2

d d.

d d 1

y y xy

x x x

+ = − +

By further differentiation, obtain the Maclaurin’s series for y, up to and including the

term in 3x . [5]

Deduce the series expansion of 11 tan

,1

x

x

−+

− as far as the term in 2

x . [2]

8 (a) The curve C has equation given by y = 2

( )

x a a

a x a

− +

−, where a is a negative constant.

(i) Write down the equations of the asymptotes of C. [2]

(ii) Sketch the graph of C, labelling clearly the asymptotes and axial intercepts. [2]

(iii) Deduce the range of values of k for which the equation 2

( )

x a a

a x a

− +

− = k has

one negative real root. [2]

(iv) State the range of values of x for which the curve is concave downwards. [1]

sgfreepapers.com 40


Page 4 of 6

(b) The diagram below shows the graph of y = f(x). The curve has a local minimum

point at A(−2, 2). It passes through the point B(5, 0) and the point C(0, 4). The

asymptotes are the lines x = 2, y = 4 and y = −3.

Sketch, on separate diagrams, the graphs of

(i) y = f ( )x− , [3]

(ii) f ( )y x′= , [3]

stating clearly the asymptotes and the coordinates of the points (if any) corresponding

to A, B and C.

Qn/No Topic Set Answers 1 Seq and Series (i) A = B = 1

(ii) ( ) ( )

1 1 12

2 ! 1 !N N

− −

+ (iii)

5

48

2 Rate of Change 1 1d 7 30

ms or 0.462 ms (3 s.f.) d 83

x

t

− −=

3 APGP (ii)

2

3 (iii) least M = 6

4 Inequalities (ii) 1 4x− < < (iii)

π0 2π,

2x x≤ ≤ ≠

5 MI (ii)

1

2 (iii) a = 1, b = 5

6 Tangent/Normal (i) 2Equations of the tangent are 0 or 4e .x x−= =

0

y = − 3

y = f(x)

A

B

y

x

x = 2

•

•

y = 4 C •

sgfreepapers.com 41


Page 5 of 6

(ii) Equation of normal:

( )2 1

2 3

e e e

or e e e

y x

y x

− −

− −

− = − −

= − + +

(iii) 1.79 (3 s.f.) t = −

7 Maclaurin’s Series/

Small Angle Approx (a) 0.166 (3 s.f.)x =

(b) 2 31 1 51

2 8 48y x x x≈ + − − ;

121 tan 1

11 2

xx x

x

−+= + +

−

8 Curve sketching/ Transformations (a)(i) Asymptotes are y =

a

1 and x = a

(a)(ii)

(a)(iii) k < a

1−1 or k >

a

1

(a)(iv) x > a

y

x a

a

1

a−a2

a

1

−1

•

•

O

y =

2

( )

x a a

a x a

− +

−

sgfreepapers.com 42


Page 6 of 6

(b)(i)

(b)(ii)

A(−2, − 2 )

y = −)(f x

0

•

y = −2

B(5, 0) x

x = 2

•

y

C(0, −2) •

y

x

y = f ’(x)

−2 •

0

x = 2

A

sgfreepapers.com 43


1(i)

( ) ( ) ( )

2

2

2

1

1 ! 1 ! 1 !

1 ( 1)

Compare coefficient of : 1

Compare constant: 1

r r A B

r r r

r r A r r B

r A

B

+ −= −

+ − +

⇒ + − = + −

=

=

1(ii)

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

2 2

1 1

1

1 1 1

2 1 ! 2 1 !

1 1 1

2 1 ! 1 !

1 1(1

2 2

11

3!

1 1

2 4!

1 1

3! 5!

......

1 1

2 ! !

1 1)

1 ! 1 !

1 1 12

2 ! 1 !

N N

r r

N

r

r r r r

r r

r r

N N

N N

N N

= =

=

+ − + −=

+ +

= −

− +

= −

+ −

+ −

+ −

+ −−

+ −− +

= − −

+

∑ ∑

∑

1(iii)

( )

( )

( ) ( )

2

6

2

4

32 2

1 1

( 2) 3

2 1 !

1 (replace by 2)

2 1 !

1 1

2 1 ! 2 1 !

1 1 11 2

2 3! 4!

5

48

r

r

r r

r r

r

r rr r

r

r r r r

r r

∞

=

∞

=

∞

= =

− + −

−

+ −= +

+

+ − + −= −

+ +

= − − −

=

∑

∑

∑ ∑

sgfreepapers.com 44

2

x

x

2 2

2 2

2

d x x

x

x

= +

=

=

2

2

2

28

2

642

xh

x

= −

= −

2

22

1

3

164 (Shown)

3 2

V x h

xx

=

= −

( )

12 2 2

2

2 3

2

d 1 12 64 64

d 3 2 2 2

1 2 64

3 22 64

2

V x xx x x

x

x xx

x

− = − + − −

= − −

−

2

7

7 642

30

h

x

x

=

⇒ = −

⇒ =

( )( )

( )

3

1 1

d d d

d d d

301 d10 2 30 7

3 2 7 d

d 7 30 ms or 0.462 ms (3 s.f.)

d 83

V V x

t x t

x

t

x

t

− −

= ×

= − ×

=

d

sgfreepapers.com 45

3(i)

( )

( )

Let the first three terms of the AP be , 24,24 [or 2 ]

and the first three terms of the GP be , 24, 24 4 [or ( 2 ) 4]

24 ....................... 1

24 28 .............

24

a d a d

a d a d

d a

dr

a

+ +

+ + + +

= −

+= =

( ) 2

2

24 2 4...(2) or

24

28 24 24

52 576 0 (Shown)

a d

a

a a

a a

+ + =

+ − =

∴ − + =

OR

( )

( ) ( )

2

2

Let the first three terms of the GP be , 24,24 [or ]

and the first three terms of the AP be , 24, 24 4 [or 4]

24 ........................... 1

24 24 4 24..........(2)

a r ar

a r ar

ra

d a r

− −

=

= − = − −2

2

or 24 ( 4) 24

24 24 24 28

52 576 0 (Shown)

a ar

aa

a a

− = − −

− = −

∴ − + =

3(ii) ( ) ( ) 36 16 0

36 or 16

24 2 24 3 or

36 3 16 2

(reject since 1 OR geometric series is convergent)

a a

a a

r r

r

− − =

∴ = =

= = = =

<

3(iii) ( )1 8

1 9 1

2 81

3 9

2 1

3 9

1lg

95.42

2lg

3

least 6.

M

M

M

a r a

r r

M M

M

− ≥

− −

⇒ − ≥

⇒ ≤

⇒ ≥ ⇒ ≥

∴ =

sgfreepapers.com 46

4(i) 2

2

2

3 9 8

3 93 8

2 4

3 5 53 0

2 4 4

x x

x

x

− +

= − − +

= − + ≥ >

4(ii)

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

2

2

2

203

4 1

20 3(4 )(1 )0

4 1

20 3(4 3 )0

4 1

3 9 80

4 1

4 1 0 (since 3 9 8 0)

1 4.

x x

x x

x x

x x

x x

x x

x x

x x x x

x

≥− +

− − +⇒ ≥

− +

− + −⇒ ≥

− +

− +⇒ ≥

− +

⇒ − + > − + >

⇒ − < <

4(iii) Replace x by − sin x in

( )( )3

14

20≥

+− xx:

( ) ( )

203

4 sin 1 sinx x≥

+ −.

Thus, 1 sin 4

4 sin 1

1 sin 1 ( sin 1)

π0 2π,

2

x

x

x x

x x

− < − <

⇒ − < <

⇒ − ≤ < ≥ −

∴ ≤ ≤ ≠

∵

5(i) Let Pn be the statement

3, .

2n

nu n

n

++= ∈�

When n = 1,

LHS of P1 = 1 2u = (given)

RHS of P1 = 1 3

22(1)

+=

∴ P1 is true.

Assume Pk is true for some k +∈Z , i.e.

3

2k

ku

k

+= .

To prove Pk+1 is true, i.e. 1

4

2( 1)k

ku

k+

+=

+.

sgfreepapers.com 47

+1 1

2

k+1

LHS of P

31

( 1)( 3)

3 31

( 1)( 3) 2

4 3 3 1

( 1) 2

( 4)

2 ( 1)

4

2( 1)

RHS of P

k k

k

u

uk k

k

k k k

k k

k k

k k

k k

k

k

+=

= −

+ +

+= −

+ +

+ + −=

+

+=

+

+=

+

=

Pk is true ⇒ Pk+1 is true.

Since P1 is true, and Pk is true ⇒ Pk+1 is true, by mathematical induction, Pn is true

for n +∈Z .

5(ii) 1 3 1lim lim

2 2 2n

n nu

n→∞ →∞

= + =

5(iii)

( ) ( )

22

1 1

2

1 1

3

2

1 3

2 2

( 1)(2 1) 3 ( 1)

12 4

( 1) 2 10 ( 1) 5

12 6

N N

r

r r

N N

r r

r rr u

r r

N N N N N

N N N N N N

= =

= =

+=

= +

+ + += +

+ + + += =

∑ ∑

∑ ∑

a = 1 and b = 5

6(i)

( ) ( ) ( ) ( )

2

2

dd e e ed

dd e 2 2e e 2

d

t t t

tt t

y

y txx t t t tt t

t

−− −= = = =

− −− +

( )

2

d d undefined (or 0)

d d

2 0

0 or 2

Equations of the tangent are 0 or 4e .

y x

x y

t t

t

x x−

=

− =

⇒ =

= =

sgfreepapers.com 48

6(ii) 2dWhen 1, e .

d

yt

x= =

Gradient of normal at 21 is et−

= − .

Equation of normal:

( )2 1

2 3

e e e

or e e e

y x

y x

− −

− −

− = − −

= − + +

Normal meets C:

( )2 2 1e e e e et tt

− − −− = − −

Solving the above equation using GC gives

1.79 (3 s.f.) or 1 (N.A.)t t= − =

7(a)

( ) ( ) ( )

( )

2

2

2 3 2

3 2

3 2

πtan cos2 10

4

tan 1cos2 10

1 tan

Given that is sufficiently small,

1 41 10

1 2

1 1 2 1 10 1

1 1 2 2 10 10

2 12 10 2 0

6 5 1 0 (Shown)

x x x

xx x

x

x

x xx

x

x x x x x

x x x x x x

x x x

x x x

− − = −

−− = −

+

−− − = −

+

− − − + = − +

− − + − − = − −

+ + − =

∴ + + − =

Solving the cubic equation with GC,

4.95(N.A.) or 1.22(N.A.) or 0.166 (3 s.f.)x x x= − = − =

sgfreepapers.com 49

7(b)

( ) ( )

( )

( )

1

2 1

2

222

2

2

22

22

22 2

1 tan

1 tan

Differentiate w.r.t. ,

d 12

d 1


d d2 2 1 2

d d

2

1

d d (Shown)

d d 1

y x

y x

x

yy

x x

x

y yy x x

x x

x

x

y y xy

x x x

−

−

−

= +

= +

=+

+ = − +

−=

+

− ∴ + =

+

( ) ( )

( )

( )

( )

22 2 23 2 2

43 2 2 2

2 2

32


1 4 1d d d d d2

d d d d d 1

1 4

1

x

x x xy y y y yy

x x x x x x

x x

x

− + + ++ + =

+

− + +=

+

2 3

2 3

2 3

2 3

d 1 d 1 d 5When 0, 1, , , .

d 2 d 4 d 8

511 8412 2 6

1 1 51

2 8 48

y y yx y

x x x

y x x x

x x x

= = = = − = −

−−

∴ ≈ + + +

≈ + − −

( )

1

11

2

2 2

2 2 2

2

1 tan

1

1 tan 1

1 3

1 1 1 2 21 1

2 8 2 2

1 3 1 1 11

2 8 2 4 8

11

2

x

x

x x

x x x x

x x x x x

x x

−

−−

+

−

= + −

− − ≈ + − + +

≈ + + + + −

= + +

sgfreepapers.com 50

8a(i)

y = 2

( )

x a a

a x a

− +

− =

1 a

a x a+

−

Asymptotes are y = a

1 and x = a

8a(ii)

8a(iii) k <

a

1−1 or k >

a

1

8a(iv) x > a

8b(i)

A(−2, − 2 )

y = − )(f x

0

•

y = −2

B(5, 0) x

x = 2

•

y

C(0, −2) •

y

x a

a

1

a−a2

a

1−1

•

•

O

y = 2

( )

x a a

a x a

− +

−

sgfreepapers.com 51

8b(ii)

y

x

y = f ’(x)

−2 •

0

x = 2

A

sgfreepapers.com 52


Page 1 of 7



Name: __________________ Total marks: ____/ 91


1 By using a graphical method, solve the inequality

2 12

2x

xx

x −− − ≥

+. [5]

2 The hyperbola with equation 2 2 0x y Ax By C− + + + = passes through the points (3, 2), (−1, 2)

and (1 2 2+ , 0). Find the equation of the hyperbola and hence find the equations of the

asymptotes. [5]

3

Differentiate the following expressions with respect to x, and simplify them.

(a) 2cos

2ln

x

x

, [2]

(b) 1 2

2ta

1n

x

x

− +

−

. [3]

4 A sequence of numbers { }nu is defined recursively by

1 02 2 for 1, 2,3, , n 1a d nn nu u n u−− = = =… .

Prove, by mathematical induction, that ( 1)2n

nu n= + for , 0.n n∈ ≥� [5]

Use a graphing calculator to find the least integer N such that 610>Nu . [1]

Evaluate 1

lim n

nn

u

u→∞−

. [1]

sgfreepapers.com 53


Page 2 of 7

5 The diagram below shows the graph of f ( ).y x= The curve has a maximum point A at

(0, 4)− and a minimum point B at (4, 3)− . The equations of the asymptotes of the curve are

2,x = − 2x = and 2y = − .

On separate diagrams, sketch the following graphs, indicating clearly the equations of the

asymptotes, axial intercepts and the coordinates of the point corresponding to A and B.

(i) ( )f ,y x= [2]

(ii) f ( ) 2,y x= − − [3]

(iii) f ' ( ).y x= [3]

6 Given that r is a positive integer and ( )2

1f r

r= , express ( ) ( )f f 1r r− +

as a single fraction.

Hence prove that ( ) ( )

4

2 221

2 1 11

1 4 1

n

r

r

r r n=

+ = − + +

∑ . [3]

Give a reason why the series is convergent, and state the sum to infinity. [2]

Find ( )

4

222

2 1

1

n

r

r

r r=

− −

∑ . [2]

y

x = 2

2y = −

x −2 2 3

3

−3

A (0,−4)

f ( )y x=

0

O

x = −2

B (4, −3)

sgfreepapers.com 54


Page 3 of 7

7 The equation of a curve is given by 3 24 5 .x xy xy+ = Find

d

d

y

x in terms of x and y.

The curve meets the line y x= at the origin and the point P. Find the equation of the tangent

to the curve at P.

This tangent cuts the y-axis at Q. Find the angle OQP. [8]

8 (a) An arithmetic progression, A , has first term, a , and a non-zero common difference,

d . The first, third and sixth term of A are equal to the third, second and first term of a

geometric progression G respectively.

(i) Show that the common ratio of G is 2

3. [2]

(ii) Given that the sum of the first 16 odd-numbered terms in A is 831 more than

the sum to infinity in G, find the values of a and d . [3]

(b) A baker wants to bake a birthday cake comprising layers of cylindrical cakes each of

height h cm. The diameter of the lowest layer is 80 cm and the diameter of each

subsequent layer is 5 % less than the diameter of the layer below.

(i) Find the radius of the nth layer of the cake. [1]

(ii) Find the least number of layers of the cake such that the total volume of the

cake exceeds 40000h cm3. [3]

sgfreepapers.com 55


Page 4 of 7

9 (a)

A rectangular piece of paper measures 12 cm by 6 cm. For an origami fold, the lower

right-hand corner is folded over so as to reach the leftmost edge of the paper as shown

in the diagram above.

The length of the resulting crease is denoted by l and x is the horizontal length being

folded over.

By considering the area of trapezium ABCD and the total area of the paper, show that

2

2 2 3

3

xl x

x= +

−. [2]

Hence, find the exact minimum value of l. [5]

(b) Sand is poured onto a growing conical pile of sand, at the constant rate of 2 m3s

-1.

Given that the side of the cone always makes an angle of 60o to the cone’s base, find

the rate of change of the exposed surface area of the sand pile when the volume of the

cone is 200 m3.

[The curved surface area of a cone of radius r and slant height l is πrl. ] [5]

10 The curve C has equation 2 1

2

x kxy

x

+ +=

−, 2x ≠ , where k is a constant.

(i) Find the range of values of k if C has 2 stationary points. [3]

(ii) Given that y x= is an asymptote of C, find the value of k. [2]

(iii) Sketch the graph of 2 1

2

x kxy

x

+ +=

− for 2k = − , stating clearly the equations of

any asymptotes, coordinates of any turning points and axial-intercepts. [2]

(iv) On the same diagram, sketch the graph of ( )22 2 29 9 3y a a x= − − , where a is a

constant, showing clearly the coordinates of the axial intercepts. [3]

D

6 cm

12 cm

x

l

A B

C

sgfreepapers.com 56


Page 5 of 7

(v) Deduce the possible values of a if the equation ( ) ( ) ( )( )4 2 229 1 2 9 3x a x x− = − − −

has exactly 3 real roots. [2]

11 1Π and 2Π are parallel planes with equation

1

2 2

1

=

r i and

1

2 8

1

=

r i respectively. The

line � passes through point A(1, 0, 1) and is parallel to the vector +j k .

(i) Verify that A lies on the plane 1Π . [1]

(ii) Find the distance between 1Π and 2Π . [2]

(iii) Show that the acute angle between � and 1Π is 60� . Hence find an inequality that the

acute angle between any line on 1Π and � satisfies. [4]

(iv) Obtain the coordinates of the point of intersection, B, between � and 2Π . [3]

P is a point between 1Π and 2Π such that the ratio of the distance from P to 1Π and the

distance from P to 2Π is 3:1. Find the equation of the plane that P lies on. [2]

If 3Π is a plane that contains line � , state with reason, the number of points of intersection of

1Π , 2Π and 3Π . [1]


1 Inequality 2.51, 2 1.92, 2.09xx x≤ − − < ≤ ≥

2 SLE 2 2

2 2

( 1) ( 2)1

2 2

x y− −− = ; 1; 3y x y x= + = − +

3 Differentiation Technique 2 1a) 2 tanx x

x−−

2

1b)

1x +

4 Mathematical Induction N =16; 2

sgfreepapers.com 57


Page 6 of 7

5 Graphing Transformation

6 Method of Difference

2

11

16n−

7 Tangent and Normal 2 2d 12 5

d 5 2

y x y y

x x xy

+ −=

−;

8 5

3 3y x= − ; 1 3

tan ( )8

−

8 AP, GP - Word Problem a) ii) 3 ; 12d a= =

b) i) 140(0.95)n− ii) least n = 15

9 Max, Min, Rate of Change a)

93 cm

2l

=

b) 2d0.963 m / sec

d

A

t=

10 Rational Function, Conics i)

5

2k > −

ii) 2k = −

sgfreepapers.com 58


Page 7 of 7

iii)

iv) v) 4a = or 4a = − .

11 Vectors - Lines & Planes ii) 6

iii) 60 90α≤ ≤� �

iv) B(1, 2, 3)

113

· 22

1

=

r

sgfreepapers.com 59


Page 1 of 11

Qn

1

At A, 2.51x = − , at B, 1.92x = , at C, 2.09x =

Thus 2.51, 2 1.92, 2.09xx x≤ − − < ≤ ≥

2 (3, 2): 3 2 5 1( )A B C+ + = − − − −

(−1, 2): 2 3 2( )A B C− + + = − − −

(1 2 2+ , 0) 21 2 2 1 2 2 3( ) ( ) ( )A C+ + = − + − − −

Solving 2A = − , 4B = and 7C = −

So 2 2 2 4 7 0x y x y− − + − =

2 21 2 4( ) ( )x y− − − =

2 2

2 2

1 21

2 2

( ) ( )x y− −− =

Equation of asymptotes: 1 2

2 2

( ) ( )x y− −± =

1 3y x and y x= + = − +

3a)

2

2

2

2

2

ln 2

cosd(ln( ))

2 (ln co )

2 sin 1

cos

12 ta

d

n

s x

x

dx xx d

x x

x x

x x

x

x

−

−−

= − −

=

=

x

y

O

2 2y xx= − − 1

2y

x

x=

−

+

A

B C

2x = −

2 1−

sgfreepapers.com 60


Page 2 of 11

Qn

b)

21

2

2 2

2 2

(1 2 ) (2 )( 2)

2 (1 2tan (

(

))

21 21 )

1 2

1 2 4 2

(1 2 ) (2 )

5

5 15

1

x x

x x

xx

x

x x

x x

x x

d

dx

−

− − + −

+ −=

+−+

−

− + +=

+

=+

− +

=+

4 Let nP denote the proposition ( 1)2nnu n= + for , 0.n n∈ ≥�

When n = 0,

LHS of 0P = 0 1u =

RHS of 0P = 0(0 1)2 1+ =

Thus 0P is true.

Assume that k

P is true for some , 0k k∈ ≥� , i.e. ( 1)2kku k= + .

Consider 1n k= + ,

LHS of 1kP + = 11 2 2k

k ku u+

+ = +

1

1 1

1

2 ( 1)2 2

( 1)2 2

[( 1) 1] 2

k k

k k

k

k

k

k

+

+ +

+

= + +

= + +

= + +

i

i

= RHS of 1kP +

Thus 1kP + is true

Therefore, kP is true ⇒ 1kP + is true.

Since 0P is true and kP is true ⇒ 1kP + is true, by Mathematical Induction, nP is true for

, 0.n n∈ ≥Z

N N

u

15 524288 < 106

16 1114112 > 106

For 610>Nu , the least integer N =16.

1

lim n

nn

u

u→∞−

= 1

( 1)2

·lim

2

n

nn

n

n −→∞

+ =

( 1) 2i

·l mn

n

n→∞

+ =

1lim 2(1 )n n→∞

+ = 2

sgfreepapers.com 61


Page 3 of 11

Qn

5 (i)

(ii)

y

x = 2

x −2 2

A’ (0, 2)

f ( ) 2y x= − −

x = −2

B’ (4, 1)

0

3

y

x = 2

x −2 2 3

3

−3

f ( )y x=

0

O

x = −2

sgfreepapers.com 62


Page 4 of 11

Qn

(iii)

6

( ) ( )( )

( )

( )

( )

( )

2 2

2 2 22

2 2

22

22

11 1f f 1

1 1

2 1

1

2 1

1

r rr r

r r r r

r r r

r r

r

r r

+ −− + = − =

+ +

+ + −=

+

+=

+

( )( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( )( )

4 4

221 1

2

2 1f f 1

1

f 1 f 2

f 2 f 3

:

f 4 f 4 1

f 1 f 4 1

11 shown

4 1

n n

r r

rr r

r r

n n

n

n

= =

+= − +

+

= −

+ −

+ − +

= − +

= −+

∑ ∑

y

2

x = −2 x = 2

A' (0, 0)

B' (4, 0)

x −2

f ' ( )y x=

0

O

sgfreepapers.com 63


Page 5 of 11

Qn

( )

( )

( ) ( )

4

222

4 4 1

2 21

2

2

2 22

As , 0,

2 1Hence, li

1 thu

m 11

2 1 2 1

1 1

1

s the series is convergent.4

11

1

6

n

nr

n n

r r

nn

r

r r

r r

r r r r

n

→∞=

−

= =

→ ∞ →

+ = +

− + = − +

= −

+

∑

∑ ∑

7 3 2

2 2

2 2

2 2

4 5 .

d d12 2 5 5

d d

d(5 2 ) 12 5

d

d 12 5

d 5 2

x xy xy

y yx y xy x y

x x

yx xy x y y

x

y x y y

x x xy

+ =

+ + = +

− = + −

+ −=

−

When the line y x= meets the curve at P 3 3 2

3 2

2

4 5

5 5 0

5 ( 1) 0

1 or 0 (rejected)

x x x

x x

x x

x

+ =

− =

− =

=

Coordinates of P (1, 1)

Gradient of tangent at P = 12 1 5 8

5 2 3

+ −=

−

Equation of tangent at P 8

1 ( 1)3

y x− = −

8 5

3 3y x= −

Angle OQP = 1 3tan ( ) 20.6 or 0.359

8

− = °

sgfreepapers.com 64


Page 6 of 11

Qn

8 (a) GP: 5 , 2 ,a d a d a+ +

2

5 2

a d a

a d a d

+=

+ +

2

2 2 2

2

( 2 ) ( 5 )

4 4 5

4 0

(4 ) 0

0(rej) or 4

a d a a d

a ad d a ad

d ad

d d a

d a d

+ = +

+ + = +

− =

− =

= =

Common ratio 4 2

2 4 2 3

a dr

a d d d= = =

+ +

odd-numbered terms in A : , 2 , 4 ,a a d a d+ + ……

sum of the first 16 odd-numbered terms in A

16[2 15(2 )]

2

16 240

16(4 ) 240 304

a d

a d

d d d

= +

= +

= + =

sum to infinity in G

5 4 5

2721

13

a d d dd

r

+ += = =

−−

Given the sum of the first 16 odd-numbered terms in A is 831 more than the sum to infinity in

G

304 27 831d d= +

277 831

3

4 12

d

d

a d

=

=

= =

(b) (i)

Layer Radius

1 40

2 0 95 x 40.

3 20 95 x 40( . )

4 30 95 x 40( . )

� � n 10 95 x 40n−( . )

sgfreepapers.com 65


Page 7 of 11

Qn

the radius of the cake at the nth layer 140 0 95( . )n−=

(ii) Volume of n layers

2 2 2 2 1 2(40) [40(0.95)] [40(0.95) ] .... [40(0.95) ]nh h h hπ π π π −= + + + + 2 2 4 2 ( 1)(40) [1 (0.95) (0.95) ......... (0.95) ]nhπ −= + + + +

( )2

2

2

1 (0.95 )40

1 0.95

n

hπ −

= −

( )251554.34098 1 0.95 nh= −

Given that ( )251554.34098 1 0.96 40000nh h− >

Using GC, least n = 15

9 (a) Area of trapezium ABCD = ( ) ( ) ( )( )2 21

6 12 12 36 122

x l x− − + − −

2 272 3 12 36 3x l x= − − − −

Area of paper = 6 x 12 = 72 cm2

( ) ( )( )( )

( )

( )

2 2 2 2

2 2

2 22 2 2

2

1 172 6 12 36 6 12 12 36 12

2 2

13 12 36 3 9

2

3 9 3

33

x x x l x x l x

x l x x x x x

x x xl x x

xx

= − − + − + − − + − −

− − = − = −

−= + = +

−−

( )

( )

( )

( )

( )

( ) ( )

22 2

22 2

2 2

2 2 3 2

2 2

3

3

3 6 3 2 3 3 18d2 2

d 3 3

2 3 3 18d 2 9

d 2 3 2 3

xl x

x

x x x x x x xll x

x x x

x x x xl x x

x l x l x

= +−

− − − + −= + =

− −

− + − −= =

− −

For max/min l, d

0d

l

x=

3 22 9 0x x− =

9 or 0 (N.A)

2x x= =

x 9

2-

9

2

9

2+

sgfreepapers.com 66


Page 8 of 11

Qn

d

d

l

x -ve 0 +ve

9 is minimum when .

2l x∴ =

2

2 29

39 9 92

3 392 2 2

32

l

= + = = −

cm

Given 3d

2 m / secd

V

t= ,

Exposed area, A = πrl

= cos 60

rrπ

°

= 2πr2

Volume, V = 21

3r hπ

= ( )213

3r rπ

= 33

3rπ

2d d3 2

d d

V rr

t tπ= =

2

d 2

d 3

r

t rπ=

2

d d4

d d

24

3

8

3

A rr

t t

rr

r

π

ππ

=

=

=

When V = 200, r = 3600

3π

sgfreepapers.com 67


Page 9 of 11

Qn

d

0.963d

A

t= m

2/sec

10 (i)

2 1

2

x kxy

x

+ +=

−

( ) ( ) ( )

( )

2

2

2 2 1d

d 2

x k x x kxy

x x

+ − − + +=

−

( )

( )

2

2

4 2 1

2

x x k

x

− − +=

−

d

0d

y

x= ⇒ ( )2

4 2 1 0x x k− − + =

2 stationary points: ( )16 4 2 1 0k+ + >

5

2k > −

Alternatively:

( )2 1 2 5

22 2

x kx ky x k

x x

+ + += = + + +

− −

( )

( )2

2 5d1

d 2

ky

x x

+= −

−

d

0d

y

x= ⇒ ( )

22 2 5x k− = +

2 stationary points: 2 5 0k + >

5

2k > −

(ii) ( )2 5

22

ky x k

x

+= + + +

−

2 0k + = ⇒ 2k = −

(iii) When 2k = − , ( )

21

2

xy

x

−=

−

y x=

x

y

( )1,0

( )3,4

( )12

0, −

sgfreepapers.com 68


Page 10 of 11

Qn

(iv) ( )22 2 23 9 9a x y a− + =

( )

2 2

2 2

31

3

x y

a

−+ =

(v) Substitute ( )

21

2

xy

x

−=

− in ( )

22 2 29 9 3y a a x= − − :

( )

( )( )

422 2

2

19 9 3

2

xa a x

x

−= − −

−

( ) ( ) ( )( )4 2 229 1 2 9 3x a x x− = − − −

From graph, the ellipse cuts the graph of C at exactly 3 points when 4a = . So

4a = or 4a = − .

11

(i). Since

1 1

0 · 2 2

1 1

=

, thus A lie on 1Π

(ii). We have

8 2dist. req d =

1

2

1

6

′

−

=

(iii). Let θ be the acute angle between � and 1Π . Thus

y x=

2x =

x

y

( )1,0

( )3,4

( )12

0, −

• ( )3,0

3 |a|

sgfreepapers.com 69


Page 11 of 11

Qn

1 0

2 · 1

1 1 3 3

21sin

0 6 2

2 1

1 1

θ

= = =

Thus 60θ = �.

Let p be a line on 1Π and α be the acute angle between p and 1Π . Since p lies on 1Π ,

thus α θ≥ .

Thus 60 90α≤ ≤� �

(iv). We have

0

:

1

0 1

1 1

λ

= +

r�

At B,

1 0 1

0 1 · 2 8

1 1 1

2 3 8

2

λ

λ

λ

+ =

⇒ + =

⇒ =

Thus

1

2

3

OB

=

��. Accordingly, B(1, 2, 3).

Since the ratio is 3:1, we have

13 13

· 2 2 (6)4 2

1

= + =

r

Since 1Π and 2Π are parallel and Π is not parallel to 1Π and 2Π , there will be no points of

intersection.

sgfreepapers.com 70


Page 1 of 7


PPrroommoo RReevviissiioonn PPrraaccttiiccee PPaappeerr 66 Name: __________________ Total marks: ____/ 77 Class: ________ Duration: 2 hours 19 minutes

1 A spherical balloon is being inflated and at the instant when its radius is 20 cm, its volume is

increasing at a rate of 50 3 1cm s− . At time t seconds, the radius of the balloon is r cm.

(i) Find d

d

r

t when 20r = , leaving your answer in exact form. [2]

(ii) Find the rate of increase of the surface area of the balloon when its radius is 20 cm. [2]

[The formulae for the volume and surface area of a sphere are 34

3V rπ= , 24A rπ= .]

2 Expand ( )3

29 2x+ in ascending powers of x , up to and including the term in 3x . [2]

State the range of values of x for which the expansion is valid. [1]

By using the substitution 2x = , find an approximation for 13 , leaving your answer as a

fraction in its lowest terms. [2]

3 On joining ABC International School, each of the 200 students is placed in exactly one of the

four performing arts groups: Choir, Chinese Orchestra, Concert Band and Dance. The

following table shows some information about each of the performing arts groups:

Performing Arts Group Choir Chinese

Orchestra

Concert

Band Dance

Membership Fee (per student per month)

$15 $20 $20 $18

Instructor Fee (per student per month)

$50 $60 $75 $40

Costume fee (one-time payment per student)

$45 ? $40 $60

No. of Training Hours (per student per week)

5 6 8 7

sgfreepapers.com 71


Page 2 of 7

In a typical month, the school collects a total of $3,721 for membership fee from the students,

and pays the instructors a total sum of $11,830 (assuming that this sum of money is fully paid

by the students). As for the training in a typical week, students from Chinese Orchestra and

Concert Band spend in total 431 hours more than their peers in Choir and Dance. Find the

enrolment in each of the performing arts groups.

Hence, find the costume fee paid by each student from Chinese Orchestra if a vendor charges

a total of $9,440 for all the costumes for the four performing arts groups. [5]

4 (i) Show that ( )

( )( )

2 2 11 2 3

1 2 1 2

r

r r r r r r

++ − =

+ + + + [1]

(ii) By using the method of differences, find ( )( )1

2 1

1 2

n

r

r

r r r=

+

+ +∑ . [3]

(iii) Hence, find the value of ( )( )2

2 1

1 2r

r

r r r

∞

=

+

+ +∑ . [3]

5 The parametric equations of a curve are 3cosx a θ= , 3siny a θ= , where a is a positive

constant and 02

πθ< < .

(i) Show that d

tand

y

xθ= − . [2]

The tangent to the curve at the point with parameter θ cuts the x-axis and y-axis at points P

and Q respectively.

(ii) Find the coordinates of P and Q, leaving your answers in terms of a and θ . [4]

(iii) Let O denote the origin. Given that OPQ forms an isosceles triangle, find the value of

θ . Hence find the area of triangle OPQ, leaving your answer in terms of a. [2]

6 The r th term of a sequence, ru , is given by 14

12 −

=r

ur for r = 1, 2, 3, 4, …

(i) Write down the exact values of the first four terms of the sequence, and hence find the

exact values of ∑=

n

r

ru1

for n = 1, 2, 3 and 4. [3]

(ii) Make a conjecture for ∑=

n

r

ru1

in terms of n, and prove the conjecture by mathematical

induction. [5]

sgfreepapers.com 72


Page 3 of 7

7 (a) Solve the inequality 1 1

2 5x x≥

− − where 2 , 5x x≠ ≠ . [3]

Hence solve the inequality 1 1

2 5x x≥

− − . [2]

(b) Solve the inequality log 3 1x > − , where x > 0.

[Hint: lg

loglg

a

bb

a= ] [4]

8 (a) The graph of f ( )y x= is given below. The graph has asymptotes 0y = , 2y = ,

1x = − and 1x = , a maximum point at (3,3) and a minimum point at the origin.

Sketch the graph of

(i) ( )f 1y x= + , [3]

(ii) 1

f ( )y

x= . [3]

1x = − 1x =

2y =

O 2

(3,3)

x

y

sgfreepapers.com 73


Page 4 of 7

(b) Two students Alfred and Betty were asked to do a transformation from f ( )y x= to

5f (2 1) 3y x= + − . The equation of the original graph was 2y x= and the resulting

graph had equation ( )2

5 2 1 3y x= + − . The following are the steps that they took in

sequence:

It was known that one person had made an error in one of the steps.

(i) Identify the person who had made an error and identify the incorrect step. [2]

(ii) Write down the equation of the final graph resulting from his/her (incorrect)

transformations. [1]

9 (i) Find, in terms of k, the vector equation of the line 1l that passes through the

point A (5, 1, 7) and is parallel to the vector

1

2

k

. [1]

(ii) Find the value of k, given that 1l is perpendicular to the line 2l whose equation is given

by

2

zx y= − = [3]

(iii) Find N, the point of intersection between 1l and 2l . Hence, find the shortest distance

from point A to the line 2l . [3]

(iv) Find the position vector of C (relative to the origin O) such that ONAC forms a

rectangle and hence find the area of ONAC. [3]

(v) Find 'O , the image of the origin O when reflected in 1l . Hence, find the equation of

the line 3l , the reflection of line OC in the line 1l . [2]

Betty 1. Translate the graph along

the x-axis by 1− unit

2. Scale the graph along the x-axis by factor ½

3. Scale the graph along the y-axis by factor 5.

4. Translate downwards by 3

units.

Alfred 1. Scale the graph along the y-

axis by factor 5.

2. Translate downwards by 3 units.

3. Scale the graph along the x-axis by factor ½.

4. Translate along the x-axis

by 1− unit.

sgfreepapers.com 74


Page 5 of 7

10 (a) Peter and Samuel visit a buffet restaurant together every day for fifty days.

Peter manages to consume 1000 grams of food on Day 1, but on each subsequent day,

he consumes 2% less food (in grams) than he did on the previous day.

(i) Day n is the first day on which he consumes less than 500 grams of food.

Find n and calculate the total amount of food he consumes, in grams, from

Day 1 to Day n (inclusive). [4]

Samuel manages to consume 600 grams of food on Day 1, but in each subsequent day,

the amount of food he consumes weighs 10 grams less than the amount of food he

consumes the previous day.

(ii) On Day m , the amount of food (in grams) that Samuel consumes is at least

61% of the amount of food (in grams) that Peter consumes. Using a calculator,

find all possible values of m . [2]

(b) The arithmetic table of m rows and n columns is an extension of the concept of the

arithmetic progression. The entries in each row of the table form an arithmetic

progression (from left to right) with horizontal common difference d , and the entries

in each column of the table form an arithmetic progression (from top to bottom) with

vertical common difference f . The term at the top left corner of the table is denoted

as a .

The following table illustrates the case where 4, 5, 2, 3, 1.m n d f a= = = = =

Column

1

Column

2

Column

3

Column

4

Column

5

Row 1 1 3 5 7 9

Row 2 4 6 8 10 12

Row 3 7 9 11 13 15

Row 4 10 12 14 16 18

Now, consider an new arithmetic table T , with

100, 200, 5, 3, 50.m n d f a= = = = − =

(i) Find the sum of all terms in Row 1 of the arithmetic table .T [2]

(ii) Find the sum of all terms in the arithmetic table .T [2]

sgfreepapers.com 75


Page 6 of 7


1 Applications of Differentiation :Rate of Change

1

32π; 2 15cm s−

2 Binomial Expansion 2 3

27 13 54 1458

x x x + + −

;

9 9 9

2 2 2x or x< − < < ;

1265

351

3 System of Linear Equations

There are 43 students in Choir, 65 students in Chinese Orchestra, 60 students in Concert Band and 32 students in Dance. the costumes fee per student is $49.

4 Method of Differences (ii)

5 1 3

4 2( 1) 2( 2)n n− −

+ +

(iii) 3

4

5 Applications of Differentiation: Tangents and Normals

(i) tandy

dxθ= −

(ii) P ( )cos ,0a θ and Q ( )0, sina θ

(iii) 2

2 units4

a

6 Mathematical Induction (i) , 1 2 3 4

1 1 1 1, , ;

3 15 35 63u u u u= = = =

3

11

1

=∑=r

ru

,

2

1

2

5r

r

u=

=∑,

3

1

3

7r

r

u=

=∑,

4

1

4

9r

r

u=

=∑

(ii) Conjecture:121 +

=∑= n

nu

n

r

r for n = 1, 2, 3, 4, …

7 Inequalities (a) (i) 2 < x < 5 (ii) -5 < x < - 2 or 2 < x < 5

(b) 1

0 or 13

x x< < >

8(i) Transformations of Graphs

2x = −

3− 1

(2,3)( 4,3)−

2y =

y

x

sgfreepapers.com 76


Page 7 of 7

8(ii) Transformations of Graphs

8(iii) Transformations of

Graphs Alfred is incorrect. The incorrect step was Step 4 “Translate along the x-axis by factor ½” The equation of the final graph resulting from the incorrect

transformations is ( )2

5 2 2 3y x= + −

9 Vectors

(i) 1

5 1

: 1 2 ,

7

l

k

λ λ

= + ∈

r �

(ii) 1

2k =

(iii) N (3,-3,6)

(iv) 21

(v) 9 14

(vi) 1

6 2

: 6 4 ,

12 1

l t t

= − + ∈

r �

10 AP/GP (a)(i) 25800 (ii) m can be any integer between 8 to 15 (inclusive)

(b)(i) 109500 (ii) 7980000

1− 1

2x =

1(3, )

3

1

2y =

y

x

sgfreepapers.com 77


1 (i)

Given 50dV

dt= when 20r = ,

dr dV dr

dt dt dV= ×

3 244

3

dVV r r

drπ π= ⇒ =

2

1

4

dr dV dr

dt dt dV

dr dV

dt dt rπ

= ×

= ×

When r = 20,

( )2

1 150

324 20

dr

dt ππ= × =

(ii)

dA dA dr

dt dr dt= ×

24A rπ= 8dA

rdr

π⇒ =

When 20r = , ( )1

8 20 532

dA

dtπ

π

= =

Therefore, the rate of increase of the surface area is 2 15cm s− .

2 33 3

22 2

2 3

2 3

2(9 2 ) 9 1

9

3 1 2 3 1 1 2

3 2 2 2 9 2 2 2 927 1

2 9 2! 3!

27 13 54 1458

x x

x x

x

x x x

+ = +

−

= + + + +

≈ + + −

…

sgfreepapers.com 78

Expansion is valid for 9 9 9

2 2 2x or x< − < <

Substituting 2x = into the equation,

( )

3 2 3

2

3

2

1

2

2 2 2(13) 27 1

3 54 1458

1265(13)

27

1265 126513

27(13) 351

≈ + + −

≈

≈ =

3 Let a, b, c and d be the number of students in Choir, Chinese Orchestra, Concert Band and

Dance respectively.

Enrolment : 200=+++ dcba

Membership fee : 372118202015 =+++ dcba

Instructor fee : 1183040756050 =+++ dcba

Training hours : 4317586 =−−+ dacb

Solve using GC: 32,60,65,43 ==== dcba

So, there are 43 students in Choir, 65 students in Chinese Orchestra, 60 students in Concert

Band and 32 students in Dance.

Let x be the costumes fee paid by each student from Chinese Orchestra.

9440)32(60)60(4065)43(45 =+++ x

x = 49

So, the costumes fee per student is $49.

4 (i) 1 2 3 ( 1)( 2) 2 ( 2) 3 ( 1)

1 2 ( 1)( 2)

r r r r r r

r r r r r r

+ + + + − ++ − =

+ + + +

sgfreepapers.com 79

2 2 23 2 2 4 3 3

( 1)( 2)

r r r r r r

r r r

+ + + + − −=

+ +

4 2 2(2 1)

( 1)( 2) ( 1)( 2)

r r

r r r r r r

+ += =

+ + + + (shown)

(ii) 2

1 1

2 1 1 1 2 3

( 1)( 2) 2 1 2

n

r r

r

r r r r r r= =

+ = + −∑ ∑ + + + +

= 1

2

1 2 3

1 2 3

1 2 3

2 3 4

1 2 3

3 4 5

.

.

.

1 2 3

2 1

1 2 3

1 1

1 2 3

1 2

n n n

n n n

n n n

+ − + + − + + − + + −

− − + + − − + + + − + +

1 1 2 3 3

1 12 2 1 1 2n n n

= + + + − − + + +

1 5 1 3

2 2 1 2n n

= − − + +

5 1 3

4 2( 1) 2( 2)n n= − −

+ +

(iii) 2 1

2 1 2 1 1

( 1)( 2) ( 1)( 2) 2

n n

r r

r r

r r r r r r= =

+ += −∑ ∑

+ + + +

2 1

2 1 2 1 1 5 1 3lim

( 1)( 2) ( 1)( 2) 2 4 2 4

n

nr r

r r

r r r r r r

∞

→∞= =

+ += − = − =∑ ∑ + + + +

5 (i) 3 2cos 3 cos sin

dxx a a

dθ θ θ

θ= ⇒ = −

sgfreepapers.com 80

3 2sin 3 sin cosdy

y a ad

θ θ θθ

= ⇒ =

2

2

3 sin cos sintan

3 cos sin cos

dydy ad

dxdx ad

θ θ θθ θθ θ θ

θ

= = = − = −−

(ii) The coordinates of the point with parameter θ are ( )3 3cos , sina aθ θ .

The equation of the tangent at this point is

( )3 3sin tan cosy a x aθ θ θ− = − −

( )

3 3

3 2

2 2

tan sin tan cos

tan sin sin cos

tan sin sin cos

y x a a

x a a

x a

θ θ θ θ

θ θ θ θ

θ θ θ θ

= − + +

= − + +

= − + +

i.e. tan siny x aθ θ= − +

When 0x = , siny a θ=

When 0y = , sin

costan

ax a

θθ

θ= =

Hence the coordinates of P and Q are ( )cos ,0a θ and ( )0, sina θ respectively.

(iii) If OPQ forms an isosceles triangle, then OP OQ=

cos sina aθ θ⇒ =

tan 1θ⇒ =

4

πθ⇒ =

Area of triangle OPQ

( )( )1

2

1cos sin

2 4 4

OP OQ

a aπ π

=

=

22 units

4

a=

6(i) Given that

14

12 −

=r

u r for r = 1, 2, 3, 4, …

When n = 1, 3

1

1)1(4

121 =

−=u

sgfreepapers.com 81

When n = 2, 15

1

1)2(4

122 =

−=u

When n = 3, 35

1

1)3(4

123 =

−=u

When n = 4, 63

1

1)4(4

124 =

−=u

3

11

1

=∑=r

ru

5

2

15

1

3

12

1

=+=∑=r

ru

7

3

35

1

5

23

1

=+=∑=r

ru

9

4

63

1

7

34

1

=+=∑=r

ru

6(ii) Conjecture:

121 +=∑

= n

nu

n

r

r for n = 1, 2, 3, 4, …

Let P(n) be the proposition 121 +

=∑= n

nu

n

r

r for all positive integers n.

When n = 1, LHS = 3

11 =u

RHS = 3

1

1)1(2

1=

+ = LHS

Hence P(1) is true.

Assume that P(k) is true for some k+∈� , ie

121 +=∑

= k

ku

k

r

r

Need to prove that P(k+1) is true, ie to prove that 32

11

1 +

+=∑

+

= k

ku

k

r

r.

LHS = ∑+

=

1

1

k

r

ru

sgfreepapers.com 82

= 1

1

+=

+∑ k

k

r

r uu

= 1)1(4

1

12 2 −++

+ kk

k

= 1)12(4

1

12 2 −+++

+ kkk

k

= 384

1

12 2 +++

+ kkk

k

= )32)(12(

1

12 +++

+ kkk

k

= )32)(12(

1)32(

++

++

kk

kk

= )32)(12(

132 2

++

++

kk

kk

= )32)(12(

)1)(12(

++

++

kk

kk

= 32

1

+

+

k

k

= RHS

Hence P(1) is true, and that P(k) is true implies P(k+1) is also true, hence by mathematical

induction, P(n) is true for all positive integers n.

sgfreepapers.com 83

7(a) 1 1

2 5x x≥

− − where 2 , 5x ≠ .

Method 1 : Graphical method :

From the graph

Ans: 2 < x < 5

Method 2 : Analytical method:

1 1

2 5x x≥

− − ⇒ ( ) ( ) ( ) ( )

2 25 2 2 5x x x x− − ≥ − −

⇒ ( )( ) ( ) ( )5 2 5 2 0x x x x− − − − − ≥

⇒ ( )( )3 5 2 0x x− − − ≥

⇒ ( )( )5 2 0x x− − ≤

⇒ 2 < x < 5 as 2 , 5x ≠

Method 3: Analytical method

1 1 1 10

2 5 2 5x x x x≥ ⇒ − ≥

− − − −

5 20

2 5

x x

x x

− −− ≥

− −

sgfreepapers.com 84

( )( )

( )( )

30

2 5

3 2 5 0

( 2)( 5) 0

x x

x x

x x

−≥

− −

− − − ≥

− − ≤

1 1

2 5x x≥

− − ⇒ 2 5x< <

⇒ 2 x< and 5x <

⇒ Ans: -5 < x < - 2 or 2 < x < 5

(b) log 3 1x > − ⇒

lg 31

lg x> −

Method 1 : Graphical

From the graph,

lg 3

lgy

x= is above 1y = − for

10 or 1

3x x< < >

Method 2 : Analytical

lg 31

lg x> − ⇒ ( )( ) ( )

2lg lg 3 lgx x> −

⇒ ( )[ ]lg lg3 lg 0x x+ >

-5 -2 2 5

y = -1

lg 3

lgy

x=

1

3

x= 1

sgfreepapers.com 85

⇒ ( )lg lg3 or lg 0x x< − >

⇒ 1

0 or 13

x x< < >

8(a)

(i)

(ii)

(b) Alfred is incorrect. The incorrect step was Step 4 “Translate along the x-axis by factor ½”

The equation of the final graph resulting from the incorrect transformations is ( )2

5 2 2 3y x= + −

1− 1

2x =

1(3, )

3

1

2y =

y

x

2x = −

3− 1

(2,3)( 4,3)−

2y =

y

x

-lg3 0

sgfreepapers.com 86

9(i)

1

5 1

: 1 2 ,

7

l

k

λ λ

= + ∈

r �

(ii)

2

zx y= − =

Converting to Vector Equation Form:

Let x µ=

Let y yµ µ− = ⇒ = −

Let 22

zzµ µ= ⇒ =

2

0 1

: 0 1 ,

0 2

l µ µ

= + − ∈

r �

Since l1 is perpendicular to l2

1 11

2 1 0 1 2 2 02

2

k k

k

− = ⇒ − + = ⇒ =

i

(iii) 5 1 1

1 2 1

7 1/ 2 2

λ µ

+ = −

5

2 1

1/ 2 2 7

λ µ

λ µ

λ µ

− = −

+ = −

− = −

Using GC, 2, 3λ µ= − =

Point of intersection, N (3,-3,6)

Shortest distance of point A to l2 = AN��

sgfreepapers.com 87

2 2 2

3 5

3 1

6 7

2

4

1

( 2) ( 4) ( 1)

21

= − −

−

= − −

= − + − + −

=

(iv)

2

4

1

OC NA AN

= = − =

��

Area of ONAC= Length x Breath =

2 3

4 3

1 6

× −

2 2 2 2 2 22 4 1 3 ( 3) 6

21 54

9 14

= + + × + − +

= ×

=

Alternative Method

Area of ONAC

A (5,1,7)

O

l2

l1 N

C

sgfreepapers.com 88

2 3

4 3

1 6

27

9

18

1134

9 14

OC ON= ×

= × −

= − −

=

=

��

(v)

3 6

' 2 2 3 6

6 12

OO ON

= = − = −

��

Alternative Method

Using mid-point theorem

( )1'

2AN AO AO= +��

A (5,1,7)

O

l2

l1 N

C

O’

sgfreepapers.com 89

' 2

2 5

2 4 1

1 7

1

7

5

AO AN AO= −

− −

= − − − − −

= −

��

' '

' '

1 5

7 1

5 7

6

6

12

AO OO OA

OO AO OA

= −

= +

= − +

= −

��

��

1

6 2

: 6 4 ,

12 1

l t t

= − + ∈

r �

(taking direction vector as OC��

)

10

a(i)

Amount of food Peter consumes each day, in grams, follows a geometric progression with first

term 1000 and common ratio 0.98.

Amount of food Peter consumes in Day x is

10.91000 8x−× .

When ( )1

1000 0.98 500x−

< ,

sgfreepapers.com 90

1(0.98) 0.5

( 1) ln 0.98 ln 0.5

ln 0.5

ln 0.98

35.3 (3s. )

1

f.

x

x

x

x

− <

>

− <

− >

The first day Peter consumes less than 500 grams of food is Day 36, i.e. 36n = .

(Students may also use GC directly to find 36n = )

Total amount of food consumed from Day 1 to Day n , in grams is

( )( )361000 1 0.98

25800 (3 s.f .)1 0.98

−=

−

a(ii) Amount of food Samuel consumes each day, in grams, follows an arithmetic progression with first

term 6000 and common difference 10− .

Amount of food Samuel consumes in Day x is

( )600 10 1x− − .

When ( ) ( )( )11000 10 1 0.61 1000 0.98

xx

−− − ≥ ×

( )

( )( )1

600 10 10.61.

1000 0.98x

x

−

− −≥

×

sgfreepapers.com 91

From GC, m can be any integer between 8 to 15 (inclusive)

b(i) The top row of T consists of an arithmetic progression of 200 terms, with first term 50 and

common difference 5.

Therefore, the sum of all terms in the first row of T is

( ) ( )( )( )200

2 50 200 1 5 109500.2

+ − =

b(ii) Let iS be the sum of all terms of row i (from the top)

The sequence 1 2 3 4 100, , , , ,S S S S S… is an arithmetic progression of 100 terms, with first term

109500 and common difference 20 63 0 00.× = −−

Therefore, the sum of all terms of is

1 2 3 100

100(2(109500) (100 1)( 600))

2

7980000.

S SS S …+

+ − −

=

+ + +

=

sgfreepapers.com 92


Page 1 of 6



Name: __________________ Total marks: ____/ 72


1 The graph of a cubic function passes through the points ( )6,0− and ( )0,4 , and has turning

points at 2x = − and 1x = .

Write down and solve a system of simultaneous linear equations to find the equation of the

graph. [4]

2 Differentiate the following with respect to x.

(i) ( )1 3cos sin where

2 2x x

π π− < < , [2]

(ii) e 1

ln1 e

x

x−

+

−. [3]

3 Expand ( )

2

3

3 2

x

x− in ascending powers of x up to and including the term in 3

x .

Find the coefficient of nx . [5]

4 (i) Show that 2 2 4 2

1 1 4

2 4 2 4 4 16

r

r r r r r r− =

− + + + + +. [1]

(ii) Find 4 2

1 4 16

N

r

r

r r= + +∑ , leaving your answer in terms of N. [3]

(iii) Deduce the value of 4 2

1 4 16r

r

r r

∞

= + +∑ . [1]

5 Without the use of a graphing calculator, solve the inequality 2

2

2 7 61

2

x x

x x

− +≥

− −. [3]

Deduce the range of values of x such that

(a)

2

2

2(ln ) 7(ln ) 61

(ln ) (ln ) 2

x x

x x

− +>

− −, [2]

(b)

2

2

2 7 61

1 2

x x

x x

− +≥

− −. [2]

sgfreepapers.com 93


Page 2 of 6

6 A curve has equation 1

2

axy

x

−=

− where a is a constant such that

1

2a > .

(i) Sketch the curve 1

2

axy

x

−=

−, indicating clearly the coordinates of any axial intercepts

and the equations of any asymptotes, in terms of a. [3]

(ii) Sketch, on separate diagrams, the curves with the following equations:

(a) 2 1

2

axy

x

−=

− [3]

(b) 1

2

axy

x

+=

+ [2]

In your graphs, indicate clearly the coordinates of any axial intercepts and the equations

of any asymptotes, in terms of a.

7 A student wants to build up his stamina for the 2.4 km run in the National Physical Fitness

Assessment (NAPFA) test. Knowing that he has a knee problem, he decides to combine

swimming and jogging on a daily basis to build up his stamina before the test. His plan is as

follows:

Day Jog Swim

1 200m 200m

2 onwards Increase by 20% of the

distance jogged the previous

day, capped at 400 m per

day

Add 50 m

to the distance swam

the previous day

What distance will he have swum and jogged in total after 20 days under this training routine?

[3]

The student hopes to be able to cover a total distance of 2.4 km in one day. On which day will

he first be able to do this? [2]

8 The diagram below shows a rectangle of height x m and width y m inscribed in an equilateral

triangle of side a m.

(i) Show that ( )3

2x a y= − . [1]

(ii) Hence find the maximum area of the rectangle. [4]

x

a y

sgfreepapers.com 94


Page 3 of 6

9 The diagram shows the graph of ( )fy x= . The curve has turning points at ( )2,5− and ( )3,0

and an asymptote 3y = .

Sketch , on separate clearly labelled diagrams, the graphs of

(i) ( )

1

fy

x= , [3]

(ii) ( )fy x′= . [3]

10 The parametric equations of a curve are 2e , 1.t

x y t−= = +

(i) Find the equation of the tangent to the curve at the point ( )2e , 1

pp

−+ , expressing y in

terms of x. [3]

(ii) This tangent meets the x- and y-axes at points A and B respectively. Find the area of

triangle OAB , leaving your answer in terms of p. [3]

11 (a) By using standard series expansions for ex and sin x , find the first three non-zero terms

in the expansion of e sin( )xx π+ . [3]

(b) The curve C is defined by the equation 3 2 3y y y x+ + = .

(i) Show that there is only one point of intersection between C and the line 2y x= − ,

and find the coordinates of the point of intersection. [3]

(ii) Find the Maclaurin’s series for y, where y satisfies the equation 3 2 3y y y x+ + = ,

up to and including the term in 2x . [3]

x

3y = ( )fy x=

y

O

sgfreepapers.com 95


Page 4 of 6

12 A sequence of real numbers 1 2 3, , ,u u u … satisfies the recurrence relation

1

( 3)

2 2

n

n

n

n uu

u n+

+=

+ +.

As n → ∞ , nu L→ .

(i) Find the exact value(s) of L. [2]

(ii) For the case where 1 3u = , prove by induction that, for 1,n ≥

2

2 1n

nu

n

+=

−. [4]

Write down the limit of this sequence. [1]

Qn/

No Topic Set Answers 1 Equations 3 22 1 4

463 21 21

y x x x= + − +

2 Differentiation (i) 1; (ii)

1 1

2 1 1

x

x x

e

e e

−

+ −

3 Binomial Theorem 2 31 4 4

3 9 9x x x+ + ;

11 2 1 2

or 3 3 2 3

n n

n n

−

4 Method of Difference Summation of Series

(ii) 2 2

7 1 1

48 4( 3) 4( 2 4)N N N− −

+ + + ; (iii)

7

48

5 Iinequality 1 or 4x x< − ≥

(a) 1 40 or x e x e

−< < >

(b)

11

4x− < ≤

6 Graphing Techniques (i)

1

2

axy

x

−=

−

1,0

a

( )0, 0.5 y a=

2x =

x

y

O

sgfreepapers.com 96


Page 5 of 6

(ii)(a)

(ii)(b)

7 Arithmetic and Geometric Series 20973.6 m; 37

th day

8 Differentiation (ii) 23

.8

a

9 Differentiation Graphing Techniques

(i)

2 1

2

axy

x

−=

−

1,0

a

10,2

y a=

2x =

x

y

O

y a= − 10,

2

−

1

2

axy

x

+=

+

1,0

a

− ( )0, 0.5

y a=

2x = −

x

y

O

x

1

3y =

3x =

O 12,

5

−

( )1

fy

x=

sgfreepapers.com 97


Page 6 of 6

(ii)

10 Parametric Equations Differentiation

(i) ( )2

2 1py pe x p= − + +

(ii)

( )4

1

4 p

p

pe

+

11 Maclaurin’s Series (a)

32

3

xx x− − − −�

(b)(i) (1, 1), (ii)

23 9y x x= −

12 Recurrence Relation Mathematical Induction

(i) 1

0 or 2

L L= = , (ii) 1

2

0y =

( )fy x′=

( )3,0

( )2,0−

sgfreepapers.com 98

1

Promo Revision Practice Paper 7

Qn Possible Solution Considerations

1 Let y = ax3 + bx

2 + cx + d .

Then

216 36 6 0 from 6, 0

4 from 0, 4

a b c d x y

d x y

− + − + = = − =

= = =

2d3 2

d

yax bx c

x= + +

12 4 0

3 2 0

a b c

a b c

− + =

+ + =

From GC, 2 1 4

, , 63 21 21

a b c= = = −

Equation is 3 22 1 44

63 21 21y x x x= + − +

“Cubic function” means the

function is of the form

y = ax3 + bx

2 + cx + d

where a, b, c and d are constants.

“Turning point(s)” so use

dy/dx = 0

“Find the equation …” so

must write “y = f(x)” form

at the end.

2 (i)

( )( )( ) ( )

( )

1

2 2

2

d 1 coscos sin cos

d 1 sin cos

3cos 0 when ,

2 2 1

cos cos

xx x

x x x

x x

x x

π π

− −= − ⋅ =

−

< < <

= ∴ = −

∵

Need to ask why the

question gave a domain for

x. In this case, you use the

domain to simplify the

answer.

2 (ii)

( ) ( )

( ) ( )

1 1ln ln 1 ln 1

1 2

1 1ln 1 ln 1

2 2 1 1

1 1

2 1 1

xx x

x

x xx x

x x

x

x x

ee e

e

d e ee e

dx e e

e

e e

−−

−−

−

+ = + − − −

+ − − = − + −

= − + −

For expressions involving

logs, try to simplify first.

Makes it easier to do latter

part(s).

3

( )( ) ( )

22 2

2

2

2 3

3 23 3 2 3 3 1

3 2

41 4

3 3

1 4 4

3 9 9

xx x x

xx

xx x

x x x

−− −

= − = − −

= + + +

≈ + +

…

“ascending powers of x

…” so want x, x2, x

3, etc.

“powers of x” so check

Maclaurin’s Expansion and

Binomial Expansion in

MF15.

Note that for n ≠ positive

integer, you must make

the first term in the bracket into “1” before

applying the Binomial

formula.

sgfreepapers.com 99

2

3 Coefficient of nx :

( )( ) ( )( )( )

( )

1

1

1

2 3 2 1 11 2

3 1 ! 3

1 2 3 4 2

3 1 ! 3

1 2 1 2 or

3 3 2 3

n

n

n n

n

n

n

n

n n

−

−

−

− − − − − + − −

⋅ ⋅ ⋅ ⋅ = −

=

……

…

For “finding coefficient”,

either

write out enough terms to

identify a pattern

or

use the general term in

MF15, ( 1) ( 1)

(1 ) 1!

n rn n n rx x

r

− − ++ = + + +

…… …

4 (i) 2 2

2 2

4 3 2 3 2 2

4 2

1 1

2 4 2 4

2 4 ( 2 4)

2 4 2 4 8 4 8 16

4

4 16

r r r r

r r r r

r r r r r r r r

r

r r

−− + + +

+ + − − +=

+ + − − − + + +

=+ +

“Verify …” means that you

must show all relevant

working. For “verify”, it is

mostly “LHS = … = RHS”

or “RHS = … = LHS” type

of working.

4 (ii)

4 2 21 1

2 2

2 2

2 2

1 1 1

44 16 2 4 2 4

1 1

3 7

1 1

4 12

1 1

7 191

...4

1 1

6 12 2 4

1 1

7 3

1 1

2 4 2 4

N N

r r

r

r r r r r r

N N N N

N N

N N N N

= =

= −

+ + − + + +

−

+ −

+ −

= + + − − + − +

+ − + + + − − + + +

∑ ∑

2 2

1 1 1 1 1

4 3 4 3 2 4N N N

= + − −

+ + +

2 2

7 1 1

48 4( 3) 4( 2 4)N N N= − −

+ + +

Finding sum involving

factorial or fraction � if

you can see a minus “−−−−” in

the expression, the working

will most likely involve

“method of differences”.

Step1: Write enough terms

to identify how the terms

cancel each other out.

Step 2: Stack up the terms

for easy checking and cancellation.

Step 3: Write enough terms

at the end to help identify

which terms are left at the

end.

Look for a pattern in the

cancellation � easier to

see whether you are correct

or not.

4(iii) 2 2

1 1As , 0, 0

2 4 3N

N N N→ ∞ → →

+ + +

Hence 4

1

7

484 16r

r

r r

∞

==

+ +∑

This is the proper

presentation!

sgfreepapers.com 100

3


5 2 2

2

2

2

2 7 6 ( 2)0

2

6 80

2

( 4)( 2)0

( 2)( 1)

x x x x

x x

x x

x x

x x

x x

− + − − −≥

− −

− +≥

− −− −

≥− +

1 or 4x x< − ≥

Step 1: For polynomials,

add or subtract terms to

make one side zero.

Step 2: Combine into a

single fraction.

Step 3: Factorise (or find

roots) and plot roots on a

number line (or x-axis).

Step 4: Test region and

write answer.

5 (a)

(b)

Replace ‘x’ by ‘ln x’ ⇒ ln 1 or ln 4x x< − >

1 4 0 e or e x x−⇒ < < >

Alternatively,

Asymptote at x = 0.

Exact intersections at lnx = −1 and lnx = 4

⇒ x = e−1

and x = e4 ⇒ overall get

1 40 e or e x x−< < >

Replace x by 1

x

1 11 or 4

x x⇒ < − ≥

1

1 0 or 04

x x⇒ − < < < ≤

Note that 0 is possible for (b), so overall answer is 1

14

x− < ≤

Alternatively,

Asymptotes at x = 0, y = 0.

Exact intersections at 1/x = −1 and 1/x = 4

⇒ x = −1 and x = 1/4 ⇒ overall get 1

14

x− < ≤

Easier to understand if you

sketch graph of lnx to

check inequality for lnx.

If lnx < −1, then 0 < x < e−1

because the asymptote is at

the left.

Easier to understand if you

sketch graph of 1/x to

check inequality for 1/x.

Look at graph to get

11 0 or 0

4x x− < < < ≤

−1 2 4

+ + − −


4


6 (i) 1 2 1

2 2

ax ay a

x x

− −= = +

− − Asymptotes are x = 2 and y = a

When x = 0, y = 1/2 (or 0.5)

When y = 0, ax − 1 = 0 giving x = 1/a

Have a fraction so check

highest power in

numerator and

denominator. Carry out long division if

highest power in numerator

≥ highest power in

denominator

Label coordinates (0, 0.5) and (1/a, 0) in the sketch,

not just mark 0.5 and 1/a

on the axes.

Label equations of

asymptotes in the sketch.

6 (ii) (a)

On rough paper, sketch

(1) y = √f(x) first (2) reflect about y-axis to

get y = −√f(x)

(3) copy graph to answer

script but smoothen the

curve at the x-

axis (cannot be a

sharp point)

In step (1), new y-values =

√(all original y-values)

6 (ii) (b)

Carry out the same process

as in 7(i):

1 1 2

2 2

ax ay a

x x

+ −= = +

+ + so asymptotes are x = −2

and y = a

7

Day Jogging distance Swimming distance

1 200 200

2 1.2(200) = 240 200 + 50 = 250

3 1.22(200) = 288 250 + 50 = 300

4 1.23(200) = 345.6 300 + 50 = 350

5 exceeds 400 350 + 50 = 400

GP for “jogging” AP for “swimming”

for days 1 to 4 only

List out a few terms in a

neat form (preferable table

format) to look for pattern.

Since there is a cap of 400

metres, you may want to

check the number of days

to reach the cap by using a

GC.

2 1

2

axy

x

−=

−

1,0

a

10,2

y a=

2x =

x

y

O

y a= − 10,2

−

1

2

axy

x

+=

+

1,0

a

−

( )0, 0.5

y a=

2x = −

x

y

O

1

2

axy

x

−=

−

1,0

a

( )0,0.5 y a=

2x =

x

y

O


5

7 From GC, the jogging will be capped at 400 m for days 5 to 20.

[ ]20total swimming distance = 2(200) (19)(50) 23500m

2+ =

4200(1.2 1)total jogging distance = 400(16) 7473.6m

1.2 1

−+ =

−

Total = 20973.6 m

The student wants to cover a total distance of 2400 metres in one

day. If he does this after day 4, we need to find which day he

covers at least 2400 – 400 = 2000 metres by swimming.

i.e. 200 ( 1)50 2000 37n n+ − ≥ ⇒ ≥

So it happens on the 37th

day of training.


8 (i)

( )

( )

( )

11 2tan 303

2 3

3

2

a y

x

x a y

x a y

−° = =

⇒ = −

⇒ = −

Equilateral triangle so all sides have the same

length, a, and the angle at

any vertex is 60°

8 (ii)

( ) 23 3 3Area

2 2 2xy a y y ay y= = − = −

Let Area = A

33

2

dAa y

dy= −

3 0 3 0

2

dAa y

dy= ⇒ − =

3,

2 4

ay x a⇒ = =

23area

8xy a⇒ = =

2

2

2

3 0 when 2

3The maximum area of the rectangle is .

8

d A ay

dy

a

= − < =

∴

∵

Note that since A is in the form A = λ y − µ y2, you may use “completing

square” to find the stationary point and you may use the “coefficient of y2

is negative” to justify “maximum” A.

Re-write to have only one

variable Area on the LHS

of the equal sign, and only

one variable y on the

RHS.

Show that this gives the

maximum value of A.

x

( )1

2a y−

30°

y a


6


9 (i)

Keep x-values the same but

replace all original y-values

with 1 ÷ (original y-value).

Check lecture notes for

summary on what to do.

9 (ii)

Move a ruler with a

transparent edge along the

original curve to estimate the

value of the gradient at

different points on the curve.

Keep x-values the same but

replace all original y-values

with the gradient.

10 (i)

2e , 1

d de , 2

d d

d 22 e

d e

t

t

t

t

x y t

x yt

t t

y tt

x

−

−

−

= = +

= − =

⇒ = = −−

( )( ) ( )( )

( )

2

2

2

2

Equation of the tangent at point e , 1 is:

1 2 e e

2 e 1 2

2 e 1

p

p p

p

p

p

y p p x

y p x p p

y p x p

−

−

+

− + = − −

= − + − +

= − + +

10(ii) ( )

( )

( )( ) ( ) ( )

2

2

2 4 4

2

When 0, 1 .

1When 0, .

2 e

1 1 11Area of 1

2 2 e 4 e 4 e

p

p p p

x y p

py x

p

p p pOAB p

p p p

= = +

+= =

+ + +∆ = + = =

Use modulus since we are not sure whether p is

positive or not.

y

x

1

3y =

3x =

O 12,

5

−

( )1

fy

x=

0y =

( )fy x′=

( )3,0 ( )2,0−


7


11 (a) ( ) [ ]

2 3 3

32

e sin e sin cos cos sin

e sin

1 ... ...2! 3! 3!

3

x x

x

x x x

x

x x xx x

xx x

π π π+ = +

= −

= − + + + + − +

= − − + +�

“standard series

expansions” so refer to

MF15 and use the

expansions given. sin(x + π) has to be re-

written as the series for sin x

works for small values of x,

and x + π is not small.

11(b) (i)

( )

( )( )

3 2

3 2

2

Curve : 3 (1)

The line : 2 (2)

Substitute (2) into (1): 3 2

1 2 6 0

C y y y x

y x

y y y y

y y y

+ + =

= −

+ + = −

− + + =

�

�

( )

( )( )

22

2

2 6 1 5 0 for all

1 2 6 0 1 0, i.e. 1

y y y y

y y y y y

+ + = + + > ∈

∴ − + + = ⇒ − = =

∵ �

Since the equation has only one solution, there is only one

point of the intersection between the curve and the line.

( )When 1, 1. The point of intersection is 1, 1 .y x= = ∴

“intersection” so solve

simultaneously.

Write statement(s) to explain

the significance or the

reasoning of your

calculation(s).

11(b) (ii)

( )

( )

3 2

2

2

22

2

3

Differentiate both sides w.r.t :

3 2 3

3 2 1 3

Differentiate both sides w.r.t :

3 2 1 6 2 0

y y y x

x

dy dy dyy y

dx dx dx

dyy y

dx

x

d y dy dy dyy y y

dx dx dx dx

+ + =

+ + =

+ + =

+ + + + =

2

2When 0, 0, 3, 18

dy d yx y

dx dx= = = = −

2

2

2

180 3

2!

The Maclaurin's series for up to and including the term in

is: 3 9

y x x

y x

y x x

−= + + +

= −

�

Space out your work so that

it is easier to keep track of

what you have written.

List out the values neatly so

that it is easier to use them in

the formula in MF15.


8


12 (i) Since as n → ∞ , nu L→ ,

2( 3) 2 2 3

2 2

n LL L nL L nL L

L n

+= ⇒ + + = +

+ +

Solving for L, 22 0L L− =

( )2 1 0L L⇒ − =

1

0 or 2

L L⇒ = =

12 (ii) Let ( )P n be the statement “

2

2 1n

nu

n

+=

−” where n +∈� .

Need to verify ( )P 1 is true:

LHS of ( ) 1P 1 3u= = (given)

RHS of ( ) 1 2P 1 3

2 1

+= =

−= LHS

Therefore, ( )P 1 is true.

Assume ( )P k is true for some positive integer k, i.e.

assume 2

2 1k

ku

k

+=

− is true for some positive integer k.

Need to show ( )P 1k + is true, so show that 1

3

2 1k

ku

k+

+=

+

LHS of ( ) 1P 1 kk u ++ =

( 3)

2 2

k

k

k u

u k

+=

+ +

2( 3)

2 1

22 2

2 1

kk

k

kk

k

+ + − =+ + + −

by assumption on ( )P k

( )( ) ( )( )

( 3) 2

2 2 2 2 1

k k

k k k

+ +=

+ + + −

( )( )3 3

RHS of P 12 2 1 2 1

k kk

k k

+ += = = +

+ − +.

Therefore ( )P 1k + is true.

Since ( )P k is true implies ( )P 1k + is true, and ( )P 1 is true, by

the principle of mathematical induction, ( )P n is true for all

positive integers n.

The limit of this sequence is 1

2.

Name the proposition first.

Write proper statement,

explain that LHS = RHS.

Show all working.

Write a conclusion.

Write out the assumption

as you will need it later.

Write out what you need to

show so that you know what

to aim for.

Use the recurrence relation

here to link uk+1 to uk.

Use the assumption here to remove uk.

Make sure this matches

your aim (written earlier).

Write a proper overall conclusion.


H2 Promo Revision Paper 1 IJC Promo 2012(2) 2 H2 Promo … · 2019-09-07 · H2_Promo Revision...

Documents

Transcript of H2 Promo Revision Paper 1 IJC Promo 2012(2) 2 H2 Promo … · 2019-09-07 · H2_Promo Revision...