1

2019 JC1 H2 Promotional Examination Suggested Mark Scheme

1y

x

2

2 a(i) 2x b x b x b

So, x b and x b are asymptotes.Hence, 2 4b b .

2

2 2

ax bx a bx ab ay ax b x b

So, y a is an asymptote.Hence, 3a .

6m (ii)y

x

3y

2x 2x

( 0.578, 0.461) 3(0, )4

3

3

7m

(i)

Since any horizontal line intersects the graph of f at most once, f is one-one and it has an inverse. (shown)

(ii) fR [ 9, )

Since fR [ 9, ) Dg, therefore gf does not exists.

(iii) 2

2

fg : (e 1) 4(e 1) 5 e 2e 8, , 0

x x

x x

xx x

(e 1)x 1)1)1), 0,,,

fgR [ 9, 8)

y

x

y

x

y = – 8(0, – 9)

4

4

(a)

12 2

22

2 2

2

2

d 1 1 1 2d 2

1 1

1 1 22 1

1 1 01

11

y x xx

x

xx x

x xx x

x(b) 5 2 2d d sin

d dd d2 1 2 cosd d

d 2 cos 2d

d cos 2d 2

x xy y xx x

y yx x y y xx x

y x y x x yx

y x x yx x y

5

5 (i) 1sin cosec

4 4 2cosec

PQPQ

QR PQ RS2 cot 4 2 cosec 2cotPS QR

1 121 4 2 cosec 2cot 4 2 cosec 24 2 cosec cot shown

A PS QR

7m (ii) 2d 2 cosec cot cosecd

A

For maximum area, d 0d

A .2

2

2

2 cosec cot cosec 02 cosec cosec cos cosec 0

cosec 2cos 1 0

2cos 1 0 or cosec 0 No solution1cos2

3

3 3 3

dd

A0 0 0

Hence, A is a maximum.

Maximum Area 4 2 cosec cot3 3

4 143 3

343

4 3

mummmm....

a 4 24 24 24 24 224 24 2224 cosecececec cococococ t33

44444 14444 1114444444444444444444444444443

4 24 24 24 24 244 24 24 24 2444 24444 ccccccccosososososososoooososososoosooooosooooosoooososoooosoosoooooooooooooooo ecececec

4444444444444444

6

6 (i) 13

11 33

13

2

2

8 3

38 1 8

1 312 8

1 41 1 3 33 31 ... 2 3 8 2 8

1 1 21 ... 2 8 64

x

x

x

x x

x x

21 1 1 ... 2 16 64

x x

3 18

8 83 3

x

x

Method 2:13

4' 3

7'' 3

'

''

f ( ) 8 3

f ( ) 8 3

f ( ) 4 8 3

1f (0)21f (0)

161f (0)

32

x x

x x

x x

123

1 1 18 3 ... 2 16 64

x x x

3 8 818 3 3

x x

2222222222222222222222222222222222222211111111111111111111111111111111111111111111 ...... 2 111111111111111116 66 66 66 66 666 66 66 66 66 66 6666 666 66666 6666 666 6444444444444

x xxxx 222222xxxx xxx xxxxxxxxxxxxx xxxxxxxxxxxxx xxxxxxxx xxxxxx xxxx

888888888888888888888 88881113333

1 xxxx1

7

7m (ii) 2

3

3

3

3

1 1 1 1 1 1 ... 2 16 16 64 1618 3

161 1 1 1 ...

2 256 163841251616 8192 64 15 16384

4128516 16384

8

7

8m

(a) 2

2

2

2

3 5 12

3 5 1 02

3 5 20

22 3 0

23 1

02

x xx

x xx

x x xx

x xx

x xx

Hence, 3 or 1 2x x .Method 2:

2

22

22

3 5 12

3 5 2 2

3 5 2 2 0

2 1 3 0

x xx

x x x x

x x x x

x x x

3 or 1 2x x .(b)

Hence, 2 1.99 or 4.94x x .

3 1 2

9

8 (i)2 2

2

12

4

1 114 4

4 114

1 11 15 2

n

nr

r

n

10m (ii) 22 1 2 3 2 1 2 3

1 2when , 112 2 32

3 2when , 132 2 12

2 1 12 1 2 3 2 1 2 3

A Br r r r

r A

r B

r r r r

1 1

1

1 1 22 1 2 3 2 2 1 2 3

1 1 1 2 2 1 2 31 1 12 3 5

1 15 7

...1 1

2 1 2 11 1

2 1 2 31 1 12 3 2 31 16 2 2

n n

r r

n

r

r r r r

r r

n n

n n

n

3n

222222222222211111111111 166666666666666666666666666666666 222222222

1 11113333333333333333333333333333333333333333333333333333333 222222 333333333333333333333333333 222222222222222 33

22

10

(iii)

4

2

1

2( 1)

1

2

11

4

2( 1)

2

1

4

142 1 2 3

4

4 4

1 1 116 6 2 2 3

1 1 15

12 1 2 3

1 16

42

2

u

2

16 6 2 2 31 1 1 1=

16(15) 6 2 2 2 31 1 (ded ced) since

240

3

4

1 111 2

nr

r

nr

r

n

n

r

r

r

n

r

n

n

r

r r

n

r

n

n

n

r

10 and 02 2 3n n

11

9 (i) Solving the cartesian equations representing planes 1 and 2 (on the GC) produces

3 2 3 22 0 2

0 1

x xy yz z

3 2 : 0 2 ,

0 1l r 0

000

(ii) 3 : ( 2 2 3) ( 2 2 6) 0 ( 2) (2 1) (2 2) 3 6

p M x y z x y zM x M y M z M

2

2 1 3 62 2

Mr M M

Mr 2222

2 2222 2, A normal vector to plane 3p is

22 12 2

MMM

.

9m (iii) (Elegant) method :

Let ( , , )x y z be a point on line l.Line l is the common line of intersection between planes 1p and 2p , ( , , )x y z

is on plane 1p as well as plane 2p .2 2 3x y z ,

2 2 6x y z .

The expression

( 2 2 3) ( 2 2 6)

( 3 3) ( 6 6)

(0) (0) 0

M x y z x y z

M

M

( , , )x y z lies on plane 3p .Any point on line l lies on plane 3p , line l lies on plane 3p .

Alternative standard method :

Direction vector for line l,22

1mm 2

111.

Normal vector for plane 3p , 3

22 12 2

Mn M

M3n3 M22

222.

pla

dard dddddd ddddddddddd memmememmemeeeethththththththththththththtthththththhththththththththtthhthththhhthhththththhhhhththhhhthhhtthhthhhhhhthhhhhththhhhhthhhthhhhhhhhhhhhhhhhhhoddododododododoododoodoooodododododododoododoododododoododoodoododododdodododdodododdododoododododododdodododododdoooooo ::

ioooooooooon n n n nnnn nn vvvevevvevvvvvvvvvvvv ctctctctctctctctctctctctctctttc orooooooooooooooooooo fffffffffffffffffffffoorororororororoororororoorooorooo lllllllllllllllllllllllinininininininninininininininininininnnnnnninnne e e e ll, m22

m

12

3

2 22 2 1

1 2 22( 2) 2(2 1) (2 2) 0 ,

Mm n M

MM M M

m n 22111

3m n3n

line l is parallel to plane 3p .Point (3, 0, 0), a point on line l.

The expression

3 2(0) 2(0) 3 2(3) 0 2(0) 6(3 3) ( 6 6)(0) 0 0 ,

MMM

.

this point is also on plane 3p .

Line l // plane 3p , and l contains a point (3, 0, 0) that’s also on 3p ,

line l lies on plane 3p .

(iv) Normal vector for plane 1 , 1

122

n1n 2222

.

Normal vector for plane 3 , 3

22 12 2

Mn M

M3n3 M222

222Let be the angle formed between normal vectors 1n1n and 3n3n .

1 3

1 3cos n n

n n1 3n n1

n n1 3n1 3n n1

1 3

1 22 2 12 2 2

( 2) 2(2 1) 2(2 2)9

Mn n M

MM M MM

1 3n n1 3n3 222222

2 2 21 1 2 2 3n1 1n1 1

2 2 23

2 2 2

2

2

( 2) (2 1) (2 2)

( 4 4) (4 4 1) (4 8 4)

9 9

3 1

n M M M

M M M M M M

M

M

3 (n3 ((

2(22

2 333333333322 22

22222222

2

2222)))))))))) (2222 1)))) ((((22 22

( 444444444 4444))2

922

222222222222))))))))))) ((222222222 1) ((((22 2

44444444 44444))))222222

2222222222222222222)))))))))))) (((((((((((((((((((((2222222222222222 1) (2122222222 2

( 4444444444444444444444 4444))))222

22

(((

13

1 3

1 3

22

cos

9 0 if 013 3 1

n nn n

M M MMM

1 3n n1

n n1 3

9n1

M3n n1

The cosine of the acute angle between planes 1p and 3p is 2 1

M

M.

14

10

(i)

1Area of triangle 2

OAB OA OBOA OBOA

01 1 12

2 1

1 21 0 22

0

11 22

2 21 1 22

21 1 52

10m (ii)By the Ratio Theorem, 3

4OA OCOB A OC3OAOB OA

011 3 14

1 2OCOCOCOC

4 44 3 14 6 2

OCOC4

4444

OBOB is perpendicular to ,OCOC 2

41 1 4 1 01 2

1 02

(iii) Exact perpendicular distance from B to OA = BNBN

From (ii), 12

21 1 1Area of triangle 1 5 1 52 2 4

OAB

1 9 12 4 2

BN OABN OA

3 1 42

BNBNBN

32 5

BN 3BN

2 2 2

21 91 14 4 5

OB ON BN

ON

2 2 2OB ON BNON

92ONON

of trrrriaiaiaiangnnn

OAAAAAAAAAAAA

4444444444444444N 11111111111111N 1 41111 41111N 111111111111NNNNNNNNN

3

15

2 95

ON2 9ON

35

65

01

023

5 1 4ON 3ON

16

11

(i)

As 21, 1x t y e e

12m (ii)3

1 d 1d1 2 1

xxtt t

2 2d 2

dt tyy e te

t

2

3

d 21d

2 1

ty tex

t

2 34 1 (shown)tte t

(iii)Gradient of normal at 5

4t is

54

1 1 0.012421d 80.50628d t

yx

When 54

t , the point on C is 25162 ,

3e or 2 , 4.7707

3

The normal at 54

t meets the y-axis at the point P:

At y-axis: 2516 20.012421 0

3y e

4.76245yP: 0, 4.76245

tsts tthe

:

on CC isis 22 ,33233

,33

ormrmmmrmmmmmmrmrmmrrrmrmalaaaaaaaaaaaaa atatatatatatatatatatatatataaaa 555555555555555555554444444444444444

ttttttttttttt

01

17

Given Q: 21 ,1

qeq

PQ =2

221 4.76245

1qe

q

Min PQ = 0.667 units when q= 1.25

ORFor PQ to be a minimum, PQ must be perpendicular to the tangent of Q [B1],

thus Q must be 54

q [B1]

Thus Min PQ =

2

225161 4.76245

5 14

e =0.667 [B1]

ORFor PQ to be a minimum, PQ must be perpendicular to the tangent of Q [B1]Thus

2 3

5 34

1 4 1 1554 14 4

qqe q

e

Solving gives q = 1.25 [B1]

Thus Min PQ =

2

225161 4.76245

5 14

e =0.667 [B1]

ORFor PQ to be a minimum, gradient of the line PQ must 0

[B1]1]1]1]

in PPPPPPPPPPPPPPPQQQQQQQQQQQQQQQQQQ ==============

22222222222222222222222222

1111111111111111 252511111111111111111111111 e555555555555555555555555555 1555555555555555 144

18

P: 0, 4.76245 , Q: 21 ,1

qeq

, thus 2

d 4.76251d

1

qy ex

q

Solving for 2

d 4.7625 01d1

qy ex

q1.25q (to 3 sf) [B1]

Thus Min PQ = 22 25

161 4.762451.2493 1

e =0.667

19

12 (a) (i) Month Outstanding amt. owed

1st day of month aft. instalment Last day of month aft. interestOct-2019 12000 12000(1.005)Nov-2019 12000(1.005) P 212000(1.005 ) (1.005)PDec-2019 212000(1.005 ) (1.005)P P

Amount owed upon paying the 2nd instalment (in $)= 212000(1.005 ) (1.005)P P

14m (a)(ii) Instal Outstanding amt. owed

-ment 1st day of month aft. instalment Last day of month aft. interest0 12000 12000(1.005)1 12000(1.005) P 212000(1.005 ) (1.005)P2 212000(1.005 ) (1.005)P P 3 212000(1.005 ) (1.005 ) (1.005)P P3 3 212000(1.005 ) (1.005 ) (1.005)P P P …… … …n 112000(1.005 ) (1.005 )... (1.005)n nP P P

Amount owed upon paying the nth instalment (in $)1

Geometric series, termsStarting from right : first term , common ratio 1.005

12000(1.005 ) (1.005 )... (1.005)

(1.005 1)12000(1.005 ) 1.005 1

12000(1.005 ) 200 (1.005 1

n n

nP

nn

n n

P P P

P

P

tric series terms

(1.005 )... (1.005) 1(1 005 (1.005)(1.005)(1 005)

) (shown)

(a)(iii)

Intended last instalment on 1st December 2021.- 2019 : 2 instalments- 2020 : 12 instalments- 2021 : 12 instalmentsTotal no. of instalments = 26

Amount owed upon paying the 26th instalment (in $)26 2612000(1.005 ) 200 (1.005 1) 0 if loan is repaid completelyP

formulate eqn.26 26

26 26 60

493.3395...

12000(1.005 ) 1.005200(1.005 1) 1.005 1

P

Minimum monthly instalment needed = $493.34 (nearest cent)

(b)(i)

Let a hertz be the frequency of the tone produced by the 1st key on the piano, and r be the common ratio of the frequency between successsive keys.

y ng tttthehehehe 2622626222222000000000000000000000000000 (((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((111111111111111111111111111111111111111111111111111111.....0000000055552626

26226226226626626666666

262626262262622222 601222222220000000000000000000000000000000000 0(0(0(00(0(0(0(0(0(0(00((0(0(00(000(1.1.1.111111.11.111.111.00000000000000000000000000000000000000005 )5 )5 )5 )5 )5 )5 )5 )5 )5 )5 )))5 ))5 )5 )5 )5 )5 ))5 )55 ))55 )2666660

20202020020200002000202000002022000(0(0(0(0(0(00(0(0(0(00((((((11.111.111.111 00000000000000000000000000000000000000000005 15555555555555555 111))))2226222222222222 11)26262666266((((((((((( )))))))))))))

20

the 49th key produces 440 hertz, 48 440ar (1)the 61st key produces 440 hertz, 60 880ar (2)

Eliminating a, 12(2) : 2(1)

r , 1

122r

112

st

48 48

4 (frequency of 1 key in hertz)

440 440From (1) : 2

440 440 = 27.5 162

ar

(b)(ii)

Frequency produced by the nth key on the piano (in hertz),1

12

112

1 127.5 (2 )

27.5 (2 )n

n nnu ar

Sequence of logarithm of the frequencies produced by successive piano keys :

1 2 3ln( ), ln( ), ln( ), ... , ln( ), ... nu u u u

The difference between consecutive terms of the sequence

112

112

1 2 3

1

1

, , , ... , , ... is in a G.P., with 2

ln(2 ) , a constant

ln( ) ln( )

ln , using properties of the ln function

ln( ) ,

n n

n

n

nu u u

u u

u

uu

r r1 21 22, ,, 1 22

1 2 3ln( ), ln( ), ln( ), ... , ln( ), ... nu u u u is in an A.P.

(b)(iii)

Ratio of ascending frequencies produced in a major triad chord, sounded by the nth, ( 4n )th and ( 7n )th keys is

1 112 12

1 73 12

1 ( 4) 1 ( 7) 1

1 1 4 1 7

4 7

4 7

: :

(1) : ( ) : ( )

1 : :

1 : (2 ) : (2 )

1 : 2 : 2

n n n

n n n

ar ar ar

ar ar r ar r

r r

,

which is independent of n. (shown)

11111 11111

11111111

( 7)7

1111111111111111 7777

4 7

44 774 77444444 1

( ) ::: (((( )))) 4 1111111111111111111111111111111 7777

: 4

1 ::::::::::: ((((((((((((((222222222222222222222 )))))))))))))))))))) : (((((((2222 )12212212121212121211212112221 12124444444444444 1212

: 2222 33

ar

( ))))))) ::::::: ((((

rrrrrrrrrr ::::::

( 7)7