2019 ACJC H2 Math Promo P1 Marking Scheme

Qn Solutions1 Let the amount invested in high risk stock, moderate risk stock and low

risk stock be H, L and M respectively.100 000 (1)

0.2 0.1 0 (2)0.12 0.08 0.04 10 000 (3)

H M LH M LH M L

Using G.C,

$65 000, $20 000, $15 000.H M L

2 (i)

2

61 , 0

6

6 0

6 0

3 20

3 2 02 or 0 3

x xx

x xx

x xxx x

xx x

xx x x

x x

(ii)Replace x with x ,

2 or 0 3

(Reject as >0 ) 33 3, 0.

x x

x xx x

3 2

2

2

2 2

2 2 2

2 1

2

d 1d 1

1 d 1 d1

1 12 12 2 d 1 d

12 11 1 1d 1 d

2 21 12 11 1 1d 1 d

2 21 3 14 2

11 1 1 2ln 1 tan2 2 3 3

4 41 ln2

m m mv m

m x mm m

mx m

m mm

x mm m m m

mx m

m mm

mm m m C

m m 11 2 11 tan .3 3

m m C

4 (i)2 (5 )nS kn k n

21

2

2

1 (5 ) 1

2 5 53 2 5 5

nS k n k n

kn kn k n kn kkn kn k n

1

2 25 3 5 5 2

2 2 5

n n nu S S

kn n kn kn kn n k

kn k

(ii)1 2 1 2 5nu k n k

1 2 2 5 2 2 2 5 2n nu u kn k kn k k k

Since 1 2n nu u k which is a constant, hence the sequence is an arithmetic progression.

(iii)Given e ru

rv , 1

1 e rurv

5

5 kkkk5 22225 2222

k whwhwhwhwhwhwhwhhwhwhwhwhwhwhhwhwhwwwwwhww iciiiiciiii h hh hh hhhh isisssisisissisisisississisissssisisisisiisisisiiiisisiiisisssssssssssssss aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa ccccononononststststanananannnt,ttt rogrgrrgrgrgrrrggrgrrgrrgggg esssssssssssssssssssssisisisisisississississsiississ ononononnonononnonnnonoon........

ru ,

1

1

1

2

ee

ee

r

r

r r

ur

ur

u u

k

vv

Since 2

1

e kr

r

vv

which is a constant, this sequence is geometric.

For sum to infinity to exist, 2

2

e 1

e 12 ln e ln1

0.

k

k

kk

.

5 (i)3y x

Translation of 5 units in the positive x-direction.35y x

Scaling by a factor of 2 parallel to the y-axis.32 5y x

Translation of m units in the positive y-direction.32 5y x m

(ii)f hR [ , ), D 1,m

Since 0m hence f hR D .hf exists.

(iii)hf fD DfD

To find the range of hf , we use the fR as the restricted domain of hand read off the corresponding range from the graph of h( )y x .

hfR [ln 1 , )m[ln 1 , ) .

6 (a)(i)

x

y

nge

yyyyyyyyyyyyx = 1

(ii)

(b)2 22 24 (1)

d d2 1 4 1 0d dd d1 2 1 0 (2)d d

x y x y

y yx y x yx xy yx y x yx x

Since the tangent crosses the positive x- and positive y-axes and makes

an angle of 45 to the horizontal, d tan 45 1dyx

1.

Sub. d 1dyx

into eqn. (2),

1 1 2 1 1 0(3)

x y x yx y

Sub. (3) into (1),2 2

2

2 24

2 24

6

Since 0, 6 and 6.

x x x x

x

x

x x yHence equation of tangent: 6 1 6

2 6.

y x

y x7 (a)

tan tan3tan

3 1 tan tan3

3 tan1 3 tan

13 1 3

x

y

tatattttatttattttttt n tn tnn tn tn tn tnnnnn tnn tnn tnnn tn ttn tttn tttanananan33333333333333

1 t1

3 1 1 3 ...

3 3 3 4 .(b)(i)

2secy x

2

d 2 sec sec tand

2sec tan2 tan

y x x xx

x xy x

22

2

2

2 2

2 2

2

d d2 sec 2 tand d

2 2 tan 2 tan

2 4 tan

2 4 sec 1

2 4 1

y yy x xx x

y x y x

y y x

y y x

y y y26 4y y .(Shown)

(ii)3

3

d d d6 2 4d d d

d d12 4d d

y y yyx x x

y yyx x

4 2 2

4 2 2

22 2

2 2

d d d d d12 4d d d d d

d d d12 12 4d d d

y y y y yyx x x x x

y y yyx x x

When 0x , 1y , d 0d

yx

,2

2

d 2d

yx

,3

3

d 0d

yx

,4

4

d 16d

yx

2 2 42 16sec 1 ...2! 4!

x x x

2 4213

x x

(iii)2 42

2 4

2sec 2 1 2 23

321 43

x x x

x x2 2Since tan 2 sec 2 1x x

2 42 42 42 42 422 42 42 42 44222 42 444

2222222222222

22222222222221 2222222

3333333333333333332324x

11

24x

2 2 4

2 4

32tan 2 1 4 13

3243

x x x

x x

8 (i) Method 1:

22

0

222 2

0 0

20 2

0

22

0

e cos 2 d

1 1cos 2 e e 2sin 2 d2 2

1 1 1cos e cos 0 e e 2sin 2 d2 2 2

1 1e e sin 2 d (1)2 2

x

x x

x

x

x x

x x x

x x

x x

22

0

222 2

0 0

20 2

0

22

0

Consider e sin 2 d

1 1sin 2 e e 2cos 2 d2 2

1 1sin e sin 0 e e cos 2 d2 2

0 e cos 2 d (2)

x

x x

x

x

x x

x x x

x x

x x

Sub. (2) into (1).

2 22 2

0 0

22

0

22

0

1 1e cos 2 d e e cos 2 d2 2

1 12 e cos 2 d e2 2

1e cos 2 d e 1 (Shown)4

x x

x

x

x x x x

x x

x x

222

0000

ee

111111111111111 11111111s 2222222222 dddd eeeeeeee1222222222222222222222

2ee 2

dddddddddddddddddddddddd22222 d 1

12 2222222222222222222222222222222222222222222222222222222222

eeeeeeeeeeeeeeeee

12222222222222222222222222222222222222222222222222222222222222222222222

8

(i) Method 2:

22

0

222 2

0 0

20 2

0

22

0

e cos 2 d

1 1e sin 2 sin 2 2e d2 2

1 1 1e sin e sin 0 sin 2 2e d2 2 2

0 e sin 2 d (1)

x

x x

x

x

x x

x x x

x x

x x

22

0

222 2

0 0

20 2

0

22

0

Consider e sin 2 d

1 1e cos 2 cos 2 2e d2 2

1 1e cos e cos 0 e cos 2 d2 2

1 e 1 e cos 2 d (2)2

x

x x

x

x

x x

x x x

x x

x x

Sub. (2) into (1).

2 22 2

0 0

22

0

22

0

1e cos 2 d 0 e 1 e cos 2 d2

1 12 e cos 2 d e2 2

1e cos 2 d e 1 (Shown)4

x x

x

x

x x x x

x x

x x

(ii)

x

y

y=1

x

222

11111111111112 dddddddddddd ((((ShShShShhhowoooo1111144444444444444444444444

ddddddddddddddd2 dddd 1111 e 1e 1e 111e 111e 1111eeee 111111111111111111111ee

8

22 2

0

2 2 2

0

2 22 2

0

2 22

0

2 22 2

0

2 2 22 2

0 0

2 2 2 2

0

Required Volume 1 d2

e cos d2

e cos d2

1 cos 2e d2 2

e e cos 2 d2 2

e d e cos 2 d2 2 2

e e c2 2 2 2

x

x

x

x x

x x

xx

y x

x x

x x

x x

x x

x x x

2

0

2 22 22

00

2 22

02

2

2

3

os 2 d

e e cos 2 d2 4 2

e 1 e cos 2 d2 4 2

1e 1 1 e2 4 2 4

e e2 4 4 8 8

3e2 8 8

4 e 3 units .8

x x

x

x x

x x

x x

9 (i)Finding x-intercepts:Using siny k , when 0,y

sin 00, (since π 0)

When 0, cos 0When , cos

Hence -intercepts are ,0 and ,0 .

k

x h hx h h

x h hFinding y-intercepts:Using cosx h , when 0,x

0,

e π 0ππ 000000π 00πππ ))))))))))0, cccc ssososs 0

,,,,,,,,,, cocococococococococococococooossss

rcecceceeeeeeeptptptptssss ar

hhhhhhhhhhhhhhhhhhhhhhhhhhcccccccccccososososoosoososossoosososssoosss 000000000000000000000e πππππππππππππππππππππ

0 ccccccccosossooosossoosoosooosoos 000000000000000000coccococccococococoocococoos

cos 0

(since π 0)2

When , sin2 2

Hence -intercept is 0, .

h

y k k

y k

(ii)cos

d sind

x hx h

When 0, cos 0 π 02

When , cos 1 0 π 0

x

x h

00

20

20 2

2

0

20

20

2

2

2 d

2 sin sin d

2 sin sin d

2 sin d

1 cos 22 d2

1 cos 2 d

sin 2=2

sinsin 00 units .2 2 2 2

hA y x

k h

kh

kh

kh

kh

kh

khkh

(iii)cos

cos (1)

x hxh

sin

sin (2)

y kyk

2 2

2 22 2

2 2

2 2

2 2

(1) (2) :

cos sin

1

x yh k

x yh k

ssssiiiinnnnk

2 22 22 22 22 22 222 222 22 22222 22 22222

2 22222222x yx yx yx yx yx yxx yyyx yx yx yx yyyx yx yyx yx yx yx yx yyyyyhhhh kh khhhhhhhhhhhhhhhh kh k2222222222222222

2222n 22222222innnnnnnnn

2 2

2 2

22 2

2

22

2

2

2

1

1

1

1 (since >0 and from diagram 0)

x yh k

xy kh

xy kh

xy k k yh

(iv)2

22 2

y m x

y x m

Since 0y , curve D is a semicircle with centre (0,0) and radius m ,lying below the x-axis.

If h k , m k .If h k , m h .

10 (i)1

1 1

1

1

1 1f ( ) f ( 1) 12 2

1 1 12 2 2

1 122 2

1 1 (Shown)2 2

r r

r r r

r r

r r

r r r r

r r

r r

r

(ii)1

1

1 1

1 1f ( ) f ( 1)2 2

1 1f ( ) f ( 1)2 2

r r

r rn n

r r

r r r

r r r11111111111111111

r

n

1r 122222222

r

111111111111111111111rrrrrrrrr rrrrrr 11111111111111111 1rrrrrrrrrrrrrrrrrrrrrrrrrrr 111111111111111111111 111111111111111111111 111111111 1rr111 1111rr2 222 22222222222222222222222222222 22 2

1))))))))))

1Consider f ( ) f ( 1) f (1) f(0)

f (2) f(1) f (3) f(2) ... f (2) f(1) f ( 1) f( - 2) f ( ) f( -1)

f ( ) f(0)

1 02

n

r

n

r r

n nn nn

n

1

1 1

1

1 1

1

1 1

1 1f ( ) f ( 1)2 2

1 1 102 2 2

1 1 12 2 2

1 1 0.5 11 0.5 2

2 1

r rn n

r r

n r rn n

r r

r r nn n

r r

n n

r r r

n r

r n

n

1

10.52

1 12 (Shown)2 2

nn

n n

n

n

(iii)

2 1

2 1 2

1

2 11 1

2 1

2 11 1 1

2 1

2 1

1

1 1g g2 2

Since ,

1 1 1g g 02 2 2

g g .

Hence g is a one to one function as g g .

g exists.

r rn n

r r

r r rn n n

r r r n

n n r r

n n

n n r r r

n n

n n

.

s aaaaaaaaaaa ooooooooooooonnnnnnneeeeeeeeeeeeeeeee ttttttttttttttooooooooooooo ooooooooooooooooooooonnnnnnnnnnnnnnnnnnnnnnnnnnneeeeeeeeeeeeeeeeeeee ffffffffffffffffffffuuuuuuuuuuuuuuuuuuuuuuuuuuunnnnnnccccttttiiiiiiooooon

s.

1111111111111111111111111112 222211111111111111111111111111111111111rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr r

11111111rrrr22222222

rrrrrrrr

1

1

11

1g2

1 1g 22 2

1 1g 22 2

rn

r

n n

n n

n r

n n

n n

1 251To find g ,128

11 1 25122 2 128

Using GC,8.

n n

n

n

11

11

(i)Using 2

1 360k kA A k ,2

1 0 0

2 2 22 1 0 0

2 2 2 23 2 0 0

360 1 360

360 2 360 1 360 2 1800

360 3 360 1 360 2 360 3 5040

A A A

A A A A

A A A A

2 2 2 20

2 2 2 20

20

1

0

20

3 20

360 1 360 2 360 3 ... 360

360 1 2 3 ...

360

360 1 2 16

60 2 3 1

120 180 60 .

120, 180, 60.

n

n

r

A A n

A n

A r

nA n n

A n n n

A n n na b c

(ii)(a)2

12 2

1 1

2 3

( ) (450 )

(450) ( )

2450 2 1 22

900 2 .

n

rn n

r r

P n nr

n r

nn n n

n n n

2

((((((((4444444444555555555555555555000000000000 ))))))))))))))))))

n2

(44444444445555555550000000000000000

00))0)

2 3( ) 900 2P n n n nUsing G.C to find maximum point of graph ( )y P n ,

12.08n (since n>0)Number of days the amount of bacteria present reduces is 13.

Alternative Method (Using G.C):11, (11) 711712, (12) 720012.5, (12) 7187.513, (13) 7137

n Pn Pn Pn P

From G.C, it can be observed that the bacteria start to reduce between the 12th and 13th day. Since n is a positive integer, the number of days the amount of bacteria present reduces is 13.

(ii)(b)

2 3

( ) 0900 2 0P n

n n nUsing G.C to solve 2 3900 2 0n n n ,

20.96n (since n>0)Number of days the amount of bacteria to be wiped out is 21.

(iii)2( ) 1617 20 7Q n n

The maximum point of graph ( )y Q n occurs when 7n .Since it only takes 7 days for the amount of bacteria present to be reduced when using the synthetic medication as compared to the 13 days required when using the antibody, hence the synthetic medication is more effective in reducing the amount of bacteria present.

12 (i)

Key into G.C,

2X

1A 1

450-XAY and press

[2nd][graph] to look at table generated.

2

2

2

be the distance run in seconds

tan10

1sec10

1010, tan 110

4

1 1 /10 10 202sec

4

Let x tdx pdt

x

d dxdt dt

when x

d p p p radians sdt

(ii)

1

1 1

2tan10 5

tan10

tan5

tan tan5 10

qt qt

qt

qt

qt qt

(iii)

2 2

2 2

3 32 2

32

2

5 10 01 1

5 10

1 1 05 10 10 5

05 500 10 250

10 50050 5 2

q qdLetdt qt qt

q qt q qt

q q q qt t

q q t

tqq

1 1

1 1

50 50 50, tan . tan .5 10

2tan 2 tan 0.3402

q qwhen tq q q

radians

222222222222222225qqqqqqqqqqqqqqq

0

3 32

2 2 22 2

22

3 3

2

2 2 2

2 2

2 2125 1000

1 15 10

50

2 50 2 50125 1000

1 2 112

1 4 502 50 0125(9) 1000(9) 1125

q t q tddt qt qt

when tq

q qq qd

dt

q q

Hence value obtained is a maximum.