I

SULIT 3472/2 Matcmatik Tambahan Kcrtas 2 Ogos/Septcmber 2016

2.1:. jam 2

3472/2

MAKTAB RENDAH SAINS MARA

PEPERIKSAAN AKHIR SIJJL PENDIDIKAN MRSM 2016

PERATURANPEMARKAHAN

MAT EMA TIK T AMBAHAN

Kcrtas 2

Dua jam tiga puluh minit

UNTUK KEGUNAAN PEMERIKSA SAHAJA

AMARAN

Peraturan pemarkahan ini SULIT dan Hak Cipta Bahagian Pendidikan Menengah MARA. Kegunaannya khusus untuk pemeriksa yang berkenaan sahaja. Sebarang maklumat dalam peraturan pemarkahan ini tidak bolch dimaklumkan kepada sesiapa.

3472/2 ©2016 l Tak Cipta Bahagian Pcndidikan Mcnengah MARA·

[Lihat sebclah SULIT

SULIT

Pl

Nl

@

~

8

KEKUNCI SKIM PERMARKAHAN PEPERIKSAAN SIJIL PENDIDIKAN MRSM 2016

MATEMATIK TAMBAHAN KERT AS 2 (3472/2)

I markah diberi berdasarkan pengetahuannya

3472/2

I markah dibcri untukjawapan daripada kacdah sah dalam bcntuk nilai atau ungkapan

Markah kacdah, I markah diberijika pelajar mcncuba buat langkah pengiraan atau tunjuk kaedah yang sah menghasilkanjawapan yang betul.

I markah dibcri bagijawapan yangjitu / tcpat scperti dalam skim

I markah diberi bagi jawapan yang didapati dengan menggunakan nilai yang kurang tepat daripada bahagian soalan. Biasanya diikuti dengan tanda--/ dengan catatan kuantiti yang salah yang diperoleh lebih awal

Jika perkara dalam kurungan ini tertinggal, bcri markah penuh scperti yang tercatat dalam skim.

PER HA TIAN : Sila ikut skim pemarkahan yang disediakan supaya perbandingan antara MRSM boleh dilakukan.

3472/2 © 201611:tk Cipta Bahagian Pendidikan Menengah MARA SULIT

No

I

.

ADDITIONAL MATHEMATICS PAPER 2 SPMRSM 2016

MARKING SCHEME

Solution Scheme -~~---~

y =3x-2 y+2 r··-PCJ Implied or x=-- or

3 3x 2y

x=--··· or x=--2+2x 3-2y

3x-2(3x-2) = 2x(x-2) or (0 Substitute value x or y to other

~+2=3 or y+2=3( -3L j or 2+2x 3--2y

3(y;2)-2y=2(y;2}

Solve quadratic -C-1)± JC-I)'- 4{' 6X,-4} Kl equation using

x= 2('6}

or formulae or completing the square

_ '7±-Fi'- 4{·2x-·6J y=

2{·2J

X=-0.737 First set

x=0.904 ' NI of values

y=0.712 y=-4.211

or ! Scw,dsc< y=0.712 y=-4.212 of values

X =0.904 '

x=-0.737

Note: OW -I if method of solving quadratic equation not shown.

--No Solution Scheme

Sub Marks

marks

2

~ (a) 30-2x or 16-2x Sub

Marks marks

3

t"sc ,oJ,mc fonacloe V =x' (30-2x)' (16- 2x) = x'(30- 2x)'(l 6-2x)

480x-92x 2 +4x 3 480;-92l +4x'

(b) dv =' 12x 2 +480-184x

Differentiate * V with

d'( respect to x

'12x 2 -184x+480=0 G)use

* 12x 2 -184x+480 = 0 and solve for x

7

5 5 10

X =]2,x=- 4 3

Values ofx

10 x=-

3 10

x=-3

Note: value ofx of suitable for the cardboard

Not· • ( J( Solution Scheme I Subk I Marks mar s

(!) \30)'·(.'30;rJ or _1_(15)'(30;rJ M Use area =_l_r'B a#d 2 180 2 180 Y 2

0 = 1.2._ x 3. l 42or equivalent 180

use of area with radius 30 •

· ! (30)2 (JO;r J-' -~ (15)' (JO;r J area with radius 15 2 180 2 180 . OR cqrnvalent

J 76.7488 area 176.7 H 171 I 4

RM2209.36 I@ price 2208.8 H 2212.5

b) I (44)'(0.5237)

23.04

~

Kl) Use s = rO with r = 44 or equivalent

NI 123.03 H 23.05 1

,l ;,

g 1:-1

No

4

(a)

b)

-·-----Solution Scheme

Sub Marks

marks

70+ 62+ 59+ 75 + 68 I Lft lJ SC X = ---;:;-

5

= 66.8 66.8

702 + 622 + 592 + 75 2 + 682

- (66.8)2

5 5

5.706 5.706

Zubaiclah 's result is Zubaiclah more consistent

7

334+x > 67_5 ~>+x Use ------ > 67.5

6 6 2 Note: accept '='67.5

X > 71 71<x'.S72

----------·--- -----------~

N"i Solution

(a~i) 7

ii) l-4

b) I x+I

2

(4x-~ 2

4x -19 ---

2

c) • ( 4\ 19) = 4x - 20

21 x=-

4

\ .

I Scheme

10 1

10-4

J ( Kt) rind(

® substitute 4x - 20

into/*

4x:tl9 ~----

2'

KI ) find /'g(x)

N~J 21 x=-

4

I

Sub marks

3

2

Marks

7

No Solution

6

(~) [m = -6-~ (i) 0 - (-3)

111 = -2

(ii)

m - -­' - 2

• 1 y - (-5)= --(x - (-4))

2

I Y = -x - 3

2

_!_10 -3 -4 01 2 -6 0 -5 -6

_!_ 10 s + 24) - (18)1 2

10.5

(b) I Jc,-..:-- 0) 2 + (y - (-6))' or

~-~03)) 2 + (y - O)'

2 _Jc; - 0) 2 + ( y - ( - 6 )) 2

-J0<.-~(-3)) 2 +(y-0) 2

3x 2 +3y 2 -t-48y-6x+l35 =0

Scheme

Kl Find gradient of l'Q

@ use (-2)m 2 = -1 and

substitute gradient into

y - Yi = m(x - Xi) I 0 1 . I y = -x-3 or equ1va ent

2

0 Find area or I\PQR

10.5

Kl Find the distance SP or SQ

0 Equate 2SP= SQ or

equivalent

1,----u----, 3x 2 + 3y2 + 48y-6x + 135 = 0

le_____-~ or equivalent

Sub. I Marks marks ----i

J

J

No

7 a)

b)

Solution Scheme

[ J I ]'

secx cosecx use secx = -- and cosec.i: = -- or

[cosx-sinx]2

cos 2 x-2cosxsinx+sin 2 x

l-sin2x

G0 cosx sinx

K"I - 1-- =cosx and ---1-- =sinx

sec x cosec x

G) 1- sin 2x

usecos2 x+sin2 x=l or

2sinxcosx =:: sin 2x

Y·

2

............ Ll ...... . \21:=~

0

Shape of sin x

1[

2

0 2 cycle for O s x s 2n0

Reflect

Shift upward (+I)

y=2-_:_ 1[

0:,-J 0,,-1 ~~

n

Sketch straight line* y=2- -~ with ~I n·

* gradient or *y-intercept property correct __

No. of solutions= 3 [NI

~

3 --1[

2

~

2n X

Sub marks

3

7

/

Marks

10

No

8 (a)

Solution

X 2 3 4 5 6 f-----;--1----1~-+--+----t--i

Iog 10y I L52 I I .44 I 1.3411.2s 11.21 11.12

See graph

(b) I Plot Jog 1oy against x

(c)

Correct axes and uniform scale All points arc correct

*6 points plotted correctly

Draw line of best fit

log,, y = -2log10 px + log, 0 q

m = -2log, 0 P c = log, 0 q

1.28-1.52 i) -2 log'° p = *m = -0.08

4-1

p = 1.096// l.0965

ii) Jog'° q = I.6

q = 38.90

(iii) log 10 20 = 1.30 I

X = 3.75

Scheme

~ All values of L.:..:..:_j log 10y correct

Note : at least 2 d.p

@Plotting log10 y against x with

correct axes and uniform scale

If table is not shown, all points correctly plotted award NI

Linc of best fit

6 log,,, y = -21og10 px+ log10 q

Use

'm=-2log 1

q = 38.90 ~ 40.74

X = 3.75 ± 0.05

Use 'c= log 10q Any value round off to 2 d.p. 1.59 ~ 1.61

Note: SS - 1 only once if i) Does not use given scale ii) Axes interchanged iii) Not using graph paper

Sub Marks

3

10

6

Iog10y

1.9 -

1.8

1.7

', 1.t'"i·,,

"-,

"'"" ', 1.5

1.4

1.3 ~-

l.2

1.1

1.0 _,

'<!~

"'·· "'·· ',,

Graph for Question 8

!

·-r

·."' :

~ ,, '""· "'ot

-~ ,,"~

~--,----+------2 3 4 5

"',. ·,,

·, ·•.

6

/

X

7 8

~· .-------------··---

No Solution Scheme Sub

Marks marks

9 (a)

i) 400-420 : 400,-o"' 50 2

0.3446 ~ 0.3446

ii) z ='.-i0.524 ~ 3

- ' 0.524 = m - 420 I . m 420 - 0 524 =-- --50

50

393.8 // 394 393.8

(b) JO

i) 4°C15 (0.25) 15 (0.75) 25 Use of

~ "C,(p)'(q)"' 2

0.0282

0.0282

ii) 24 or n =25 or r = 9

~ 24 or n =25 or r = 9 Seen or implied

3

"c, co.25)' (0.75)"

~ Use of

"Cr(p )' (q)" '

0.0781 Bo.01s1

_, ___________ i___ ________

No Solution

10

(a) J (i)OP=OA+AP

(b)

3 = IOx+-y - 2-

(ii) BA= BO+ OA = -10y+8-':

(i) OR= hOP 3 = lOhx+-hy

- 2 -

(ii)(JR=OB+BR

=lOy+k(BO+OA)

= 8k-':+(l0- l0k)y

I Oh = Sh and '}___ h = IO - I Ok 2

h=~ 7

,/: :t~

k=H r.~

.,-)) (;'\.

Scheme

Use triangle law to find - ~

OP or BA

NI 1-!0y+8~

~C]Ioh~+%h~

I Use - -· -BR= k(BO + OA)

&k! + (10-1 Ok),)'

lNl_]

G~

Equate the coefficients

of x and ofy for OR

@solve simultaneous · linear equation in h & k

5 . h = - or first value

7

25 h =--;-,or second value

)i~(

Sub marks

3

6

'\

Marks

10

No Solution Scheme Sub

marks I Marks

ta~ fc2x'+4)d.x-+(k)(4) I ~lntcgratcf(2x'+4)d:c-,

[ 2k

1

]

3 -+4k-O

·[2;' + 4k ]-·[T(k)4]

"[2k1 ] ·[1 J 8 ~3-+ 4k - 2(k)4 = 3k

k

OR I

f( f- 2 )'r1y

(~\~=-~)% 3

4

[~:'lj 3

• l (k2)'i "k(2k2 +4)- -(k)(4)- -~

2 3

4

• 3

'k(2k2 +4)- 2l·(k)(4)- (k')2 =~ -~ 3

k 4

OR

0 Substitute Limit

Kl

@Find area of

' Triangle _!__(k)(4) 2

C0 I .[2t +4k]-

'[t(k)4 J

an<l solve for k

[ntcgrate J( f- 2 )i dy

G.0 Find area of rectangle & Triangle

k(2k' + 4) & _!__(k)( 4) 2

+ 4) --

J

I (k ')' -(k)(4)- ----2 J

G0 84

Equate *area = ~ k 3

and solve fork

GQk=l

IO 6

(b)

18 y-4

ff ~-dx I 2

2

]

8

a·i[~--2y :! -4 4

[( • 8

2 J ( • 4 T{ ~4 -2(8) - 4

41!

~·.,·---~-

~-2(4)JJ

--·--

·- --

~] Limit 4 or 8

Kl Integrate TC f ( y; 4 )d¥

@ Substitute Limit *4,*8

4n

- - -·-

-

-- -. _ ____,_____

\,

No

12 (a)

(b)

(c)

Solution Scheme - Sub I Marks ~ ~~ '

sin / !~CD sin 75 ° -------·-] .8 5

L EC!) ·• 47 .23 ° // 47 ° 14 '

L"ACB=132.77°or L"EDC=57.77°

EC 4.6 --~-=~--sin· 57.77° sin· 47.23

= 4.38

AC=4.6+4.38

=8.98

AB' =82 +· 8.982 -2(8)(8.98)Cos·l32.77

= 15.56

L\CDE=_l_(3.8)(5)sin· 57.77° 2

8.04

or

MB(> 1 C 8.98)( 8)sin' 132.77° 2

26.37

. 26.37 +. 8.04

34.41

G ~

sin L'.f;;CD sin 75° ----=~

3.8 5

4 7.23° // 47' 14'

~ L'.ACB=l32.77°or

~ L'.lWC=57.77°

4.6+' r,;c

EC 4.6

sin' 57.77° = sin' 47.23

G s' +· 8.98 2

-

2(8)( 8.98)Cos' l 32.77°

Use area of

L\ =_l_absin LC 2

I ~26.37 +. 8.04

~4.41

2

5

10

3

No I Solution -~~- Scheme I Subk r mar s

13 (a) 2t -6 = 0

t=3

V='32 -6(3)+8

=-I

OR

V = (t - 3)2 - 9 + 8

V=(t-3)2-1

= -I

(b) 1,'-61+8=0

I= 2,t = 4

s = J (t 2 - 61 + 8)dt

l f-31 2 +81]

( ~ -3(4)2

+8(4))-(231

-3(2)2

+8(2))

4

3

-I

~Substit'.1te o.,=3 mlov

Complete the square

@), Substitute , I= *2&t = *4 into *s

Kl

(I - 3) 2 - I

Equate v to O to find t

® Integrate v

61

3

4

Marks (c)

15 ·-·-·-·-·-·-·-·-·-·-·-·-·-

CE] Correct shape

0 Points (0,8), *(3, -1), (7,15) 8 -

~ are correct

3

.. 7

*(3,-1) ~3<t5c7

Note: Accept correct range without graph Range 3 </Sc 7

10

"------~------~-~ ·----------~--

,--.-----r-------~--

No 1 ----Solutio~

14 (a) x =

60 x I 00 = 120

50

Xx 100 = n, § 20

33 X 100 = !J()'

z

Scheme

® Use !l_ x I 00

nd x,y and z

Sub

-----f marks

3

~

Po

! 1\ : ( - ii-,

X = 120,y = 25, z-=-30

(b) 'J 20(25) + I 25(20) + 140_(1_5)___:t:_1__1_()_(_30) + 125(10.

25 + 20 + I 5 + 30 + I 0

Use

_ I1w I= LW

121.5

(c) I p ~xlOO = *121.5 150

Kos schari = I 82.25

Kos Januari 2014 = * 182.25 x 31

5649.75

(cl) [ 11s.2 ·121.s ---x----xlOO

[00 JOO

139.97 // 40

NI ] 121.5

I' Use ~xlOO

150

182.25

[NJ -15649.75

I I 115.2 '121.5 Use -100

-x-100x 100

I N~J 139.97//40

2

3

I 2

arks

I I

10

I

No. I- - -Solution

15

(a) I 00 x + 60 y S 6600

8() X + 120 )' 2 4800

y22x

(b) Refer graph

c)

(i) I 30 ::;y s ss

ii) I (30, 60)

50 (30) + 25 (60)

3000

Scheme

EJ 8 CttD @] Drawing correctly at least one

straight line from *inequality which involves x and y

10 All three *straight lines are

correct

cfu Note: Accept dotted lines

Region shaded correctly

0 30:Sy:S85

~l (30,tjO)

~ ~ Use 50x I 25y for any

~ poi'.1t in the *shaded

cclJ region

3000 l

Note: SS - l oucc if in

(a) i) the symbol '=' is not used at all

ii) more than 3 inequalities given

(b) i) does not use given scale

ii) axes interchanged

ii) not using graph paper

Sub Total

Marks Marks

I 3

I 3

4

10

---L--~-----~----------------------1---- --- -- ----~------t---~---~

0 1£

)

<n

""' .g

~

0

"' ..,.

=

'-' g

u [/]

'-~

¢::

er::: .c

<

0

.. 0

~

r~

M

(.!.. '-'

0 :::::: z ~

~