ACJC 2013 JC2 H2 Mathematics REVISION SET G

COMPLETE SOLUTIONS

Part I

FUNCTIONS

1 (i)

Since Rf = (0, 2] = Dg, gf exists.

Method 1 (mapping method)

(−1, 1) (0, 2] (−, ln 2] = Range of gf

Method 2 (graphical method)

gf(x) = 2ln 2 1 x , −1 < x < 1

Range of gf = (−, ln 2]

1 (ii) A translation of 1 unit along the positive x-axis [so y = f(x 1)]

followed by a scaling of factor 1

2 along the x-axis. [so y = f(2x 1)]

OR, “A scaling of factor 1

2 along the x-axis [so y = f(2x)]

followed by a translation of 1

2unit along the positive x-axis.” [y = f(2{x

1

2})]

Note the different values for translation depending whether you scale or translate first.

2 (i)

(ii)

Greatest value of k is – 2

2 2 22 2

e e 2 ln( )

2 ln( ) 2 ln( )

Since 2, 2 ln( )

x xy y x y

x y x y

x x y

This gives 1f : 2 ln( ), , 1x x x x

−1 1

y

x

2

y = f(x)

f g

−1 1

y

x

ln2 y = gf(x)

2 (iii)

2 (iv) 1 gf , 2 andR D ℝ

1

1

gf Since , gf exists.R D

1f 1 1 1

g

gf f gf , 1 , 2 0,D D R

3 (i)

(ii) 1 2

let 2

xy

x

1

2 1 2

2 1 2

1 2

2

1 2f : , 2

2

xy y x

x y y

yx

y

xx x

x

3 (iii) ,g

R ,2f

D

Since , ,2g f

R D , fgdoesnot exist

2,f

R 3,g

D

Since 2, 3,f g

R D , gf exists

1 2 1 2 1 2 3 6 5

gf g ln 3 ln ln , 22 2 2 2

x x x x xx x

x x x x

x

y

O

(0, 1/2)

(1/2, 0) O

4

5 (a) (i)

1

2

1f '( ) ( 1) sin

1x x x

x

For 1

02

x , f '( ) 0 f ( )x x is a strictly increasing function

f is a one-one function the inverse function f -1

exists.

Let f -1

(1) = x, then f(x) = 1

1 1

1

( 1)sin 1 1 ( 1)sin 0

1 0 or sin 0

1 (rejected since 0) or 0

x x x x

x x

x x x

1f (1) 0

(ii) At 110, f '(0) 1 sin 0 1 and f (0) 1

1 0x y

Equation of tangent to the curve y = f(x) at x = 0 is y = x +1

At x = 1, using symmetric property of the curve of f and f -1

along y = x,

equation of tangent to the curve y = f-1

(x) at x = 1: x = y +1 y = x – 1

5 (b) Range of g = [11, 5]

6 (i) Range of f is {y ℝ: y ≤ 3}

(ii) The line y = k (k ≤ 3) cuts the graph only once, so the function f is one-one.

Hence f-1

exists.

Let 4 secy x

1

sec 4cos

x yx

-1 1 1

f : cos ( ) , 34

x xx

(iii) Graph is shown on the right.

(iv) Using GC,

x = 2.14 (3 sf) or x = 2.99 (3 sf)

7 (a) 1

1 1

fff is one-one f exists. R D ff exists.

(b) (i) 2gh : ln 2 2 , 0x x x x

ghR [0, )

7 (b) (ii) Domain of g-1

g = Domain of g = [1, )

and y = g-1

g(x) = x

Domain of gg-1

= Domain of g-1

= Range of g = [0, )

and y = gg-1

(x) = x

x

11

5 y=g(x)

0

(π, 3)

x =

(3, π)

y = 2

(iii) x ≥ 1

8 (a)

(b)

(c)

2

1

Let 2 1 2. Then 2 2 1

1 1Since 1, 2

2 2

1 1f : 2, 7

2 2

y x y x

x x y

x x x

,7fR , ,1gD . Since gf DR , gf exists.

f g, 1 7, ln 7 ,a

Domain is x > 1 and range is y > 0 1 a = 1 a = 2

9 (a) (i) 2f ( ) 2 f '( ) 2 0 x x x x x min. point at

2x

Inverse of f exists when 2 42

Alternatively, by completing the square,

2 22f ( ) 2 2

2 4x x x x

Inverse of f exists when 2 4.2

(ii) 22Let 4 2 2 2. Then 2 2y x x x x y

1f : 2 2, 2x x x

x

y

x O

(1, 0)

(1,1) (0, 1) g( )y x

1g ( )y x

1g g( )y x

8

6

4

2

-2

-4

-6

-8

-15 -10 -5 5 10 15

x a

1 a

gy x

9 (b) (i)

2 22g( ) 1 1

2 4

k kx x kx x

2

R 1 ,4

g

k

g hhg exists if R D

2

2

Thus, 1 0 4

4 0

k

k

k

(ii) g

3For 1, R ,

4k

3( , ) , [4, )

4

hgR [4, )

Alternatively,

hgR [4, )

10 (i)

2f

xx

x

12 2 2

'

3 22 2

12

2f 0

x x x x

xx x

f is an increasing function f(x) increases as x increases.

(ii) domain of 1f range of f = ,0 .

2

2 2 2

2 2 2

2

2

1

1

xy

x

y x x

x y y

yx

y

Since 0,x 0y . Therefore, 21

yx

y

.

1

2f ( )

1

xx

x

, for 0.x

(iii)

1 1

2f f ( )

1 1

xx

x

2

2

0.50.5

1 1 0.5

4 5 0

4 5 1 0

Since 0, 5

4

Alternatively,

1 1

1

f f ( 0.5) 0.5

f ( 0.5) f 0.5

2 2

2

0.5 0.5

1 0.5 0.5

4 5 0

4 5 1 0

Since 0, 5

4

11 (i)

(ii)

(iii)

fR ( , 4 ]a a b

2

1

Let f ( ) . Then

1Since ,

2

f ( ) , 4

b bx y a y x

x y a

bx x

y a

bx a x a b

x a

2

1

2g( ) gf f

bx x

b

x a

g( ) , 4x ab bx a x a b

12

(i) Least value of k = 1.5

(ii) Let 22 3y x x

21 25

2( )4 8

y x 1 8 25

4 16

yx

1 1 8 25f ( ) , 0

4 16

xx x

(iii) 22 3x x x

1 7

2x

13 Let (1 )xy a e

1

ln 1

x

y

a

ye

a

x

h 1 (x) = ln 1xa

and 1D R ( ,2 ]hha a

0D R ( ,2 ]h h a a

Since 1h ( )y x can be obtained by reflecting h( )y x about the line y = x, the

point of intersection of h( )y x and 1h ( )y x is at y = x = b.

0 0 (1 ) d (1 ) d

b bx ya e x a e y I .

Area bounded by 1h( ), h ( )y x y x and the axes

2

0 0 (1 ) d (1 ) d ( ) 2

b bx ya e x a e y b b I b

Alternatively, Area OAB = 2 2

0 00

h( ) d d2 2

bb b x b

x x x x I I

Since h( )y x and 1h ( )y x are symmetrical about y x ,

Area OAB = Area OBC = 2

2

bI .

Area bounded by 1h( ), h ( )y x y x and the axes

222 2

2

bI I b

14 (i)

h( )y x

1h ( )y x

( , )B b b

O

(0,2 )A a

(2 ,0)C a

y a

y xx a

14

15 (i) fh x exist h f , 2,R D least 2

(ii) For 1g x , let lny x y yx e x e

1g xx e x

(iii)

16 (i)

(ii)

(iii)

Since any horizontal line, y = k (k ≠ 1) cuts the graph of f only once, f is 1-1.

f 1

exists.

5 4

f 11 1

xx

x x

. Let

4 41 1

1 1y x

x y

.

1 4 4f 1 1 , \{1}

1 1x x

x x

f 1

= f .

51 50 2 1f 4 = f f 4 = f 4 f f f

5 4 1

1 4 3

x x x

22g 2 4 = 2 1 2 2x x x x

For fg to be defined, Rg Df = \{1} 2 + > 1 > 3.

g f 7

2, 2 , , 13

y =0

x =0

17 (a)

(i)

(ii)

(iii)

(iv)

1f ( ) 2 tan 2x x x '

2

4f ( ) 1

1 4x

x

3 3

2 2x 2 2

2 2

3 1 1 40 1 1 4 4 1 1 4

4 4 1 4 1 4x x

x x

Since 2

41 4

1 4x

, we have

2

41 0

1 4x

giving 'f ( ) 0x

So f ( )x decreases when x increases (shown).

f

3 2 3 2,

2 3 2 3R

or ( 1.23, 1.23)

For ff to exist, f fR D

3 2 3 2,

2 3 2 3fR

or f ( 1.23,1.23)R

3 3,

2 2fD

or g ( 0.866,0.866)R

f fR D . This means that ff does not exist.

Let 12tan 2y x x

Then, after reflection, 12tan 2x y y

17 (b) 2h( ) 2x x

2

4 2 2gh 2 5 1 4x x x x

2

2g 2 2 3 4x x

2

g 3 4x x

Alternatively, 2h( ) 2x x and

2 2

4 2 2 2 2 2 2gh 2 5 2 4 4 2 5 2 6 2 13x x x x x x x x

2g 6 13x x x

18 (a) (ii) If 1f f , then f (intersection at y = x)

2

2 4 2 4 4 4

2 24 4 4 0 3 0 2 3 0 (shown)

18 (a) (i)

Include the line y = x (in dashes) to show symmetry, and label the equation of the

line. Label the two end-points (2, 4) and (4, 2) and label the equation of both curves.

18 (b) Maximum value of k = 7

(i) Let 2

1

( 7) 4y

x

2 1( 7) 4x

y

2 17 4x

y

17 4x

y

x 1 1

7 4 or 7 4 rejected 7xy y

1 1 1g : 7 4 , , , 0

4x x x x

x

or 1g

1D ( , ] (0, )

4

(ii) gf

1R 0,

5

Method 1 By mapping, (0, 2] f (−, 4] g 10,

5

Rgf = 10,

5

.

Method 2 By using the graph of y = gf(x) with (−, 2] as domain to get

Rgf = 10,

5

.

19

19

(iii)

20 fg(x) = gf(x) f( )2ln( x ) = g( )2ln( x )

ln [ )2ln( x + 2 ] = ln [ )2ln( x + 2 ] )2ln( x = )2ln( x

y = )2ln( x y = )2ln( x

Consider )2ln()2ln( xx

2

1)2(

xx 1)2)(2( xx 14 2 x

32 x 3x (reject 3x )

Hence for fg(x) = gf(x), 0x or 3x

g(x) = )2ln( x

Note that both 1x and 1x are in the domain of g.

Clearly 1 1, but g(1) = g(1) = ln3.

Therefore, function g is not 1-1 and inverse of function of g cannot be formed.

Greatest a = 0.

Let y = g(x) = )2ln( x .

Since 0x , y = )2ln( x yex 2 yex 2 yey 2)(g 1

Hence xex 2:g 1 , ),2[ln x

GRAPHING TECHNIQUES

1. DHS 2011/P1/Q7

(i)

(ii)

(b)

O x

x = 1

y =

y =

y

O

x = 2

x

y

2

x

y

O

y

O x

y = –2

2. HCI 2011/P1/Q10

a(i)

(ii)

b(i)

The curve 2

21 1

yx

h

is an ellipse with centre at (1, 0). The horizontal

distance from its centre is 1 and its vertical distance from its centre is |h|.

The curve 23( 1) 1y x is the part of the curve in (i) that lies above the x-axis.

From observation, if 1(i.e. 1)h h , there will be exactly one point of

intersection. In order for the curves to intersect at two points, 1 or 1h h .

2

2

3( 1)y x

3( 1)y x

1,1

1, 1

x

y

3

3

1,1

1,0

x

y

O

O

x

y

1 O

1

|h|

3( 1)y x

3( 1)y x

b(ii)

3. NYJC 2011/P1/Q5

(i)

2

f ( )c b ad dax bx c

x ax b adx d x d

a = 1, d = 2 b – ad = – 1 b = 1

(ii)

2

2 2f ( ) 1 f ' 1

2 2

c cx x x

x x

.

If the graph of f ( )y x has turning points, then

2

21 0

2

c

x

2

2

21 2 2 2 2

2

cx c x c

x

Since there are no turning points, 2c .

(iii)

4 3 2 2 0x x x x 2 2 1 2 0x x x x 2 2 1 2x x x x

2

2

1 1

2

x x

x x

and sketch the curve

2

1y

x

onto the graph of f ( )y x .

There are no points of intersection of the two graphs. number of roots = 0

x

y

1 O

(0, 0.5)

(0.268, 0.464)

(3.73,6.46)

y = x – 1

x = – 2

x

y

4. PJC 2011/P1/Q10(a)(b)

(a)(i)

(a)(ii)

Let 2 2

4 4h

4 4 1 (2 1)x

x x x

.

Before C,

2 2

4 4h

1 22 1y x

xx

Let

2

4p

1 2x

x

Before B,

2 2

4 4p

2 11 2

2

xy

xx

Let

2

4g

1x

x

Before A,

2 2

4 4f =g 1

1 1y x x

xx

Note : If you do not want to complete the square for the denominator, then the

working is shown below.

Let 2

4h

4 4 1x

x x

.

Before C,

2 2

4 4h

4 4 14 4 1y x

x xx x

Let 2

4p

4 4 1x

x x

O (3, 1/3)

1

O 1 1

Before B, 2 2

4 4p

2 2 14 4 1

2 2

xy

x xx x

Let 2

4g

2 1x

x x

Before A,

2 2

4 4f =g 1

1 2 1 1y x x

xx x

5. RJC 2011/P1/Q10

(i)

3 2139 399 ,

10y x x x so 2d 3 3

26 133 7 19 .d 10 10

yx x x x

x

Thus d

0d

y

x when 7x or 19,x and the stationary points are 7,122.5 and

19,36.1 .

2

2

d 313 ,

d 5

yx

x so

2

2

d0

d

y

x at 7,122.5 and

2

2

d0

d

y

x at 19,36.1 .

Thus 7,122.5 is a maximum point and 19,36.1 is a minimum point.

(ii)

x

y

O

(iii)

(iv)

The volume of water in the vase when the depth is ,h ,V is given by

3 2

039 399 d ,

10

h

V x x x x

so

3 2 3 2

0

d39 399 d 39 399 .

10

d

1d d 0

h

x x x x h hV

hh

h

Since d d d

,d d d

V V h

t h t we have

3 23 2 39 39939 399

10

d 10 100d d.

d dd h h hh h

h V V

t hth

When 19h we get d 100

0.0882d 361

h

t (3 s.f.).

6. RVHS 2011/P1/Q10

(i) vertical asymptote at 1 0x x c when 1x so 1c 2 5

1

ax bxy

x

2 2

2 2

1 2 5 1 2 5

1 1

x ax b ax bx ax ax bdy

dx x x

Since there is a turning point on the y-axis, we have 0dy

dx when 0x

2 2 5 0ax ax b when 0x 0 0 5 0 5b b

(ii) 2 5 5

1

ax xy

x

C has no x-intercept2 5 5

01

ax x

x

has no real roots

2 5 5 0ax x has no real roots 25 4 5 0a

25 20 0a 5

4a (shown)

x

y

O

(iii) 2 5 55

1 1

ax x ay ax a

x x

2

2

2 5 5

1

ax axdy

dx x

20 2 0 0 or 2ax ax x x

(iv) Add the line 1y ax . It has the same gradient as the oblique asymptote of C, but

with a smaller y-intercept.

Solving for intersection between C and 1y ax :

5 1 4 0 4 11 1

44 4 4 4

4

a aax a ax a a x a

x x

x ax a a a x xa

Hence set of values of x is4

: 1 or 4

x x ,x xa

R .

(v)

O x

y

y

O

x

7. TJC 2011/P1/Q11

(a)

(b)(i) Horizontal asymptote: y = 0

Vertical asymptotes: x = 0, x = −3

(ii) 2

2 2

3 2 3d

d ( 3)

x x x p xy

x x x

2 2

2 2

3 2 3 2 3

( 3)

x x x x px p

x x

2

2 2

2 3

( 3)

x px p

x x

= 0

2

2 2

2 30

( 3)

x px p

x x

2 2 3 0x px p

Since C has two stationary points, discriminant > 0 24 12 0p p

3 0p p

0 or 3p p (shown)

(iii)

1 3 5 7

−2

x

y

O

x = 1 x = 7

y = 0

y = 0

x = 0 x = −3

( −6, ) ●

● ( −2, − 1)

(−4, 0)

y

x

●

O

8. VJC 2011/P1/Q13

EQUATIONS and INEQUALITIES

1. HCI 2011/P1/Q1

1 Let x , y and z be the number of red, blue and green matchsticks respectively.

88 1

6 6 0 26

4 6 2 0 33 6 4

x y z

zx y x y z

y z xx y z

Solving the inequality, 16x , 24y and 48z

Number of blue matchsticks = 24

2. MJC 2011/P1/Q6(b)

(b)

Let 2f x ax bx c substitute 13

1,2

and 1

2,2

into fy x to obtain:

13

...... 12

a b c 1

4 2 ...... 22

a b c

3 2

1

12

0

03 2

19 19 d

6 6

ax bxcxax bx c x

19

...... 33 2 6

a bc

Using G.C. to solve equations (1), (2) and (3): 1

2, 4,2

a b c

Therefore equation of the curve is 2 1

2 42

y x x

3. NJC 2011/P1/Q1

3 Let $x, $y, and $z be the sales price of a bottle of soy sauce, oyster sauce and chilli

sauce respectively.

600 400 350 3990

450 320 250 3051

400 360 280 3164

x y z

x y z

x y z

From GC, x = 2.50, y = 4.30, z = 2.20.

Revenue collected per week from the Shop and Spend supermarket

= 300 2.50 220 4.30 180 2.20 $2092

4. HCI 2011/P1/Q3

4

2 2 15 1

5 1 2

x x

x x

5 2.4031 or 1 10.403 5 2.40 or 1 10.4x x x x

ALTERNATIVELY (using analytical method)

2 2

2

2 15 1 8 250

5 1 2 4 5

x x x x

x x x x

Roots to 2 8 25 0x x are 2.4031 and 10.403.

Roots to 2 4 5 0x x are 1 and 5.

5 2.4031 or 1 10.4031

5 2.40 or 1 10.4

x x

x x

Replace by 2x x

2

2

2 2 2 15 1

2 5 1 2 2

2 15 1

7 1 2

x x

x x

x x

x x

5 2 2.4031 or 1 2 10.4031

7 4.40 or 1 8.40

x x

x x

5 2.4031 1 10.4031

+ + +

5. JJC 2011/P1/Q2

From GC, x-coordinate of Points of Intersections:

1

4 and

14

2

Solution: 1 1

or 44 2

x x

Alternatively

93 2 1 5, Ans:

2

3 2 1 5, Ans: No Soln

13 2 1 5, Ans:

4

73 2 1 5, Ans:

2

x x x x

x x x

x x x x

x x x x

Solution:

6. RJC 2011/P1/Q1

6(a)

From the graph, x > 1

4.

1 1 or 4

4 2x x

y

4

x 0

y = x + 5

y

x O 1

4

y = 2x

y = x

6(b)

2

2

( 1)( 3)0

( 2)

Since ( 2) 0 for all ,

( 1)( 3) 0, 2

1 3, 2

x x

x

x x

x x x

x x

OR

2

4

2

( 1)( 3)0

( 2)

Multiply both sides by( 2) ,

( 1)( 3)( 2) 0

1 3, 2

x x

x

x

x x x

x x

7. TJC 2011/P1/Q2

To find intersection point, 2 ( 2 )x a x a =>

3

ax

From the graph, for 2 2x a x a , 3

ax

Replace x by –x and let a = 2 in the above inequality,

( ) 2(2) 2( ) 2x x becomes 4 2 2x x

Thus 2 2

3 3x x

1 3 x

1 3 2 x

2 3

2a x

y

2a

O

a

SEQUENCES and SERIES

1

2 2 22 1 1 2

1 1

1 12 2 4 2 4 4 ... 4 2 1 ... 2

2 2

4 4 1 4 4 11 12 2 3 ( 1)

2 4 1 2 6

n n nr r n n n

r n r n r n

n n n n

r r n n n

nn n n n

2 Required sum

1 2 3 100 7 17 27 77 87 97 70 71 72 77 78 79 77

100 10 10((1 100) (7 97) (70 79) 77

2 2 2 3862

3 First Method 5(1 )

931

a r

r

----- (1) and

3 5(1 )744

1

ar r

r

------ (2)

Divide (2) by (1): 3 744

93r 2r

Substitute 2r into (1) to obtain 3a

Thus 8

9 768u ar .

Second Method

Observe that 3

4 5 6 7 8 1 2 3 4 5u u u u u r u u u u u

Thus, 393 744r 3 744 / 93 8r 2r

From 5

1 2 3 4 5

(1 )93

1

a ru u u u u

r

substitute 2r

5

93(1 2)3

1 2a

8

9 768u ar

4(i) 22

1

1

2 2r r

rr

2 2

1 1

2 2 1

2 2r r

r r r

2

1

2 1

2r

r r

2

11

2 1

2

n

rr

r r

22

11

1

2 2

n

r rr

rr

2

2

22

2 3

2 2

3 4

22

1

1 22 2

32

2 2

3 4

2 2

( 1)

2 2n n

nn

21 ( 1)

2 2n

n

4(ii)

11

( 2)

2

n

rr

r r

2

1 11 1

2 1 1

2 2

n n

r rr r

r r

2

1 1222

12

11 ( 1)

2 2 1

n

n

n

2

1 12 2

1 ( 1)1

2 2

n

n

n

2

1

( 1) 11

2 2n n

n

5(i) 1 81

1

uS

r

---------------(1) and

41

4

180

1

u rS

r

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

Substituting (1) into (2), 4 4 8081 1 80 1

81r r 4 1

81

r 1

3

r .

Check by substituting into (1):

when 1

3

r , 1

481 108 100

3u

.

when 1

3

r , 1

281 54 100

3u

.

Hence 1

3

r and 1 108u . Therefore

11

1083

n

nu

.

5(ii)

3 4 5 nu u u u is a GP with 2n terms, first term

22

3

1108 12

3u ar

, and

common ratio 1

3 .

Hence

2

2

3 4 5

112 1

3 19 1 .

1 31

3

n

n

nu u u u

9 and 2p q

6(a)

AP: first term, 1100a ; common difference, 2.7d

Let the nth be the first negative term.

0 1 0 1100 1 2.7 0

408.407

nT a n d n

n

409n

409 1 1100 408 2.7 1.6T a n d

Sum of all positive terms, 408

4082 1100 408 1 2.7 224624.4

2S

6(b)

(i)

(ii)

72

n

nS c n

1

1 7 147 7 1

2 2 2n n n

n n c nU S S c n c n

1

7 14 1 7 14 147

2 2 2n n

c n c nU U

6(c)

(i)

1 1 1 1 1 1 11, , , , , , ,..., ,...

3 3 9 9 9 9 729

2 2 2 2 6

1 1 1 1 1 1 11, , , , , , ,..., ,...

3 3 3 3 3 3 3

m = 1 + 2 + 4 + 8 + 16 + 32 + 1 = 64

Sum of m terms, Sm

(ii)

62

131 1 1 1 1 1 1 2 4 8 1 1 538

1 ... 1 ... 223 3 9 9 9 9 729 3 9 27 729 729 729

13

1 1 1 1 1 1 2 4 8 11 ... 1 ... 3

23 3 9 9 9 9 3 9 271

3

S

7(i) 2 2 24 5 ( 2) 4 5 ( 2) 1n n n n =

7(ii) 2 2

3

2 2

3

2 2

2 2

2 2

1 4 5

1 ( 2) 1

3 1 1 1

4 1 2 1

5 1 3 1

N

n

N

n

n n n

n n

2 2

2 2

2 2

2 2

( 2) 1 ( 4) 1

( 1) 1 ( 3) 1

1 ( 2) 1

1 ( 1) 1 5 2

N N

N N

N N

N N

7(iii) 2 2

2 2

2 2

1 ( 1) 1 5 3

2 1 ( 1) 2( 1) 1 (since >1)

( 1) ( 1 1) 1 (since >0)

2 1

N N

N N N N N

N N N N N

N

8(a)

Since 1 2u , 2nu for all n , i.e. 22 2

22 1

a

6 6a 6 36a 30a

8(b)

(i) 0a

21 1

21 1

u uu

u

For 2u to be defined,

2

1 1 0u u and 1 1u

1 1 1 0u u

1 11 or 0u u

Since 1 1u , 1 11 or 0u u

(ii) As n , nu and 1nu ,

2

1

21

22 21

22 1 1 0

21 1 0

21 1 4 1 10 or 1

2

1 5

2

Since 0.618 , 1 5

2

9(i) On the 31st Oct 2012 (at the end of 1

st month), the amount John owes the bank

$ 10000 1.05x or $ 10000 1.05 1.05x

9(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

1.05 1.05 1

$ 10000 1.051.05 1

n

n x

$ 10000 1.05 21 1.05 1n nx

9(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.

10(i)

1

1

sin and sin 6 6

n n

n n

r r

OC OC

1

1

1 1

1 1Therefore and

2 2

2 and 2

n n

n n

n n n n

r r

OC OC

OC r OC r

1 1

1 1 1 1

Observe that =

12 2 3 (shown)

3

n n n n

n n n n n n n n

OC r r OC

r r r r r r r r

10(ii) 2 2

1 1

1 1 and

2 2n n n nA r A r

2 2 21

1 1

2

11 12

1 3 9

2

nn n

n nn

rA r

A rr

which is independent of n

{ }nA follows a geometric progression.

Area of the first semicircle = 2

1

1

2r

Sum to infinity of the areas of semicircles =

2 221 1

2 11

1 199 12 2 . (shown)

1 8 8 2 161

9 9

r rr

r

10(iii) From diagram, we can deduce that

1

n

n

S

+ total area of the semicircles = area of the sector + area of the triangle

= 2

1 1 1

1 2 1tan

2 3 2 3r r r

= 2 2

1 1

1 3

3 2r r

1

n

n

S

= 2 2

1 1

1 3

3 2r r total area of semicircles

= 2

211

3

3 2

rr

2

19

16

r= 2

1

3 11

2 48r

(Answer)

MATHEMATICAL INDUCTION

1.

[YJC 2011 PRELIM P1 Q2b]

Let nP be the statement “ 2

nu n n ”, 0.n

Consider 0n . LHS = 0 0u (given)

RHS = 20 0 0 = LHS

0P is true.

Assume kP is true for some 0k , i.e. 2

ku k k

Need to show 1kP is true, i.e. 2

1 1 1ku k k

1

2

2

2

LHS 2 1

2 1

2 1 1

1 1 RHS (Shown)

k ku u k

k k k

k k k

k k

Since 0P is true, and 1 is truek kP P is true, by induction nP is true for all 0.n

2.

[IJC 2011 PRELIM P1 Q6]

Let Pn be the statement

1 1 1 1 1 1, 1

3 4 5 4 5 6 5 6 7 2 3 4 24 2 3 4n

n n n n n

.

Consider n =1. LHS =

1 1

3 4 5 60

RHS =

1 1 1

24 2 4 5 60 = LHS

P1 is true.

Assume that Pk is true for some 1k ,

i.e.

1 1 1 1 1 1

3 4 5 4 5 6 5 6 7 2 3 4 24 2 3 4k k k k k

Need to prove Pk + 1 is true,

i.e.

1 1 1 1 1 1

3 4 5 4 5 6 5 6 7 3 4 5 24 2 4 5k k k k k

LHS =

1 1 1 1

3 4 5 4 5 6 5 6 7 3 4 5k k k

=

1 1 1

24 2 3 4 3 4 5k k k k k

=

1 1

5 224 2 3 4 5

kk k k

=

1 1

24 2 4 5k k

= RHS

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

1n .

3.(i)

[TJC 2011 PRELIM P1 Q4]

2 3 4

1 1 1 1 2 2 1 3

2(1) 2 2 3(2) 3 3 4(3) 4S S S

(ii) 1n

nS

n

, n 2

(iii) Let Pn be the statement

1n

nS

n

where n , n 2.

When n = 2, LHS = 2

1

2S from (i) RHS =

2 1 1

2 2

= LHS

P2 is true.

Assume Pk be true for some k , k 2, i.e. 1

k

kS

k

.

Need to prove that Pk+1 is true.

LHS = 1

1

2 2

1 1 1

( 1) ( 1) ( 1)

k k

k

r r

Sr r r r k k

2

1 1=

( 1)

( 1)( 1) 1

( 1)

( 1) 1

( 1) 1 1

k

k k k

k k

k k

k k k

k k k k

Since P2 is true and Pk is true Pk +1 is true, by the method of mathematical induction, Pn

is true for all n , n 2.

As n , 1 1

1n

nS

n n

1 . Thus

2

1

1r r r

converges to 1.

4. [NYJC 2011 PRELIM P2 Q3b]

Let nP be the statement 1

1

1

n

r

nr r

for all n

When n = 1, LHS = 1

11 and RHS = 1 = LHS. 1P is true.

Assume kP is true for some k , ie 1

1

1

k

r

kr r

.

We need to show 1kP is true, ie 1

1

11

1

k

r

kr r

LHS = 1

1 1

1 1 1

1 1 1

k k

r rr r r r k k

1 11

1 1

k k kk

k k k k

1 11 1

1 RHS1 1

k k kk k kk

k k k k

Since 1P is true and kP is true 1kP is true, by mathematical induction, nP is true for all

n .

1 1 1

1 1 1

1 2

n n n

r r rr r r r r

1 1

1 12

2

n n

r r

n nr r

Hence A = 2

5.(i) [DH 2011 PRELIM P2 Q4b]

1 1

1 3,

( 1)( 2) 2n nu u u

n n

n nu 1nu

1 32

53

2 53

74

3 74

95

4 95

(ii) Conjecture:

2 1for .

1n

nu n

n

(iii)

Let Pn be the proposition 2 1

for .1

n

nu n

n

When n = 1, LHS = 1

3

2u

RHS = 2 1 3

1 1 2

= LHS

P1 is true.

Assume Pk is true for some , i.e. k 2 1

1k

ku

k

.

We want to show that Pk+1 is also true, i.e. 1

2 1 1 2 3

1 1 2k

k ku

k k

LHS = 1

1

( 1)( 2)k ku u

k k

2 1 1

1 ( 1)( 2)

k

k k k

2(2 1)( 2) 1 2 5 3

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

k k k k

k k k k

( 1)(2 3)

( 1)( 2)

k k

k k

= 2 3

2

k

k

= RHS

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

for all n .

6.

[VJC 2011 PRELIM P1 Q3]

7. [NJC 2011 PRELIM P1 Q3]

1 2 3 42 2 2 2 2 2

1 2 1 1

22 2 2 2 2 2 2

2 2 2 2 2 21

3 3 4 3 4 5

2 2 2 2

3 4 .... ( 1) 2 3 4 .... ( 1) ( 1)!

n n n

n

u u u u

un n n

Let Pn be the statement “

1

2

2

( 1)!

n

nun

, for all n .”

When 1n : LHS: 1 1u RHS: 2

2

21

(2!) = LHS

P1 is true.

Assume that Pk is true for some k i.e.

1

2

2

( 1)!

k

kuk

.

Need to prove that Pk+1 is true.

1

1 22 2

( 1) 1

2

2 2 2LHS

( 2) ( 2) ( 1)!

2RHS

( 2)!

k

kk

k

uu

k k k

k

Since P1 is true, and Pk is true Pk+1 is true, Pn is true for all n .

8. [RI(JC) 2011 PRELIM P1 Q3]

Let nP be the statement “2

1

1 1( 1) ( 1) 1

! !

nr n

r

r r n

r n

, n ” .

When n = 1, LHS = 21 1 1

( 1) 31!

, RHS =

1 1( 1) 1 3

1!

= LHS

1P is true.

Assume kP is true for some k i.e. 2

1

1 1( 1) ( 1) 1

! !

kr k

r

r r k

r k

.

Need to prove 1kP is true i.e.

211

1

1 ( 1) 1( 1) ( 1) 1

! ( 1)!

kr k

r

r r k

r k

21

1

2 21

1

21

21

2 21

1

1( 1)

!

1 ( 1) ( 1) 1( 1) ( 1)

! ( 1)!

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

! ( 1)!

( 1) ( 2) 1( 1) 1

( 1)! !

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

( 1)!

(( 1)

kr

r

kr k

r

k k

k

k

k

r rLHS

r

r r k k

r k

k k k

k k

k k k

k k

k k k

k

2)1

( 1)!

kRHS

k

Since 1P is true and kP is true 1kP is true, by mathematical induction, nP is true for all

n .

The series converges as 1

( 1) 0!

n n

n

when n , which implies

1( 1) 1 1

!

n n

n

when n . Hence the sum to infinity is 1 .