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3472/2 ADDITIONAL MATHEMATICS Kertas 2 Ogos

21 •

2 Jam

MAJLIS PENGETUA SEKOLAH MALAYSIA CAW ANGAN PULAU PINANG

MODUL LATIHAN BERFOKUS SPM 2015

ADDITIONAL MATHEMATICS

KERTAS2

Dua jam tiga puluh minit

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

I . Kertas soalan i11i adala/J dalam dwiba/Jasa.

2. Soalan dalmn balwsa !11ggeris menda/111/ui soala11 yang sepoda11 dalam ba/Jasa Meh~1111.

3. Ca/011 dike/Je11daki 111e111baca 111ak/11mat di hala11w11 belaka11g kertas soala11 ini.

4 . Ca/011 dike/Je11daki 111e11ceraika11 lw/a111a11 19 dan ikat sebagai 11111ka /Jadapm1 bersama-sama dengan kertas jmvaµcm.

Kertas soalan ini mc11ga11du11gi 19 halaman bcrcctak clan I halaman tidak bcrcetak.

[Lihat halaman selJ elah

3472/2 © flak cipta AIPSM P11ft111 Pi11a11g 2015

tutormansor.wordpress.com

z

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

I. I

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2.0

2. 1

2.2

2.3

2.4

2.5

2.6

2.7

2.8

2.9

3.0

2

THE UPPER TAIL PROBABILITY Q(z) FOR THE NORMAL DISTRIBUTION N(O, 1) KEBARANGKALIAN HUJUNG ATAS Q(z) BAGI TAB URAN NORMAL N(O, 1)

0 I 2 3 4 5 6 7 8 9

0.5000 0.4960 0.4920 0.4880 0.4840 0.4801 0.4761 0.4721 0.4681 0.464 1

0.4602 0.4562 0.4522 0.4483 0.4443 0.4404 0.4364 0.4325 0.4286 0.4247

0.4207 0.4168 0.4 129 0.4090 0.4052 0.4013 0.3974 0.3936 0.3897 0.3859

0.382 1 0.3783 0..1745 0.3707 0.3669 0.3632 0.3594 0.3557 0.3520 0.3483

0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3228 0.3192 0.3 156 0.3121

0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810 0.2776

0.2743 0.2709 0.2676 0.2643 0.26 11 0.2578 0.2546 0.25 14 0.2483 0.245 1

0.2420 0.2389 0.2358 0.2327 0.2296 0.2266 0.2236 0.2206 0.2 177 0.2148

0.2119 0.2090 0.206 1 0.2033 0.2005 0.1977 0. 1949 0.1922 0.1894 0. 1867

0.1841 0.1814 0. 1788 0.1762 0.1736 0.1711 0.1685 0.1660 0.1635 0.161 1

0.1587 0.1562 0.1539 0.15 15 0. 1492 0.1469 0.1446 0. 1423 0.1401 0.1379

0. 1357 0.1335 0.1314 0.1292 0. 127 1 0.1251 0. 1230 0.1210 0.1 190 0.1170

0.1151 0.1131 0.1 112 0.1093 0.1075 0.1056 0.1038 0.1020 0. 1003 0.0985

0.0968 0,095 1 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823

0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.072 1 0.0708 0.0694 0.0681

0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559

0.0548 0.0537 0.0526 0.0516 0.0505 0.0495 0.0485 0 .. 0475 0.0465 0.0455

0.0446 0.0436 0.0427 0.0418 0.0409 0.0401 0.0392 0.0384 0.0375 0.0367

0.0359 0.0351 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307 0.0301 0.0294

0.0287 0.028 1 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 0.0239 0.0233

0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0 192 0.0188 0.0 183

0.0 179 0.0174 0.0170 0.0166 0.0 162 0.0158 0.0154 0.0150 0.0146 0.0143

0.0139 0.0136 0.0132 0.0129 0.0 125 0.0 122 0.01 19 0.0116 0.0113 0.01 10

0.0107 0.0 104 0.0102

0.00990 0.00964 0.00939 0.00914

0.00889 0.00866 0.00842

0.00820 0.00798 0.00776 0.00755 0.0073<1

0.007 14 0.00695 0.00676 0.00657 0.00639

0.00621 0.00604 0.00587 0.00570 0.00554 0.00539 0.00523 0.00508 0.00494 0.00480

0.00466 0.00453 0.004•10 0.00427 0.00415 0.00402 0.0039 1 0.00379 0.00368 0.00357

0.00347 0.00336 0.00326 0.00317 0.00307 0.00298 0.00289 0.00280 0.00272 0.00264

0.00256 0.00248 0.00240 0.00233 0.00226 0.002 19 0.002 12 0.00205 0.00199 0.00193

0.00187 0.00181 0.00175 0.00169 0.00164 0.00159 0.00154 0.00149 0.00144 0.00139

0.00135 0.0013 1 0.001 26 0.00122 0.00118 0.00114 0.001 11 0.00107 0.00 104 0.00100

Example I Co11to/i:

/(z) = k exp(- ~z 2 ) f(z)

~

Q(z ) = J /(z) dz ~

I

4

4

4

4

4

3

3

3

3

3

2

2

2

2

I

I

I

I

I

I

0

0

0

0

3

2

2

2

2

I

I

I

0

0

2 3 4 5 6 7 8

Minus I To lak

8 12 16 20 24 28 32

8 12 16 20 24 28 32

8 12 15 19 23 27 31

7 I I 15 19 22 26 30

7 II 15 18 22 25 29

7 10 14 17 20 24 27

7 10 13 16 19 23 26

6 9 12 15 18 21 24

5 8 II 14 16 19 22

5 8 10 13 15 18 20

5 7 9 12 14 16 19

4 6 8 10 12 14 16

4 6 7 9 II 13 15

3 5 6 8 IO I I 13

3 4 6 7 8 10 II

2 4 5 6 7 8 10

2 3 4 5 6 7 8

2 3 4 4 5 6 7

I 2 3 4 4 5 6

I 2 2 3 4 4 5

I I 2 2 3 3 4

I I 2 2 2 3 3

I I I 2 2 2 3

I I I I 2 2 2

5 8 IO 13 15 18 20

5 7 9 12 14 16 16

4 6 8 II 13 15 17

4 6 7 9 II 13 15

3 5 6 8 9 II 12

2 3 5 6 7 9 9

2 3 4 5 6 7 8

I 2 3 4 4 5 6

I I 2 2 3 3 4

I I 2 2 2 3 3

ff X ~ N(O, I), then

Jika X ~ N(O, 1), maka

P(X > k) = Q(k)

9

36

36

35

34

32

31

29

27

25

23

21

18

17

14

13

II

9

8

6

5

4

4

3

2

23

21

19

17

14

10

9

6

4

4

P(X> 2. 1) = Q(2. 1) = 0.0179

0 k z

34 72/2 © I/ale cipta M PSM />11/a11 Pi11a11g 2015


3

The fo llowing formulae may be helpful in answering the questions. The symbols given are the ones commonly used. R11111us-m11111s bel'i/wt bo/eh 111e111bantu anda 111enja111ab soa/an. Si111bo/-si111bol yang diberi adalah yang biasa dig11nakan.

- b ±-Jb2 - 4ac x =

2a

2 c/11 x a" - 111 + 11 - a

3 a'" + a'' - 111 - 11 - a

5 log,, 11111 = log,., /11 + log.., n

Ill 6 log0 - = loga Ill - log0 n

n

2

3

dy dv du y = uv, -= 11 - + v -

dx dx d"

du dv II c/y

v-- 11 y = - , -=

v dx

dy c(y du - = - X -clr du dx

dr dr v2

34 72/2 © Hale cipta MPSM P11/a11 Pi11a11g 2015

ALGEBRA

8

9 T,1 = a + (n - l ) d

I 0 S11 = !!. [2a + (n - J)d) 2

I I 'T' JI - I l II = QI

ak1 - 1) a(l - r 11'

12 Su = = ~' r -:1= r - 1 1- r

13 S'-" =_!!_ ,!ri < I I - ,.

CALCULUS KALK UL US

4 Arca under a curve

Luas di bawah /engkung

b

= J y dx or (atau) (I

h

= Jx dy (I

5 Volume of revolution Jsi padu kisaran

" = J ~ y 2 dx or (atau)

" b

= J ~ x 2 c(>'

a

[Lihat hal:iman scbclah


- I> x =--N

2

3

4 (J =

5111 = L + [~N - F]C .!,,,

6 I = i1_ x 100 Qo

Distance /Jarnk

2 Midpoint /Titik /e11ga/J

( x ' y) ~ (''· : x, / • : y,) 3 A point dividing a segment of a line

4

STATISTICS STATISTJJ(

7 - ""W.1. ! =£_, I I

L:wi

8 llp - JI! /' - ( ) JI - /'!

9 Ile = n! /' (11 - /') ! /' !

10 P(A u B) = P(A)+P(B) - P(AnB)

) "C ,. II ,. l 11 P(X = I = ,. p q - ' p + q =

12 Mean I Min , µ = 11p

13 (]" = ;;;pg 14 Z = X - µ

(]"

GEOMETRY GEOMETRI

s Id = ~x2 + >'2

6 " xl + yj /' = -

Jx2 + y 2

Titik yang 111e111ba/Jagi s11at11 te111bere11g garis

( ) (

l1X1 + ll1X 2 llY1 + 111Y2) x , y = '

111+11 111 + 11

4 Area of triangle I l11as segi tiga

34 72/2 ({) Ha/1 cipta MPS M Pu/au Pi11a11g 20 15


Arc length, s = r0

Pa11ja11g le11glwk, s = j0

2 Area of sector A = _!_,. 2e , 2

Luas sektor, l = _!_ /e 2·

3 sin2 A + cos2 A = l

si112 A + kos2 A = I

4 sec2 A = l + tan2 A

se!! A = 1 + tmi2 A

5 cosec2 A = l + cot2 A

kosel! A = I + kot2 A

G sin 2A = 2 sin A cos A

sin 2A = 2 si11 A kos A

7 cos 2A = cos2 A - sin2 A

2 cos2 A - 1

l - 2 sin2 A

kos 2A = kos2 A - si112 A

2 kos2 A - l

I - 2 sin2 A

5

TRIGONOMETRY TJUGON01l1ETJU

8 sin (A ± B) = sin A cos B ± cos A sin B

si11 (A ± B) =sin A kos B ± kos A si11 B

9 cos (A ± B) = cos A cos B +sin A sin B

kos (A ± B) = kos A kos B +sin A sh1 B

10 (A B) tan A ± tan B tan ± = J+tan AlanB

11 2 tan A tan 2A =

1- tan 2 A

{/ b c 12 --=--= --

sin A sin B sin C

13 a 2 = b 2 + c 2 - 2bc cos A.

a 2 = b 2 + c 2 - 2bc kos A.

14 Area of triangle I L11as segi tiga

= _!_ab sin C 2

[Lihat halaman sebclah

3472/2 © Hale cipta MPSM P11/a 11 Pi11a11g 2015


1

6

Section A Ba/Jagian A

(40 marks] [ 40 111arkah]

Answer all questions. Jmvab semua soalan.

Solve the simultaneous equations p + 2k = 3 and -1- _ _!_ = I .

p - 1 k Give your answers correct to three decimal places.

1 1 Selesaikan persa111aa11 serentak p + 2k = 3 dan --- - = I .

p - 1 k Beri jmvapan anda betul kepada tiga te111pat perpuluhan.

2 Given that/: x ~ 4x + 3 and g: x ~ 2x - 5, find

Diberi f: x ~ 4x + 3 clan g: x ~ 2.\· - 5, cari

3

(a) .r-1 (x) ,

(b) g/-1 (x),

(c) h (x) such that hg (x) = 8x - 9.

'1 (x) de11ga11 keadaan hg(x) = 8x - 9.

P I cos A sin 11 2cos(A + B)

(a) rovetrnt -----= . sinB cosB sin 2B

B , ., kosA sin A 2kos(A + B)

lf1(//J((/I/ -- - -- = . sin B kosB sin 2B

[5 marks]

[ 5 111arkah]

[ l mark] [ 1 111arkah]

(2 marks] [2 111arkah]

(2 marks] [2 markah]

(2 marks]

(2 IJWJ'kafl]

(b) Given that tan A = ~ and sin B = ~, where A and Bare acute angles, find the value of "3 5

Diberi tan A= ~ dan sin B = ~, de11ga11 keadaan A dan B adalah sudut tims, cari nilai "3 5

(i) sin 28,

(ii) 2cos(A +B)

sin 2B

2kos(A +B)

sin 2B

3472/2 © /lak cipta MPSM P11/a11 Pil/{111g 2015

[5 marks]

[ 5 111arkah]


7

4 Diagram 4 shows the graph of a quadratic function .f(x) = 3x 2 + I 2x + c . The graph has a minimum point at A and intersects thefi.x)-axis at B (0, 1).

Rajah 4 J11e111111j11kkan graffu11gsi kuadratik f(x) = 3x2 + 12x + c. Gm/ itu J11eJ11p1111yai titik J11i11iJ1111J11 pada A da11 J11enyi/ang paksi -j{x) pada B (0, 1).

j(x) = 3x2 + 12x + c

f{x)

A

Diagram 4 Rajah 4

B(O, 1)

x

(a) State the value of c. [ 1 mark]

Nyataka11 ni/ai c. [ l 111arka'1]

(b) By using the method of completing the square, find the coordinates of A. [ 4 marks]

De11ga11 J11e11gg1111aka11 kaedah pe11ye111pumam1 Iwasa dua, cari koordi11at A. [ 4 //larkah]

( c) Determine the range of values of x, if /(.r);::: 16. [2 marks]

Te11tuka11 ju/at ni/ai x,jika.f(x) ::::: 16. [2 111arkah]

(cl) If the above graph is refl ected about the x-axis, write down the equation of [1 mark] the curve.

Jika gra.f di alas diprmtu/ka11 pada paksi-x, tu/iska11 persm11am1 /e11gku11g itu. [I 111arkah]

3472/2 © flak cipta MPSM P11/a11 Pim111g 2015


8

5 Diagram 5 shows a triangle PQR. The straight line PB intersects the straight line QC at point A.

Rajah 5 men1111j11kka11 sebuah segi liga PQR. Garis /urns PB bersila11g de11ga11 garis /urns QC pada titik A.

Diagram 5

Rajah 5

1 - -It is given that PC = 2 CR , QB = 3 QR , PQ = 6~ and PR = 8t

Diberi balwwa PC = 2 CR , QB = .;- QR , PQ = 6~ don PR = 8y.

(a) Express, in terms of l and y

U11gkapka11, dalam seb11ta11 ~ da11 y

(i) QR , (ii) PB.

[3 marks] [3 J/IOl'ka/J]

(b) Given that PA = kPB and PA = PC+ hCQ , where k and h are constants, find the value of k and of h .

[5 marks]

Diberi PA = k PB da11 PA = PC + hCQ, de11ga11 keadaa11 k dan '1 iala'1 pemalar, cari 11ilai k da1111ilai '1 .

G The gradient function of a curve that passes through A (5, 8) is 2 - x .

Fungsi kecen111a11 Slfaf11 lengkung yang 111ela/11i A (5, 8) ialah 2 - x.

Find

Cari

(a) the equation of the normal to the curve at point A,

persa111aa11 11or111al kepada lenglw11g itu pada titik A,

(b) the equation of the curve.

persa111aa11 le11gk1111g itu.

3472/2 © Hak cipta MPSM Pu/au Pi11r111g 201 5

[ 5 111arkah]

(4 marks]

[ 4 111arkah]

[3 marks]

[3 111arka'1]


9

Section B Balwgim1 B

[40 marks] [ 40 markah]

Answer any four questions from this section. Jawab 111a11a-111a11a empat soala11 daripada bahagia11 i11i.

7 Diagram 7 shows triangle AGE and triangle BCH. The straight line AC intersects the straight line GH at point B.

Rajah 7 111e111111jukka11 segi tiga AGB da11 segi tiga BCH. Garis !urns AC bersila11g de11ga11 garis lums GH pada titik B.

y

0 Diagram 7 Rajah 7

It is given that Bis the midpoint of CH and AB : BC = l : 2.

Diberi bahawa B ialah titik tengah CH da11 AB : BC = I : 2.

(a) Find

Cari

(i) the equation of the straight line GH,

persamaan garis !t11·11s GH,

(ii) the coordinates of C,

koordi11at C,

(iii) the area, in unit2, of triangle AGB.

luas, dalam unit2 ,segi tiga AGE. [8 marks]

[8 mark ah]

(b) Determine whether the straight lines AC and GH are perpendicular to each other.

Te11tuka11 sama ada garis /urus AC dan GH bersere11ja11g a11tarn satu sc1111a lain.

3472/2 © Hak cipta MPSM P11/a11 Piuaug 2015

[2 marks] [2 markah]


10

8 Use the graph paper to answer this question.

G1111a kertas gmf 1111t11k menjawab soalan ini.

Table 8 shows the va lues of two variables, x and y obtained from an experiment. The variables p+qx

x and y are related by the equation y = 2

, where p and q are constants. x

Jadual 8 me111111j11kka11 11ilai-11ilai bagi dua pe111bo/e/111bah, x dan y, yang dipero/eh daripada p+qx

s11at11 eksperimen. Pe111bo/e/111bah x da11 y di/111b1111gka11 of eh persa111aa11 y = , dengan

keadacm p dan q ia/ah pe111alar.

x 1.5 2.5

y 0.33 0.64

3.5

0.58

Table 8 Jadua/ 8

4.5 5.5

0.52 0.45

1 (a) Based on Table 8, construct a table for the va lues of ,\y and - .

x

Berdasarkc111 Jadual 8, bina sat11 jad11a/ bagi 11i/ai-11i/ai ,\y dcm

x2

6.5

0.40

[2 marks]

[2 markah]

(b) Plot ,\)' against J_ , using a scale of 2 cm to 0. 1 unit on the J_-ax is and 2 cm to 0.4 unit on x x

the ,\)'-axis.

Hence, draw the line of best fit. [3 marks]

Plot ,\y 111e/awa11 J_, de11ga11 111e11gg1111aka11 ska/a 2 cm kepada 0. J 1111it pada paksi- J_ dan x x

2 cm kepada 0.4 unit pada paksi-,\y.

Sete/'lls11ya, lukis garis /11/'lls pe11y11aia11 /erbaik.

(c) Use the graph in 8(b) to fi nd the va lue of

G1111a gm/ di 8(b) 1111t11k mencari nilai

(i) p,

(ii) q.

34 72/2 «) Hak cipta MPSM P11/a11 Pim111g 2015

[3 111arkah]

[5 marks] [ 5 markah]


11

9 (a) In Bandar Mu tiara, 55 % of the residents are adults aged above 25 years old.

Di Bandar Mutiara, 55 % daripada peng/111ninya adalah dewasa yang benm1ur 25 ta/11111 ke atas.

If 8 res idents of Bandar Mutiara are chosen at random, calculate the probability that

Jika 8 penghuni Bandar Mutiam dipi/ih secam rnwak, hit1111g kebarangka/ian bahawa

(i) exactly 3 of them are adults above 25 years old,

tepat 3 omng daripada 111ereka ada/ah dewasa yang bem111ur 25 ta/11111 ke atas,

(ii) at least 3 of them are adults above 25 years old.

sekurang-kurangnya 3 ornng daripada 111ereka ada/ah dewasa yang bem111ur 25 ta/11111 ke afas.

[5 marks] [ 5 mark ah]

(b) The marks in a Mathematics test in a certain school follow a normal distribution with a mean of 60 and a variance of 64.

Markah bagi suatu ujian Mate111atik di sebuah seko/ah tertentu ada/ah mengikut tab11J'C111 nor111a/ dengan min 60 clan varians 64.

(i) If the passing mark is 50, find the probability that a student chosen at random from the school passes the test.

Jika markah hr/us ia/ah 50, cari kebarangkalian bahawa seorang pelajar yang dipi/ih secam rawak daripada seko/ah itu lu/us 11jia11 tersebut.

(ii) Find the number of students who pass the test if 280 students took the test.

Cari bi/a11gan pe/ajar yang /11/us ujian jika 280 pelaj ar 111enga111bil 11jia11 itu.

3472/2 © flak cipta MPSM P11la11 Pi11a11g 2015

[5 marks] [ 5 111arkctl1]


f •'

r

12

10 Diagram 10 shows the curve y = 6x - x2 and the tangent to the curve at Q(2, 8).

Rajah I 0 111en111!jukkc111 lengkung y = 6x - x2 dan tangen kepada lengkung itu pada Q(2, 8).

y

0

Calculate

Hi tung

Diagram 10

Rajah 10

y = 6x - x2

x

(a) the equation of the tangent to the curve at Q,

persamaan tangen kepada lengkung itu pada Q,

(b) the area of the shaded region P,

luas rantau berlorek P,

[3 marks]

[3 111arkah]

[4 marks]

[ 4 111arka'1]

(c) the volume of revolution, in terms of n, when the region bounded by the curve, x-axis and the straight line x = 2, is rotated through 360° about the x-axis.

[3 marks] lsi padu kisaran, dalam sebutan rr, apabila ran/au yang dibatasi oleh lengkung it11, paksi-x dan garis lums x = 2, diputarka11 360° pada paksi-x.

[3 markah]

3472/2 © !Tri!< cipta MPSM P11/a11 Piuaug 2015


13

11 Diagram 11 shows a circle with centre 0 and radius 9 cm. PR is a tangent to the circle at P.

Rcy"ah I I 111e111111j11kka11 sat11 b11/ata11 be1p11saf 0 da11 be1jejari 9 cm. PR ialah ta11ge11 kepada b11/ata11 it11 pada P.

9cm 0

R p

Diagram 11

Rajah I I

The length of PR is equal to the length of arc PSQ.

Pm!ja11g PR sa111a de11ga11 pmlja11g le11glwk PSQ.

[Use I G1111a n = 3. 142]

Calculate

Hit1111g

(a) the length, in cm, of the arc PSQ,

pa11ja11g, dalam cm, Ieng/wk PSQ,

(b) the perimeter, in cm, of segment PSQ,

peri111eter, da/0111 cm, bagi te111bere11g PSQ,

(c) the area, in cnl, of the shaded region.

/11as, dalam cm2, kawasa11 berlorek.

34 72/2 © llak cipta MPSM Pu/au Pi11a11g 2015

[2 marks]

[2 11wrka'1]

[3 marks]

[3 111arkah]

[5 marks]

[ 5 11wrka'1]


14

Section C Balwgirm C

l20 marks] [20 markah]

Answer any two questions from this section. Jawab 11w11a-111a11a dua soalan daripada bahagian ini.

12 A particle moves in a straight line and passes through a fixed point 0. Its velocity, v m s·1, is

given by v = - t 2 + 1111 - 7, where tis the time, in seconds, after leaving 0 and /11 is a constant. The particle stops instantaneously for the I st time at point P and the 2 nd time at point Q.

S11at11 zarah be1gemk di seµanjang s11at11 garis lurus dan 111ela/11i sat11 titik tetap 0. Halqj1111ya, v m s · 1, diberi oleh v = - I 2 + 111/ - 7, dengan keac/aan t ialali 111asa, dalam saat, se/epas 111ela/11i 0 da11 111 ialah pe111alar. Zarah berhenti seketika di titik P pada kali per/ama da11 di litik Q pada kali kedua.

[Assume motion to the right is positive]

[Anggapkan gerakrm ke arah kanan sebagai posit if]

Find

Cari

(a) the value of m if the particle is at P, one second after leaving 0 ,

11ilai 111 jika zarah bernda di P, sa/11 saa/ selepas 111e11i11ggalka11 0,

(b) the time, in seconds, at which the particle stops instantaneously for the 2 nd

time, 111asa, dala111 saa/, ketika zarah berlienti seketika pada kali kedua,

(c) the distance, in m, between P and Q,

jamk, dalam m, di cmtarn P dc111 Q,

(d) the maximum velocity, in m s·1, of the particle.

lialaj11111aksi11111111, dalam m s·1, bagi zarnh i/11.

3472/2 © Hal< cipta MPSM P11/a11 Pi11a11g 2015

[2 marks]

[2 111arkah]

[2 marks]

[2 111arkah]

[3 marks]

[3 mark ah]

[3 marks]

[3 111arkah]


15

13 Diagram 13 shows a quadrilateral ABCD and L BCD is acute.

Rajah J 3 111e111111j11kka11 sisi e111pat ABCD dan L BCD ialah fi/'lls.

c

(a) Calculate

flit1111g

A

Diagram 13

Rajah 13

(i) the length, in cm, of BD,

/JCll/fang, dala111 cm, bagi BD,

L DB. (ii) c

(b) The straight line DA is extended to A ' such that BA = BA'.

Garis /11/'lls DA dipm!ja11gka11 ke A 'dengan keadaa11 BA = BA '.

(i) Sketch the triangle BA 'D and state L BA 'D.

Lakarka11 segi tiga BA 'D clan 11yatakc111 L BA 'D.

(ii) State the type of triangle BA 'A.

Nyataka11 je11is segi tiga BA 'A.

(iii) Find the area, in cm2, of triangle BA 'D

Cari /11as, dala111 cm2, segi tiga BA 'D

3472/2 © l frtlc cipta MPSM P11fa11 Pi11a11g 2015

[5 marks] [ 5 111arka/J]

[5 marks] [5 marka/J]


16

14 Table 14 shows the prices, price indices and percentage of usage of four ingredients A, B, C and D used in the making of a particular type of cake.

Jadual 14 111e111111jukka11 lwrga, i11deks lwrgo da11 peratus pe11gg111wa11 e111pat baha11 A, B, C da11 D 1111/uk 111e111buat seje11is kek.

Price index in the year 2014 based on Percentage of usage Ingredient the year 2011 (%)

Baha11

A

B

c D

(a) Find

Cari

J11deks harga pada ta/11111 2014 bef'{fsaska11 ta/11111 20 I I

90

120

130

250

Table 13 Jadual 13

Pef'{fflfS pe11gg111wa11 (%)

20

16

34

30

(i) the price of ingredient A in the year 2014 if its price in the year 20 11 is RM 2.50,

//{/rga ba//(/11 A pada talu111 2014 j ika lwrga11ya pada talu111 20 I I ialah RM 2.50,

(ii) the price index of ingredient C in the year 2014 based on the year 2008 if its pri ce index in the year 20 I l based on the year 2008 is 110.

i11deks lwrga balw11 C pada ta/11111 2014 bef'{fsaska 11 talu111 2008 jika i11deks lwrga11ya pada ta/11111 2011 /Jef'{fsaska11 talu111 2008 ialah 110.

(b) Calculate

Hitu11g

[ 4 marks] [ 4 111arkah]

(i) the composite index for tbe cost of making the cake in the year 2014 based on the year 2011,

i11deks gubalw11 bagi kos 111e111buat kek pada ta/11111 2014 bef'{fsaslw11 ta/um 2011 ,

(i i) the price of a cake in the yea r 2011 if its corresponding price in the year 20 14 is R_M 95.10.

//{/rga sebiji kek pada ta/um 2011 jika harga11ya ya11g sepada11 pada ta/um 2014 ialah RM 95.10.

[4 marks] [ 4 111arkah]

( c) The cost of making the cake is ex pccted to increase by 9 % from the yea r 2014 to tbe year 2016, fi nd the expected composite index for the year 20 16 based on the yea r 20 J I.

Kos 111e111bual kek d!ja11gka 111e11i11gkat 9 % daripada ta/11111 20 l 4 ke ta/11111 2016, cari i11deks gubalw11 pada ta/11111 20 l 6 bef'{fsaska11 ta/11m 201 1.

3472/2 © flak c:ipta MPSM Pu/au l' i11a111: 2015

[2 marks] [2 11wrkah]


17

15 Use graph paper to answer this question.

Gw10ka11 kerlas graf 1111111k 111e11jawab soala11 i11i.

A factory produces two types of lamps, A and B. In a particular day, the factory produced x lamps of type A and y lamps of type B. The profit from the sales of a lamp of type A is RM 12 and a lamp of type B is RM 10. The production of the lamps per day is based on the fo llowing constraints:

Sebuah kila11g 111e11ge/11arlw 11 d11aje11is la111p11, A clan B. Dal(//}/ s11al11 /Jari lerle11l11, kila11g ilu 111enge/11arka11 x la111p11 je11is A da11 y la111p11 je11is B. Ke1111l1111gc111 daripada j11ala11 selmah la111p11 j e11is A ialah RM 12 da11 sebuah la111p11 je11is B ialah RM l 0. Pe11ge/11ara11 la111p11 selwri adalah berdasarkc111 keka11ga11 berikul:

I The tota l number of lamps produced is not more than I 00.

J11111/ah la111p11 ya11g dihasilka11 lidak lebih daripada 100.

II The number of lamps of type A produced is at most three times the number of lamps of type B.

Bila11ga11 /a111p11 j e11is A ya11g dilwsilka11 adalah selebi/J- /ebi/111ya liga kali bila11gm1 la111p11 j e11is B.

111 The minimum total profit for both types of lamps is RM 480.

)11111/ah ke111111111ga11111illimw11 bagi ked11a-d11aje11is /a111p11 adalah RM 480.

(a) Write tltree inequali ties, other than x 2:: 0 and y 2:: 0, which satisfy all the above constraints. [3 marks]

Tulis liga kelaksa111aa11, selai11 x 2:: 0 da11 y 2:: 0, yang 111eme1111hi semua keka11ga11 di alas. [3 marlw h]

(b) Using a scale of 2 cm to I 0 units on both axes, construct and shade the region R which sat isfies all of the above constraints.

[3 marks] lvfe11gg11110kc111 ska/a 2 cm kepada l 0 unit pada ked11a-d11a paksi, bi11a da11 /orek m11la11 R ya11g meme1111/ii semua kekm1ga11 di alas.

(c) Using the graph constrncted in 15(b), find

Me11gg1111aka11 gl'Clj)1a11g dibi11a di 15(b ), cari

[3 11wrkah]

(i) the minimum number of lamps of type B if l 0 lamps of type A is produced,

bila11ga11 mi11i1111m1 lamp11 j e11is Bjika 10 b(ji /m11p11je11is A dihasilka11,

(ii) the maximum tota l pro fit per day.

j umlah ke1111l1111ga11111aksin111111 dalam sehari.

END OF ASSESStvlENT MODULE

3472/2 © I/ale l'ipta MPSM Pu/au Pi11r111N 2015

[4 marks] [4 markah]


19

TING KA TAN : ---

NO. KAD PENGENALAN

I I I I I I I - I I I -I I I I I ANGKA GILIRAN

I I I I I I I I I I Arahan Ke1>ada Calon

Tulis nombor Irnd pengenalan clan anglrn giliran ancla pacla petak yang disecliakan.

2 Tanclakan ( >/ ) untuk soalan yang dijawab.

3 Ceraikan helaian ini clan ikat sebagai muka hadapan bcrsama-sama dengan helaian jawapan.

Kod Pemeriksa

Bahagian Soalan Soalan Mark ah Markah Diperoleh

Dijawab Penuh ( Untuk Keg1111aa11 Pemeriksa)

1 5

2 5

3 7 A

4 8

5 8

6 7

7 10

8 10

B 9 10

10 10

11 10

12 10

13 10 c

14 10

15 10

Jumlah

3472/2 © Hak cipta MPSM Pu/au Pimmg 2015


20

INFORMATION FOR CANDIDATES MAKLUMAT UNTUJ( CALON

1. This question paper consists of three sections: Section A, Section Band Section C.

Ker/as soala11 i11i me11gm1cl1111gi liga bahagim1: Baliagia11 A, Ba'1agia11 B clan Balwgit111 C.

2. Answer all questions in Section A, any four questions from Section B and any two questions from Section C.

Jawab semua soalan dalam Balwgia11 A, ma11a-1J1ww empat soalr111 claripada Balwgir111 B clan ma11a-ma11a dua soalan daripacla Balrng ia11 C.

3. Write your answers on the papers provided. If the papers arc insufficient, you may ask for extra papers from the invigilator.

Jawapan ancla hendaklah clit11/is di dalam kertas ja111apa11 yang disediakan. Sekiranya kertas ja111apm1 tidak menc11k11pi, silo dapatlw11 he/aia11 tambalwn daripada pengawas peperiksac111.

4. Show yo ur working. It may help you to get marks.

T11nj11kkan langkah-/angkah penli11g dalam ke1ja me11gira anda. !ni boleh memban/11alllla11nt11k mendapatlwn markali.

5. The diagrams in the questions provided are not drawn to scale unless stated.

Rajah yang mengiringi soalan tidak di/11kis mengik11t skalr1 kec11ali clinyatakm1.

6. The marks allocated fo r each question and sub-part of a question, a re shown in brackets.

lvlarkah yang dipem11t11kka11 bagi setiap soalan dan ceralaii ,soalan dit11njukkan dalam lwr11ngan.

\ . ~ 7. The Upper Ta il Probabi li ty Q(z) For The Normal Distribution N'((),. l) Table is provided on

page 2. Jad11al Kebarangkalian I luj11ng A las Q(z) Bagi Tab11ran Normal N(O, I) disediakan di lwlm1wn 2.

8. A list o f fo rmulae is prov ided on pages 3 to 5. Sa/11 senarai rn11111s disuiiakan di lwlaman 3 hingga 5.

9. Grapb papers are provided.

Ker/as grnfdisediakan.

10. You may use a scientific calculator.

A11da dibenmkan menggunakan kalk11/ator sai11t!flk.

11. Tic lhe papers and the graph papers together and hand in to the inv igilator al the end of the examination.

I/wt kerlas ja111apa11 bersm11a-sama dengan (\er/as grnf dan serahka11 kepacla pe11gmvas peperiksaan pada akhir peperiksaan.

3472/2 © l/11k cipta MPS,\! P11la11 l'i11a11g 2015

I I tutormansor.wordpress.com

ADDlTlONAL MATHEMATICS Kertas 2 Ogos

2 I . - Jam 2

MAJLIS PENGETUA SEKOLAH MALAYSIA CA WANGAN PULAU PINANG

MODUL LATIHAN BERFOKUS SPM 2015

MARK SCHEME

ADDITIONAL MATHEMATICS

Paper 2

Two hours and thirty minutes

3472/2


Qucs

1.

3472/2

3 - p l'J = 3 - 2k or k = --

- 2

Eliminate p or k

I _I_ =J *{3-2k)- l k

or

1 1 -- = l p-I *(3;p)

or

equivalent

k =- 1.281, 0.781 or

p = 5.562, 1.438

OR

p = 5.562, 1.438 or

k= - 1.281, 0.781

Note:

2

SIC:CTJON A ( 40 MARKS)

Mark Scheme

Solve the Quadratic Equation

Using formula

-1±~(1)2 -4(2)(-2) k=-------

2(2)

7 ± ~(7)2 - 4(1)(8) p = 2(1)

OR

Completing the square

OW-1 if steps to solve quadratic equation are not shown.

3tl72/2

Sub Total Mark Mark

5 5


" .)

Ques Marl< Scheme

2 (a) I 1(x) = x - 3 ~ 4

(b) -I t-3) gf (x) = 2 -4- - 5

x -13 =

2

(c) (y+5) h ()1) = 8 * -2- - 9

h(x) = 4x + 11

3 (a) Use

(b)

(i)

(ii)

3472/2

cos (A + B) = cos A cos B - sin A sin B or

sin 2B = 2 sin B cos B

LHS = RHS

cosB = 3

~ -5

24

~ 25

-J3 . A 1 ~ cosA = - or sm = -2 2

·( ~) ·m m -·m or

2[·( ~)*(;)-{i)(~)] *(~~)

__?__ (3J3 - 4) 24

3472/2

Sub Total Mark Mark

2

2 5

2

2

3 7


Qucs

4 (a) c = I ~

(b) 3[x2 + 4x] + 1

(c) (x-l)(x+5)~ 0 or

• -5 • 1

Marl< Scheme

Complete the square 3[(x + 2)2

- (2)2] + 1

3(x + 2)2 - 11

A(-2, - 11)

x~ - 5,x:::::l

(d)

5 (a)

3472/2

f (x) = - 3(x + 2)2 + 11

(i) & (ii)

QR= - 6~+ 8y

Use Triangle Law for QR or PB

8 PB =4x+-y

- 3-

3472/2

Suh Total Mark Ma rk

4

2

8

3


Qucs

5

5

Marl< Sch eme

Use Triangle Law to find CQ

and use PA= PC+ hCQ

4 6 h = - or k = -

7 7

6 4 k= - or h = -

7 7

6 (a) 1111 =- 3 B

(b)

3472/2

Use m 1 x 1112 = - 1

I 1112 = -

3

3y=x+ 19

Integrate 2 - x wrt x ------ -- --------- ----------

x2 y=2x--+c

2

x2 21 y=2x--+-

2 2

Compare coefficients

! and y and solve

simultaneous linear equations

Use y - 8 = * _!_ (x - 5) 3

or 8 = *_!_(5)+c 3

Substitute (5, 8) into *y to find c

3472/2

Sub Total Mark Mark

5 8

4

3 7


6

SECTION fl ( 40 MA RKS)

Qucs Mark Scheme

7 (a) (i)

I lllGH = -

2 mnc = -Yi , lllAD = 2

2y=x + 3

(ii) B (5, 4)

Use x + 2.(2) = *5 or 3

y + 2(5) = *4 3

C( l l , 2)

(iii)

1 1 *5 Use

2 2 * 4

until l 1c 2

(b) Find the gradient of AC and use mAc X maH = - I

3472/2

2 1

5 2

)- ( ~

Q y - 2 = * ~ (x - 1)

~ or

2 = *J_ (l)+ c 2

OR I

y - 6 = * - (x - 9) 2

or

6 = * J_ (9) + c 2

5

Not Perpendicular

3472/2

Sub Tota l Mark Mark

3

3

2

2 IO


Ques

8 (a)

(b)

&

(c)

3472/2

7

Mark Scheme

~

1 0.40 0.29 0.18 0.15 - 0.67 0.22

x

xy 0.50 l.60 2.03 2.34 2.48 2.60

Plot -"JI against x

6 *points plotted correctly

Line of best fit

xy=p+q ~ x ~

Use * c = q

Use *m = p

p =-4 3.2

Note : SS-1 if part of the scale is not uniform or not using the scales given or not using the graph paper

3472/2

Sub Total Mark Mark

2

3

5 10


,\)' J [\

0 I' I l: I I I•

11 _[_[J

, ; I : I

I i ! l

! ! : : ii

! 1: !

i I

I!: Ii , • : 11 : ! i 11

•-t-~t-+-,..1+-I .. :: +-! f-.--t-1 ,..I +-+-.,...i n i I

3472/2

Li I I j · I I

_[ §

! ! I I ! I !

8

' ' I i , I! ,......._ 1 '

.......,-, !

! I ; i ! I ! ~ r-j· 1 j ! l

l ! ;

I j ! j j

l I i ! I i I j j I

I :

f I l I i I I !

Ji\72/2

' ' ' -0 I! '

i ' ' !! I ! ! ; I ; j ! ;

' ! : i l I I ; ; . ; ' l ---l ! i 11 ! !

: I i i '

i I I ! i I ~ i I I ! I. I I

I

I ! . ; i l l ; l ! l I I ! .


9 3472/2

Q ucs Mark Scheme Sub Total Mnrk Mark

9 (a) (i) p = 0.55 and q = 0.45 ~

~~~ --~~r_(~·~?J:'_(O_.~?)~:_'~ 8c3 (0.55)3 (0.45)8- 3

0. 1719 Nl

(ii) Write P(X?. 3) = 1 - [P(X= 0) + P(X= I)+ P(X= 2)]

OR = [ P(X= 3) + P(X= 4) + P(X= 5)

+ P(X= 6) + P(X= 7) + P(X= 8)]

5 0.91 15

(b) (i) z = 50 -60

J64

- 1.25 Use 1 - P(Z?. * - 1. 25)

NI 0.8944

. (ii) *0.8944 x 280

250 5 10

3472/2 tutormansor.wordpress.com

10

Ques Ma rk Scheme

I 0 (a) c6) Find - and subst x = 2

dx

y = 2x+4

(b) Integrate ( 6x - x 2) wrt x

6x 2 x 3 A1=----

2 3

2 Use limit f in A1

0

OR

Find the area of trapezium

A2= ..!_ x(*4+8) x(2) 2

[2x2 ]

2

or -+4x 2 0

~ 112'!:.112.667 3 3

Cc) Integrate 7t( 6x - x2) 2 w'rt x --------- --------------

v ~ n[l2x3 -3x4 + x:]

272 2 -n 1154-n II 54.4n

5 5

3472/2

Q Use y - 8 =*1111 (x - 2)

~ or

8 = *m1(2) + c

3 ~ 72/2

Sub Total Mark Mark

3

4

3 10


11 3472/2

Qucs Sub Total Mnrl< Mark

Mnrk .Scheme --- --t------- - ----------- -----------1---1-- --1

I J (a) Use s = re to find arc PQ

9 x 1.3

11. 7

(b) PQ = 2(9 sin 1h (1 .3)

(c)

3472/2

OR Use cosine rule in MOQ to find PQ

PQ = 10.89

22.59

L QPR = 0.65 rad or equivalent

Use A = 1h /e to find area of sector POQ

24.912

2

*11.7 + *10.89

3

Use .!.. x bxh OR .!..absinCto 2 2

find A2 =11 POQ or AJ = /1 PQR

A2 = 39.024; AJ = 38.55

5 10


12

SECTION C ( 20 MAHKS)

Qucs Mark S cheme

I 2 (a) Substih1te I = I into v = 0

v = - (1 )2 + m(l) - 7 = 0

II/ = 8

(b) Solve v = 0 to find I (i ::1>-c=--- / +- n-~ ·o

I= 7

(c) f-t 2 +St-7 dt

13 2 s =-- +41 -71

3

(c) dv Find - and equate to 0

8 dt Kl to find t.

t = 4 ~

v = 9

3472/2

Substitute t = I or t = *7 into *s OR

f •7 Use

1 ins

Substitute t = *4 into v

3472/2 '

Sub Total Mark Mark

2

2

3

3 10


13

- --.------ -----

Qucs Mark Scheme

13 (i) Use cosine rule in D. ABD lo find BD

.3472/2

BD 2 = 6 2 +7 2 - 2(6)(7) cos l20°

BD = 11.27

(ii) Use sine rule in t:JJCD to find L BCD --- - - - -s-i~-~- -- -~i~1-;~- ------------0

- - - =-- \J * 11.27 8

c = 41.4°

8 180° - 28° - • 41.4

110.6°

(i)

D

L BA 'A = 60°

(ii) Equilateral triangle G (iii) Use .!_ (7)(6 + 7) sin *60

2

39.4

3472/2

S ub Total Mark Mark

2

3

2

2 10


3472/2 ·,

--- - - - - - -----------.----,-----,

Qu cs l\'fa rk Scheme

14 (a) (i) Q Use - 1 x 100 Qo

(b)

RM 2.25

(ii) I _ 1lO x 130 14/08 - 100

143

Ci) I =* 90(20) + 120(16) + 130(34) + 2so(30)

20 + 16 + 34 + 30

(ii) 95 . lO x 100 = * 156 .4

Q11

156.4

RM 60.80

(c) l09 x* l56.4 100

170.48

3472/2

Sub Tota l Mark Mark

4

4

2 10


. '. 15 3472/2

~----,--------- -- -- - - -- --

Qucs Mark Scheme Sub Tola) Mark Mark

-- --------- - -----1-----t-- --1

15 (a) x + y ~ I 00 or equivalent

x ~ 3y ..QI. equivalent

12x + I Oy ~ 480 or equivalent

(b) Draw correctly at least one straight line from the *inequalities which involves x and y

Draw correctly all the three *straight lines Note : Accept dashed lines

The correct region shaded

(c) (i) 36

3472/2

(ii) Maximum point = (75, 25)

Use 12x + lOy for point in the *region R

RM 1150 Note: SS-I if in (a), the symbol "=" is not used at all or more than three inequalities are given

in (b), does not use the scale given or does not use graph paper or interchange between x-axis and y-axis

3

3

4 10


.,,

40

30

20

Jtl72/2

I . !

11

1ol:t:1::4J:l+i:Jt+P:::Q:l::ttt~:ttl::i:l:t:~~~~~~:l:tt:tti::tt!:+t:l:;P.::):~:l:t:t:~4:Q~!+i:l.I

if

~~f-+-+T-++1~·-rl-t-H--+-+-1-+-i--+..:-H-+-r+-+-H--H1 -1~--;-~t-+-+-+~1-+-1-M-t-t-t-+++--l,~

END OF MARK SCHEME

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