Add Math Spm Trial 2015 Penang p1ans

8/17/2019 Add Math Spm Trial 2015 Penang p1ans

1/31

N M

T JNGK T N

M J LI S P

ENGETU

SEKO

L H

I L YS

I

C W

NG N

PU L AU

PI

N

N

G

MODUL LATIHAN BERFOKUS SPM 2015

ADDITIONAL MATHEMATICS

Ker ta

s J

Og

os

2 jam

3472/1

D ua j

am

J

NG N

B

UK

KE R

T

S SO L N

INI SEHI

N

GG

A

DIB

E

RIT HU

1 T11/iska

nama

da11 tin

gka tan

rmda pada

m a

ga11 y ng disediaka

2

Ker/as soalan i11i adalah da/am dwibalws

a

3: Soala

da/0

bahasa /11gg

r

is e

da

/111/

i

w ala

yang sepada

dalm

balwsa Me/ayu

4

Ca/0 dibe arka e j mvab kese/11mhr

atau sebahagia soa/a

sama ada da/am

bahasa 11ggeris a/au bahasa J

t/

eloy

u

5

Ca/0

dikehe

daki e baca m

k

lumat di

/J

ala w belaka g kertas soa/m1i i

U t k Kef{u1

wm

1Pemeriksa

Soalan

Ma rka h

Ma

rka h

Penuh

Dipe

rol

ehi

2

2

2

3 2

4

3

5

4

6

3

7

4

8

4

9 3

10

3

1L

4

12

2

13

3

14

3

15

4

16

3

17

3

18

4

19

4

20

4

21 3

22

4

23 3

24

2

25 4

J1JMLAH

80

Kerl

as

soalan i

ni

mengandungi 24 halaman bercetak dan

l

halamau tidak bcrce ta

k

3472/ J [Li hat ha l

am a

n sc belah

tutormansor.wordpress.com


2/31

0.0

0.1

0

.2

0.3

o..i

)5

0.6

0.

7

0.8

0.9

1.0

1.1

1.

2

1.3

1.4

1.

5

1.6

l. 1

1.8

1.9

2.0

2.1

2.2

2.3

2.4

2.5

2.6

2.7

2.

8

2.9

.l.O

2

3472/1

TlIE lJPPl


3/31

3 3472 1

Th

e fo

ll

owing

formul

ae may

be

helpful in answering the questions. The symbols given are

the ones commonly used.

R1111111s-n11n11s berikut boleh e

111b

c111tu

c111da

e

1?imvab soa/m1.

Si111bol-si111bol yang diberi

adalah yang biasa digunakan.

ALGEBRA

X = - b±Jb

2

- 4ac

8

2a

2

a

111

xa

11

= a

111

+

11

9

3

a '

+a

11

=

/

111

-

11

10

4 a

'

)

=

a

11111

5 l

og mn

=

og

m+

log

11

n

12

6

I

111

- 1 - I

g

- OK, m

og

n

n

7

log

111

11

= n log m

13

CALCULUS

K LKULUS

dy dv

du

y = uv,

- +v -

clr dx dx

4

du dv

v u-

u dy dx dx

2 y

= -

= = - - - - = - -

\/

dx

v

2

3

dy dy du

X-

dx du dx

5

3472 1

log b =

log

eb

a

lo

ge

a

7;, =

a+ n - l )c/

n

S

11

= [2a +

n - l)d]

2

T

=

ar

11

-

1

II

Su=

a r

11

1

=

a l

-

r

11

)

.

- I

,

,.

a

Ir < I

W=--

1- 1

Area under a curve

Luas di bawah /e

gk

1111g

b

= dx

or

ata

u)

b

=Jxdy

I

Volume of revolution

Isi padu ki.rnm11

h

= J

r

r or /011)

t i

f

C

X

2

l

}

I

t i

,

/ ;f;

l

fLi hat halam: schclah



4/31

.\- =

2

- L

fx

y

.

I:

.r

3

4

STATISTICS

STATISTJJ

7

8

9

34

7

- L, w. J.

I =

I I

L, ;

n

P =- -

. -

r

)

n

C

,. 11

- r) r

~ ~ x , ; ~ ~

L ; ;

_

1

P(Au B) =

P(A)

+P(B) - P(An

B)

4

11

P X

=

r

)= C,.p''q' _,.

,

p

+

q =

I

5

6

2

[

1

J

N - F

= L

+

2

.

c

J

l O O

Qo

Di stance I Jared<

=Jex, -

x2

)2

+ Yi -

Y2 )

2

Midpoint I Titik tenga

li

.

1

_

(x, + X 2

Yi

+ Y2

J

(

.\,

- 2 ' 2

12

13

14

GEOMETRY

G OM TRI

5

6

3 A

po

int di viding a segment

of

a line

Titik yang

111

e

111b

ahagi

:m

atu te

111b

ere

11

g garis

r

1

= (nx

+

111x

2

11

y

1

+ 111y

2

J

.

J )

) Ill + 1

'

Ill + 11

4 Area

of

triangle I Luas segi tiga

= + X2Jl3 +

X3Y1

1

) -

X

2

.Y1

+ X3.Y2 +

X1

1

3

1

3472 1

Mea n / Mi11,11

= llj

a = J

npq

X - p

Z

CY

I

:

I=

Jx 2+

i

x +

y j

,. =

- J x 2

+

y 2



5/31

2

3

4

5

6

7

Arc leng

th

,

s

=

r

B

Pm1jr111

g lengkok,

s

= j

B

l

2

Area of sector , A = -

r

B

2

i ·2e

L11as sektor L = - J -

, 2

sin

2

A

cos

2

A = 1

sin

2

A+kos2A = 1

sec

2

A = I tan

2

A

sek

2

A = I tan

2

A

cosec

2

A = 1 cot

2

A

kosek

2

A =

l kot

2

A

sin 2A = 2 sin A cos A

s

in 2A

= 2 sin

A

kos

A

cos2A = cos

2

A-sin

2

A

= 2cos

2

A 1

= 1- 2s in

2

A

kos

2A =

kos2

A- sin

2

A

=

2kos

2

A-1

= l

s

in

2

A

3472/1

5

3472/1

TRIGONOMETRY

TJUGONOMETRI

8

9

JO

12

sin

A±

B)

=

si

n

A

cos

B

±c

os

A

s

in

B

s

in A

±

B

=sin

A kos H± kos A sin B

cos A = cos A

cosB

s

in A s in B

kos A±B)=

kosAkosB

+s

i11

A

sin B

_ A B tanA± tanB

tan

±

= -

1

+ tan

A

tan

B

2A

2 tanA

tan - -

l - tan 2 A

a b c

= =

si

n A s

in

B sin C

13

a

2

= b

2

+c

2

- 2bc

cosA

a

2

= b

2

c

2

- 2bc kosA

14

Area

of

triangle I L11as seg

itiga

=_ _ab

s

in

C

2

I

i h

at

lrnlaman scbclahtutormansor.wordpress.com


6/31

For

Examiner s

Use

[[ij

2

[[ij

0

3472 1

2

6

Answer

all

questions.

Ja1vab semu

soafa11

.

Diagram l shows the relation be

twe

en set

P

a

nd

set Q.

Rajah

J

111

e

111111j11kka11

lwb1111

ga

11

a11tam

set

P

da11

set

Q

State

Nya

taka11

z

Set P

Diagram 1

Rajah I

Set Q

a) the type o relat ion between set

P

and se t Q

Jeni.

· //llb1111

ga

11

a

11tam

set

P da11

set

Q

b) the range o the relation.

ju/a/ bagi

lu1b1111

ga11 itu.

An swer I

Ja

wa

pa11:

a)

b)

3472 l

[2 marks]

[2 111arkah]

The fun ction

g

is defined by

g :

x --- ~ x

.

Find the va lue o i

g -

1

6) = 4.

x - 111

[2 marks]

F

. / f n b .

2

C . { . , - I

6) 4

, 1111gs1gCi

1t

a npw11 se

aga1

g : x --- - - , x 111. n 111

m

1111 w g

= .

x -

111

[2

111ar

kcil1]

Answer I Jawapa11:



7/31

7

J 1

172

l

3 In Diagram 3, the fun ction

g

maps x onto

y

and the fun ction maps

y

ont o z.

Do/am Rl fali 3,j

111

gsi g

111

e

111

eta/wn x kepada

y

da

11 f11n

gsi

f111e111etaka11yk

epada 1.

Determine

Te 11t11ka11

(a)

r-

I (5),

.r

8=

Y·- 'J z

1

(b)

g -

1

- 1(5

).

Answer

I

Jmvap

c111:

(a)

(b)

Diagram 3

Rlljali 3

i.

2

marks

l

[/. II

c lah



8/31

.

/,. 5

xaminer s

Use

5

w

0

3472 1

8

347 1

Diagram

5

shows

th

e curves y

= 2 .r - I)

2

-

and y

= x

m)

2

-

11

- 4 ,

where

/11

and

are constants. Both

th

e cur

ves

intersect the x-

ax

is at x

= -

I.

Rajah 5 111e111111j11kka11

/e11gk1111g

y =

2 .Y

I

2 -

da11 y = x

-

111

)2 - 1 1 -

4), de

11ga11

ke

adaa11111

da11

ada/ah pemalar. Ked11a-d11a le11g

k1111

g it11111e11yila11gpaksi-x pada

x

1.

Find

Cari

y

Diag

ram

5

Rajah

5

a) the

va

lue

of

m a

nd of 11

,

11i/ai 111 da1111ilai ,

y

=

x 111)

2

- 11

-

4)

2

y =

2 .r - l ) -

11

b)

the coordinates

of

the minimum point

of

each cur

ve

.

koordi11a

t titik

111i11i11111111

bagi setiap le

11

g

k1111

g i/11

Answer I Jawa 11:

a)

b)

[4m

arks]

[4 111ar

ka/J ]



9/31

9

3472 1

6 Solve the equat i

on

.16 2 x

2

) =

8 x

[3 marks]

For

Examine

r

s

. ,. 2 ,.

Se esa ka persa

aa

16 2 · )

=

8 ·

[3 markah]

Use

An swer Jawapa11

:

Given

th

at log

3

x

= lz and log

3

y

= k

expre ss log

9

[ ]

in terms of

i

and k.

7x

Diberi

log

3

x

= i clan log3

y

= k

g

kapka11 log9

Answer Jawapa11:

3472 1

[

Y

2

]

da am seb ta

i

da

k.

27x

[4 marks]

[4

markah]

7

m

0

Lih at halaman sebelah



10/31

Fvr

Use

8

[Gj

0

347 1

8

1

3472 1

The price of a medium cost hou se in Butterworth

on

I

t

April 2014 is RM 120,000. At

that time Hassan

lrn

s a saving of

RM 1

,000. The price of

th

e

hou

se is ex pected

to

increase by 6 every year while Hassan's saving is expec ted to increase by 8 every

year. On the l

st

April

of

which year Ha ssan manage to purchase the hou se in cash?

Harga sebuah

n111w/J

kos seder/1011a di Butterworth pada I April 2014 adalah

RM 120,000. Pada ma

s

it11

Ha

ssan

111e111p1111yai

s

i111pa11a11

seba

11yak

RM 100,000.

Harga m111a/J itu d iw1gka 1wik seba11yak

6

setiap ta/11111

111a11akala

s

i111pana11

Hassan

dijm1gka 1wik seba11y k

8

setiap ta/11111 Pada

I

April ta/11111 ke b

emp

kah Hassan

dapat e111beli mma J terse mt secara t1111ai?

Answer Jawapa11:

[4

marks]

[ 4 markah]



11/31

3472/1

9

t is g

iv

en that

k 8 k

and k 2 are the first t

hree

terms of a geometric

progression.

Diberi bahawa k 8 k dc k 2 adalali tiga seb ta µertama bagi s

wt

j{IJ1ia11g

geo111etri.

Find

Cari

a)

th

e

va lu

e of

k

11ilai k

b) the common rat io of the progress ion.

nisbah

sep1111ya

jm1ja g

i i

.

An swer

I Jawapan:

a)

b)

[3 marks]

[3

111arkali]

10

An arithmetic progression bas 18 te

rm

s The first term is J

5

Given that

th

e s

um

of

the last 8

term

s

is

552, fi

nd

the co

mm

on differen

ce

of the progression

S11at

j{IJ1ia11g

aritmetik memµ1myai

18

seb ta . Seb11ta11 perlama ialolt

J5.

Diberi

sil tambah 8 se

b t

c

ternkhir adalalt 552, cari beza sep1111yajc111jm1g itu.

Answer

I

Jmvapa11:

[3 marks]

[3

markali]

For

xn

er s

U1e

10

j

3472/1

Lihat halaman

se o



12/31

12

3472/1

or

11

The points

P l

,-2), Q R(5,4) and

Sare

the vertices of a rhombus. Find the equation of

the diagonal

QS.

4 marks]

xa i er s

Use

[8j

12

m

3472/1

Titik-titik P l

,-2),

Q, R(5 ,4)

da

S adalah bucu-bucu bagi sebuah rombus. Cari

persamaa11

pepe11juru

QS.

[4 markah]

Answer I Jmvapa11:

12 After finishing his lesson, Amin looked at the clock on the wall. He thought he could

find

the equation of the movement of the minute hand if he knew the length of the

minute hand from the centre.

Selepas selesai pembelqjarc111, Amin melilwt jam di di11di11g. Dia te1jikir dia boleh

me cari persamaa11 pergerakan }arum mini jika dia tahu pa11ja11g }arum mini

daripada pusat.

Please help Amin to determine the equation of the movement of the minute hand

if

the

length is 5 cm from the centre. 2 marks]

To/011g ba11tu Amin me

e tukc persamaa11 pergeraka11 }arum mini

jika

panja11g11ya

ia/ah 5 cm daripada pusat. 2 arkah]

Answer I Jawapa11:



13/31

13

3472/1

13 Diagram a) shows the curve

y

= -

2x

3

4.

Diagram l l b) shows the straight

line graph obtained when y = - 2x

3

4 is expressed in the

lin

ear fo rm Y= 4X c

14

Rajah

a)

e

n1111j11

kka

fengk1111g

y

= -

2x

3

4. Rajah 11

b)

e111mj11kka11 grqf

garis /urns y

an

g diperoleh apabila y = -

2 x

3

+

4

di1111gkap dala111 be11t11k linear

Y

=

4X

c.

y

0

Diagram a)

Raiah 11 a)

Ex

press

X

a

nd Y

in terms ofx and mor y

Ungkapkan X dc Y dala seb11ta11 x da11 atau y.

An

swer Jawapa :

y

Di

agram I l b)

Raiah

11

b)

x

3

marks]

3 111ar

kah]

Given ABCD is a parallelogram, where BC = t 2L and

CD=

- 3£ 4L. Find the

uni t vector in the direction of AC

Diberi ABCD ialah segi empat selari, denga keadaa BC = l

-

2j dan

CD = 3£ 4j . Cari vector 1111it

da/0111

arah AC

Answer

Jawapa11:

3

mark

s]

3

111arkah]

For

T:xmniner s

Us e

13

m

14

m

3472/1

[Lihat lrnlaman sebclah



14/31

For 15

Exam

iner s

Use

5

cm

0

3472 1

14

Diagram 15 shows the graph of y

=

2 cos 11x .

Rajah 15 me111111j11kka11 g

r f

y = 2 cos

11x

y

k

0

- k

a) State the va lue

of

Nyatakan 11ilai

i) k,

ii) .

Diagram 15

Rajah

5

3472 1

y

=

2 cos nx.

b) Hence on the axes in the answer space, sketch the graph ofy = 2

cos

nx - t

for

0 X

7L

Sete

r11

s ya pada paksi yang diberi da/am

r11a11

g jawapa11, lakarka11

gr f

y

=

2

cos

nx

-

ll

k

0

x

n.

Ans

wer Ja111apa

:

a) i)

ii)

b)

y

0

[4 marks]

[4 markah]



15/31

15 3472/1

16

g_ = + y

= 4 - Y

£ =

h

+ (

k

+

3h

y

where h and k are constants

de11ga11 keadacm h

da11

k ialah pemalar

Use the above information to find the value of hand of k

if

2a

3b

- 4c.

- - -

Gu11akc111 maklumat ya11g di atas w1tuk e11cari

11ilai

1 dan nilai k jika

2a =

3b 4c.

- - -

Answer

I

Jawapa11:

3

marks]

3 111arkah]

17 The equation

of

a curve which passes through the point M is

y

= x

-

3) x + 2).

The gradient of the normal

to

the curve at point Mis - .

3

Find the coordinates

ofM.

Persa111aa11

s11at11 le11gku g yang e

la/11i

satu titik M ialah y

=

.r - 3) x

+

2).

Kecenman 11or111al kepada lengku11g it11 pada titik M ialah - .Cari koordi11af M

3

Answer I

Jmvapcm:

3 marks]

(3 111arkali]

For

t:xa111i11er

s

Use

16

m

17

m

0

3472/1 [Lihat halaman sebeJah



16/31

For 18

E.w1111i11rr s

Use

•

8

,,;

[8j

0

3472 1

16

3472 l

A circle that is formed from an iron wire

ha

s a di ameter o I0 cm. When the wire is

hea ted, its radi us

in

creases by 0.04 cm.

Seb11ah

b11/ata11

yang di/Jasilka11 daripada

suat11

logct111 111e111p1111 i diameter 10

cm.

Apab;/a dipa11aska11 jejari b11/ata

11

terseb11t bertambah

seba11yak

0.04 cm.

Fi

nd

Cari

(a) the small change in the a

rea

o the circle,

pernba/Jan kecil pada 11as b11 atc111 terse/mt

(b) the approximate va lue

o

the new area

o

the circle.

a11gga

ra11

nilai bagi luas

b11/ata11

yang barn.

Answer Jawapa11 :

(a)

(

b

[4 marks]

[4 markah]



17/31

I

I

17

3472 1

19

Diagram 9 shows the shaded region bounded

by th

e cur

ve

y = h x)

and

the

straight linex = 7.

Rqjah

9

me

11111 i11kk

a

rn11ta11

berlorek ng dibatasi oleh le

g

lw11

g y

=

h \)

da11

garis 11ms x

= 7.

y

Diagram 9

Rajah

9

.l =

/i .r)

t

is given that

th

e area

of th

e shad ed region is 8 unit

2

.

Diberi balwwa

11as

ra11ta11 berlorek ialah 8

i

t1.

Find

Cari

a)

b)

J x)

r,

7

x

+ 4/i x)] dx

7

Answer

I Jawapm1:

a)

b)

[4 marks]

[4

markah]

For

1:·.rn111i11er

0

.1

Use

19

[Ej

0

3472

[Lih t ha

lmn

n sebclah



18/31

For 2

w i er s

Use

20

J

3472

1

8

3472 1

Diagram 20 s

ho

ws a

ri

g

ht

angled triangle ABC and sector XCY with centre C

Rajah

0

e f11kka11 segi tiga bers

uc/111

tepat ABC da11 sector X Y berpusat C

x

Diagram 20

Rajah 20

ll is given that AB = BC = 12 cm and the ratio CY: CB = 3 : 4.

Diberi bahmva AB = BC =

12

cm da1111is

ba/J

CY:

CB

=

3 : 4

[U se G1111a n = 3.142]

r

ind

Cari

a)

L.

ACB in radi ans Give yo ur answer correc t to 4 dec

im

al pla ces.

L.

ACB da/a1JJ radian Beri jmvapa11 anda betul kepada 4 te111pat

perp11

/11/Ja11.

b)

the area, in cm

2

, o

the shaded region.

lua

s

clala111 cm

2

, km

vasa

berlorek.

Answer I

Jawapa11:

a)

b)

[4

ma

rk

s]

[4 arka/J]



19/31

19

3472/1

21 Table

2

shows the

fr

equency

di

stribution of

th

e talk time during a

ca

ll of40 peopl

e

Jadual 2 e1111nj11kkm1 tab11rn11 kekernpttn masa bercakap da/am telefo

bagi 40

orn g

Talk

t m ~

minut

es

)

Masa Bercakap minit)

3.00 - 3.40

3.50 - 3.90

4.00 - 4.40

4.50 - 4.90

5.00 - 5.40

5.50 - 5.

9

Table 2

Jad11al

2

Find the median of the talk tim e.

Ca ri edia11 bagi masa bercakap

Answer Jawapc

:

Number

of

p

eo

ple

Bi/a11ga11 orang

6

4

7

8

2

3

[3 marks]

[3 arka

1]

or

E rn111i11l r s

Use

21

[Jij

0

3472/1

I

Li

hat halaman sebelah



20/31

or

Examiner s

Use

m

0

3472 1

22

20

47 1

A c

lass o

30 s

t

ents took a sp

ec

ial tes

t.

Their scores are shown

in

Table 22.

Seb11alr kelas yang terdiri daripada 30 orang pe/ajar e g

a111bi/

11jia11 klras.

Kep11 sm1 skor ereka adalalr seperti da/0111 Jad11al

22.

Score

skor

I

2

3

4

5

a) State the modal score.

101afalw11 skor 111od

Number

o

s

tud

e

nt

s

Bilc111ga11 pelqjar

Table

22

Jad11a/

22

9

6

5

2

8

b) Calculate the standard deviation.

Hit1111g sisilrm1 piawai

Answcr

Ja111ap

a

11:

a)

b)

4 marks]

[4 markalr]



21/31

23

21

3472 1

There are 6 different flavours o ice cream : Carrot, Mango, Honey Dew, Van illa ,

Strawberry and Chocolate

Terdapal 6 perisa ais krim yang berbeza : Lobctl<

Ma11g

ga

Te

mbilmi Susu,

Va11i/a

Strawberi da11 Coklat.

Find

Cari

a)

th

e number

o

ways

3

different

fla vo

ur

s

o

ice cream can be chosen,

bila11gan

cam 3 perisa ais krim yang berbeza bole/ dipilih ,

b)

th

e number o ways at least 5 differe

nt

navours o ice cream ca n be c

ho

sen.

hila11ga11

cara sekurang-kura11g ya 5 perisa ois krim yang berbeza boleli

dipilili .

An

swer

I

Jawapa11:

a)

b)

[3 marks]

[3 111arka/J]

or

li.rn111i11er s

Us

23

m

3472 1

ILihat halaman scbelah



22/31

fo , · 24

Examiner s

Use

24

[Jij

3472 1

22

Two fair

di

ce are rolled toge ther.

Dua

dad11

adil

dila111b1111

g bersa111a.

F

ind

the probability that the num bers on them

Cari ke

bara11

g

ka ia11

balta1va 0

111b

or pada dad11 itu

(a) have a sum

of 11

e111p11nyai hasil ta111bah

b) have at least one number

6

e

i sek11ra11g-k11ra11g ya satu

11 111bor

6 .

An swer Jawapa11:

(a)

(b)

3472 1

[2 marks]

[2

111arkah]



23/31

23

25 In a gam e o guess

in

g, the probabi

li

ty o guessing correctly is p The mean and the

standard deviation o success are 15 and

l15

respectively.

Dala111 per111ai11a11 tekaan kebara11gkalim1 tekaa11 bet11 ialah p.

Min

dan sisiha piawai

kejayaan

111a

sing-

111a

si g adalah 15

clan

}__

J5

.

2

F

ind

Ca

ri

a)

the value

o p

11ilai p

b) the number o trials required.

bilangan c11baa11 yang diperluka11.

Answer

I

Jmvapa11:

a)

b)

END

O

QUESTION PAPER

KERT SSO L N T M T

[4

marks]

[4 111arka

h]

or

Exa111i11er

0

s

Use

25

[8j

3472/1

ILihat halaman sebelah



24/31

24

INFORMATION FOR CANDIDATES

MA/ LUMAT UNTU/

CALON

1. This question paper consists

of

25 questions.

Ker/as soalan ini mengand11ngi 25 s

oa/a11.

2. Answer all qu estions.

Jm vab semua soala11.

3. Write your an swers

in th

e spaces provided

in

this qu estion pape

r.

.Jawapa11 hendaklah dit11/is pada

ma11

g ya11g

disediaka11

dalam kertas soalan ini.

4. Show y

om

working.

t

may he

lp

y

ou

to get marks.

472/1

T111 j11kkm1 langkah-

la11

gkah pent ng do/al/I kerja l/lengira 011da. iii boleh l/le111ba11t11 anda

11nt11k mendapatlwn

111arkali.

5. If you wish

to

change your answer, cross out

th

e work that you have done.

Then write clown

th

e new

an

swer.

Sekiranya anda he

11dak

en11kar ja111apa11, batalkm1 jm vapa11 yang telah dib11at. Ke1n11dia11

t11/isjawapa11 yang bam.

6. The diagram s in the questions provided are not drawn to

sc al

e unl ess state

d.

Rajah yang me g iri11gi soalan tidak di/11kis meng ik11t ska/a kec11ali dinyataka11.

7. The marks allocated for each qu estion are shown in bra ckets.

Markah yang d

ip

e/ l/11/11kkan bagi setiap soalan

dit111u

'

11kkan

dalmn lwmng

an.

8. The Upp er Ta

il

Probability Q z) For The Normal Di stribution N O, 1) Table is provided on

page 2.

Jad11

al Ke barangkalian f-111j11ng Alas Q z) Bagi

Tab11ran

Normal N O , 1) disediakan di

hala111an 2.

9.

A list

of

formulae is prov

id

ed on pages 3

to 5.

Sa t se11C1mi I JI II

.\ di

sediakan di halamm13 hingga 5.

10. Yo

u

ma

y use a scien i e calcu

a

tor.

Anda dibe

11ar

ka11 me gg

1111alw11

kalk11/ator suint({tk.

11. Hanel in

thi s ques tion paper to

th

e invigilator

at

th

e end of the ex amination.

Semhkan kerfas soalan ini kepada pengmvas peperiksaan di akhir peperiksaan.

3472/1

Lihat hala man sebela h



25/31

I

ADDITJONAL MATHEMATICS

Kertas I

Ogas

2 jam

M JLIS

PENGETU SEKOL H

MALAYSIA

CAWANGAN PULAU PINANG

MODUL LArflHAN BERFOKUS SPM 2015

MARJ SCHEME

ADDITIONAL M THEM TICS

Paper 1

Two hours



26/31

2

347 2 /J

ADDITIONAL MATH EMJ\TI

CS

P

I

-

Question

Solutio

and

Mark

Schc

c

Sub Total

Marks Mark

a)

Many

lo one

1

I

2

b) {c, g}

2 3

2 2

Bl:

-3 _ =6

g I (x)

=

2

x

or

4 m

x

3 a) 3

I

.

I 2

b)

1

..

> ·

4

6

3

3

B2:

-

6

- 4 1) 3 + t

=

0 or

y

= -

x - 3 2

+ 6

B I :

x

2

-

6x

3 +

I

= 0

or y = - [x

2

- 6x+ -3/ - -3)2]-3

5 a)

m = 1 and

n = 8

2

Bl: m =

l

b)

1, -8 )

1

l, -4)

1

4

6

3

3 3

B2:

4

x

2 =

3x

Bl

:

16

=2

4

or

8 =

2

3

x

3472 1 tutormansor.wordpress.com


27/31

7

8

9

(a)

(b)

3472 1

3

k 3 -

2

I

o r -- 2k 3 h) 1 k

2

3 h

2 2

2 log

3

y - log

3

27 - log

3

x

B3 :

- - - ~ ~ - = - - ~ ~ - - - -

2log

3

y

log3 27 log

3

x

or

-

- log3 9 Jog3 9 log3 9

Jog

3

y

2

- log

3

27

x

B2 :

log

3

9

or log

9

y

2

- log

9

27 - log

9

x

or

l

og9

/ - l

og9

27x

2024

B3: n

= 11

B2:

100 000(1.08)1'-

1

>

120

OOO(l.06y-

1

131 : a

=

120000, r

=

1.06

or

a

=

100000,

r

=

1.08

or

R

2024

133

:

n

=

11

B2:

120000,127200, 134832, 142922, 151497, 160587,

170222 180435 191262, 202737, 214902

nd

100000 108000, 116640, 125971, 136049 146933.,

158

687

171382,

185093

199900, 215892

Bl :

120000 ,127200 134832,

01

100000 ' 1 8000, 116640, . . .

8

3

Bl :

k

k 2

-

k+8 k

1

-

4

3472

1

-

4

4

4

4

2

1

3



28/31

-

-

-

-

Jtl7211

~ -

- -

1

u 4

3

n 2 :

·I

IOd

+

7d )= 552

2

or

5 ) +

18- 1

ct

) -

. _Q

[

2{15)+

0 - I}I]

=

552

2 2

18

l :

T

1

=

15 IOd or

- [

2 1

5)+ 8- l

)ct

]

or

2

2

11

3y

=

- 2x 9

2

4

4

r y = x

3

3

B3:

y

-*

l

or

*I =

-

j

*3)

+c

B2:

3,

1

and

2

ll lQS

= - -

3

B

l :

3,1

3

or

111

R

-

-

2

2

x

2

y

2

- 25 =

0

2

2

Bl :

J

x -

0)

2

y

0)

2

= 5

·

No

te : Accept any coordinat

es

for the centre.

13

Y= L

X

x3

)

XJ

3 3

B2:

Y = L

1

or·

X

X J

X3

B

l :

y

4

2+

-x3 x3

14

-

1

-

4i

- 6j )

3

3

52

B2: + -6 2

AC{

46J

l

AC = 4i -

6j

or



29/31

5

3472 J

-

15

a)

i)

k =

2

ii )

II

=

3

b)

)

3 -

-

2

4

0

7r

Bl : Y

16

h

=

2

and

- 33

k

= -

or k

= -

8.25

3 3

4

B2.:

4 =

12

- 4h

or

6 = - 3 - 4k - l2h

Bl :

2

J

3

i - k : J

7

2, -4)

3

3

B2: x =2

Bl:

dy

=

2x l

dx

or

1111 =

3

18

a)

2

0.4n 1.257

2

T r

or

or

5

Bl

:

b A

=

lOn

x

0 04

b)

25.4

7r

127

79.80

r

7

or

2

4

Bl :

o =25n

3472 1



30/31

6

3472/l

-

-

- --

19

l

-8

J

3 4

(b)

4

82:

[

11

2 7

2

]

2 +4 -8)

Bl:

x i

2 [__

4

- 8)

20

(a)

0.7855

l

3

4

(b) 40.19

B2 :

_ _x 12x 12- ]_(9)

2

* 0.7855)

2 2

Bl :

1

_ _ (9

)2

(0.7854).

x

12 x 12

or

2

2

21

4 .638

371

3

3

or -

80

. . [ 40 17J

B2 : .

4.45+

2

8

0.5

Bl : L

4 45orF=l7orf=8or

c

0.5

(a)

(b)

1.579

3

4

B2 :

131 - (2.8)2

30

-

Bl :

x

=2 .8



31/31

7

34

7

J

-

- - -

- -

-

23

a)

20

I

b)

7

2 3

Bl:

c

6

C

5

6

01

6Cs I

or 6

6

C

6

24 a)

I

I

-

18

b) I I

l 2

36

25

a)

l

-

3

4

B :

3

q=

4

Bl:

np =

15

or

Jl5q

=1_ Js

2

b)

60

I 4

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