add. maths pahang 2011 - tutormansor.files.wordpress.com · The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used. Rumus-rumus

347212347212AT}DITIONALMATTIEIT{ATICSPaper 2Aug / Septz0fi2 % hours

AD DITIONAL MATHEMATICS

Tingkatan 5

Kertas 2Dua jam tiga puluh minit

JANGAIT BTTKA KERTAS SOALAI\I INI SEHINGGA DIBERITAHU

1. Kertas soalan ini adalah dalam dwibahasa.

2. Soalan dalam bahasa Inggeris mendahului soalan yang sepadan dalam

Bahasa Melayu.

3. Calon dikehendaki membaca maklumat di halaman belakang kertas soalan

ini.

Kertas soalan ini mengandungr 20 halaman bercetak.

@ 2011 Hak cipta MP9M Pahang [Lihat hataman sebetah

http://banksoalanspm.blogspot.com/http://banksoalanspm.blogspot.com/



http://tutormansor.wordpress.com/

http://banksoalanspm.blogspot.com/



user

Text Box

user

Text Box

The following formulae may be helpful in answering the questions. The symbols given are theones commonly used.Rumus-rumus berilu.t boleh membantu anda meniawab soalan. Simbol-simbol yang diberiadalah yang biasa digunakan.

ALGEBRA

u*rlu'-*_ g log,b=!.og,blx=To-log,a

2 e^Xar:a.^*n 9 Tn=a+(n-L)d,

3 a^ + an:qm-n 10. g = tlrr+(n-Ddl

4 ( a^)n : a^n ll T, = arn-l

o( r^ -l\ o(l- r'\5 logomn=logom+logon 0 S,=f =}/,r*l

6 togo:=logo m-logon ^ 13 g = *, lrl .tn

7 logomn= nlogom

CALCI]LUSKALKULUS

4 Area under a curve

. dy dv . du Luas di bawah lenglamg1 y--tN, +=ttT *v:chc "d)c"dJc b b: I, e or(atau) !- at

du dv

., u dY ' a* - uA 5 Volume of revolution

J=-rT:v ca v' Isi Padu kisaranbb

A dy dy du - lo Y'hc or (atau) = lo x'dY

J dx du dx

@ 2011 Hak Cipta MPSM Pahang








user

Text Box

user

Text Box

Xx

N

I' fxv-

2f

3 347212

STATISTICSSTATISTIK

= zWiI i7 I--2W,

8 '1= +t-' (n- r)!

g nCr= nl

(n - r1t vl3o=

4 oi=

ffi{l"-t

1 Distance liarak

-- J6r-*,Y +(v,-v,Y

10 P(Au B) = P(A) + P(^B) - P(A^ B)

1l P(X =r) ='C,P' Q'-', P * q -l

Meanlmin, P= nP

o = ,lnpq

E x- Pt-

o

GEOMETRYGEOMETRI

4 Area of a triangle I Luas segitiga:

|lt,',r, * xzlt * \!r)-('rY, * xtlz+ x,/,)l

n xi+yiol-ffi

t2

13

6 I=30x100t4

2 Mid point I Titik tengah

G,v)=(ry,ry)3 A point dividing a segment of a line

Titik yang membahagi suatu

tembereng garis

(",r)=( ffi,W)

EG-7\,\/=Nw_=w

1l >,f 1l ,' f

x'+y'

@ 2011 HaR CiPta MPSM Pahang fLihat halaman sebelah








user

Text Box

user

Text Box

4

TRIGONOMETRYTRIGONOMETRI

347212

I Arclength, s:rg 8 sin(r4tB) =sinr4cosB*cosAsinBPanjanglenglmk, s: je sin(,at.B) = sin A ftosBtkos lsin B

2 Areaofasector. A = !,'e' 2 g cos (r4 t .B) = cos .,4 cos B T sin Asin B

Luassektor,L: lft kos(AtB) = ftos AkosBT sin Asin B2'

3 sinzA+cos'A=IsinzA + &os2 A=L

4 sec'A=1+ tan2 A

sekz A -1+ tan2 A

5 cosec'A=l+cotzAkosek2A=l+ kotzA

6 sinZA - 2 sin A cos Asn2A : 2 sin Akos A

to tan(Arr) - ,Yi.n*!|-t' I TtanA tanB

1r tan2.A= ffi1., a b c

sin r4 sin B sin C

13 a' =b2 +c' -zbccosr{A' =b2 +c'-\bc kos A

7 cos 2A

: l4:;T^ u Area ortriangre/ Luas segitiga

: ! obsin C2

kos 2A = kos2 A - sir{ A

: ?yfl;)









user

Text Box

user

Text Box

TIIE UPPER TAIL PROBABILITY 0(e) FOR TIm NORIVIAL DISTRIBUTIONIT(0, 1)KEBARANGKALIAN HATUNG ATAS OKI BAGI TABARAN NORMAL I

z 0 32I 654 98734567

Minus /Tolak

II27

0.0

0.1

0.2

0.3

0,4

0.5

0.6

0.7

0.8

0.9

1.0

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

r.9

LO

2.1

2.2

2.3

2.4

2.5

2.6

2.7

2,8

2.9

3.0

0.50w

0.4602

0.4207

0,3E21

0.3446

0.3085

0.2743

0,2420

0,21r9

0.1841

0.1587

0.1357

0.1 151

0.0968

0.0808

0.0668

0.0548

0.M16

0.0359

0.02E7

0,02n

0.0179

0.0139

0,0't07

0.00820

0,00621

0.00466

0,00347

0.m256

0,00r87

0.00135

0.4960 0.4920 0.46E0

0.4562 0.4522 0,4483

0.4168 0.41A 0.4090

0.3783 0.3745 0.3707

0.3409 0.3372 0.3336

0.3050 0.3015 0.2981

0.2709 0.2676 0.2643

0.2389 0.2358 0.2327

0.2090 0.2061 0.2033

0.1814 0.1788 0.1762

0.1562 0.1539 0.151s

0.1335 0.1314 0.1292

0.1131 0.1112 0.1093

0.0951 0.0934 0.0918

0.0793 0.077E 0.0764

0.0655 0.066 0.0630

0.0537 0.0526 0.0516

0.0436 0.0427 0,0418

0.0351 0.0311 0.0336

0.02Er 0.0274 0.0268

0.02n 0.02,17 0.0212

0.017{ 0.0170 0,0166

0.0136 0.0132 0.0129

0.0101 0.0102

0,00776 0.0075s

0.00604 0,00587 0.00570

0.00153 0.00440 0,00427

0.00336 0,00326 0,00317

0.0021E 0.00210 0,00233

0.m181 0.00175 0.00169

q.00122

0.00990

0.4E40 0.4601 0.4761

0.4443 0.4404 0.4364

0.4052 0,4013 0.3974

0.3669 0.3632 0.3594

0.3300 0.3264 0,3228

0,2946 0.2912 0.28n

0,2611 0.2578 0.2546

0.2296 0,2266 0.2236

0,2005 0.1977 0.,1949

0.1736 0.1711 0.1685

0.1492 0.1469 0.1446

0.1271 0.1251 0,1230

0.1075 0.1056 0.103E

0,0901 0.0885 0,0869

0.0749 0.0735 0,0721

0.0618 0.0606 0,0594

0.0505 0.0495 0.0485

0.0409 0.0401 0.0392

0.0329 0.0322 0.0314

0,0262 0,0256 0.0250

0,0207 0,0202 0.0197

0.0162 0.0158 0,0154

0.0125 0.0122 0.0119

0,00964 0,00939

0.00714 0.00695

0,00554 0.00539 0,00523

0.00415 0,00402 0.00391

0.00307 0,00298 0.00289

0.00226 0,00219 0.N212

0.00161 0.00159 0,00154

0,00114 0.001r.|

0.4721 0.4681 0.4641

0.4325 0.4286 0,4247

0,3936 0.3897 0.3859

0.3557 0.3520 0.3483

0.3192 0.3156 0.3121

0.2843 0.2810 0.2776

0.2514 0.2483 0.2451

0.2246 0.2177 0,2148

0j922 0.1894 0.1867

0.1660 0,1635 0.1611

0.1423 0.1401 0,1379

0,1210 0.1190 0.1170

0.1020 0.1003 0.0985

0.0E53 0.0838 0.0E23

0.0708 0.0694 0.0681

0.0582 0.0571 0.0559

0,,0475 0,0465 0.0455

0.0384 0.0375 0.0367

0.0307 0.0301 0.0294

0.0244 0.0239 0.0233

0,0192 0,0188 0.0183

0,0150 0.0146 0,0143

0.01'16 0.0113 0.0110

0.00889

0.00676 0.0065i 0.00639

0,00508 0,00494 0.00480

0,00379 0,00368 0.00357

0.00280 0.00272 0.00264

0,00205 0,00199 0.00193

0.00149 0,00144 0.00139

4

4

4

4

4

3

3

3

3

3

2

2

2

2

1

1

1

I

1

1

III7

7

7

7

b

5

5

5

4

4

3

3

2

2

2

1

1

1

1

1

1

5

5

4

4

3

2

2

1

I

12

12

12

11

11

10

10

III

7

6

0

5

4

4

3

3

2

2

1

I

I

1

I7

6

6

5

3

3

2

I

1

0

0

0

0

3

2

2

2

2

1

1

I

0

0 1

16 20 24

16 20 24

15 19 23

15 19 22

15 18 22

14 17 20

13 16 19

12 15 18

11 14 16

10 13 15

I8

7

b

6

5

4

4

3

2

222212ttr0 13

I 12

12 14

10 12

9 11

81078

7

6

I

4

4

6

5

4

4

3

Io

5

4

2

I7

6

4

3

3

2

2

2

15

14

8

7

11 13

I 11

6

5

4

3

2

2 22

28 32 36

28 32 36

27 31 35

26 30 34

2s 29 32

24 27 31

23 26 29

21 24 27

19 22 25

18 20 23

16 19 21

14 16 18

13 15 17

11 13 14

10 11 t3

81011I7

6

5

34433423322218m2316 16 21

15 17 l9

13 15 1l

11 12 11

I7

5

3

3 4

IE

6

5

7

6

5

4

I t0

I6

4

8

6

4

3

Example / Contoh:

r(z)=#*o(-;r)z)f

3472t2

If X-N(0, 1), then

JikaX- lf(0, 1), malw

P(X> k) = Q&)

P(X> 2,r) = QQ.D = 0.0179

QQ) = [n4a,k








user

Text Box

user

Text Box

1. Solve the simultaneous equations x - 2y = = 8.

Selesailran persamaan serentak x - 2y = = 8.

(x+l) cmxcmE

2x cm

Diagram 2Rajah 2

A wire of length 7.5 m is cut into 10 pieces. Each piece is then bent to form arectangle. The dimensions of the first three smallest rectangles are as shown inDiagram 2.

Segulung dawai yang panjangnya 7.5 m dipotong kepada fi bahagian. Setiap bahagiandibengkolrlrnn untuk membentuk sisiempat tepat. Ulcuran tiga sisiempat tepat yangterkecil ditunjuklmn dalam Rajah 2.

FindCari(a) the value of x,

nilai x,

(b) the area of the largest rectangle.luas sisiempat yang terbesar.

2.

6 3472t2

Section A

Bahagian lt| 40 marksl

[ 40 marlmh ]

Answer all questions.

Jawab semua soalan.

[ 6 marks ]

[ 6 markah ]

(x+2) cm

| 4 marks l| 4 marknh I

[ 3 marks J

| 3 markah l

t2xy

812

Z(x+t) cm 2(x+2) cm









user

Text Box

user

Text Box

3. Sketch the graph of y - -3 sin 2x for 0 < x < 2n .

Lalrarkan graf y = -3sin 2x untuk 0 <; <Ztr.

Hence, using the same axes, sketch a suitable straight line to find

of solutions for the equation x + 3n sin 2x = 0 for 0 <x 32n .

State the number of solutions.

the value of k,nilai k,

the gradient of the curve al A,kecerunan lengkung itu di A,

the equation of tangent to the curue at A.persamaan tangen kepada lengkung itu di A,

347212

14 marksl

| 4 marlmhl

the number

[3 marks ]

l1 mark lI t marlwh]

[ 3 marks J

| 3 markah I

12 marksl[2 marlmhl

Seterusnya, dengan menggunakan palai yang sama, lalcar satu garis lurus yang sesuai

untuk mencari bilangan penyelesaian bagi persamaan x + 3a sin 2x = 0 untuk

0 <r 4r.Ny atalran bil angan p eny ele s aian itu. 13 marlahl

4. The curve ky = ft, where k is a constant, passes through the point A (2, - 6),

Lenghmg y = +, dengan keadaan k ialah pemalar, melalui titik A (2, - 6).4 - 3x' <

FindCari

(a)

(b)

(c)

@ 2011 Hak Cipta MPSM Pahang [Lihat halaman sehelah








user

Text Box

user

Text Box

347212

5. Table 5 shows the cumulative frequency distribution for the marks of 40 students ina test.

Jadual 5 menuniulckan taburan kekerapan longgokan bagi markah 40 orang murid dalamsatu ujian.

Marks(Markah) <20 <40 <60 <80 <1 00

Number of students(Bilangan murid) 3 13 27 35 40

Table 5Jadual 5

(a) Based on Table 5, copy and complete the table b.1 below:

Berdasarknn Jadual 5, salin dan lengkapkan Jadual 5.1.

Table 5.1Jadual 5.L

(b) Given that the minimum mark to obtain an A* is 80,

find the percentage of students who score A*.

Diberi bahawa markah minimum untuk memperoleh A+ ialah 80,cari peratus murid yang memperoleh A*.

(c) Without drawing an ogive, calculate the median mark.Tanpa menggunakan ogtf, hitungkan markah median.

| 1 markl| | marlwh 7

12 marksl

12 marlwh I

[ 3 marks ][3 marlwh)

Marks(Markah) 0-19 20-39 40-59 60-79 80-99

Number of students(Bilangan murid)









6.

34722

Diagram 6

Rajah 6

ln Diagram 6, OPQR is a trapezium such that 3RQ = 4OP. S lies on PR such

that PS : SR = 3 : 1, Giventhat G = 6-r and Ofr.= 4y.

Dalam Rajah 6, OPQR ialah se;buah trapezium dengan keadaan 3RQ : 4 OP.

SterletakpadaPRdengankeadaan P,S; ^SR - 3 : l. Diberi OF = 6r dan

oF. - 4y.

(a) Express in terms of x and I of ,

Unglmpkan dalam sebutan 4 dan y bast

(i) m,(ii) B

(b) Show that PQ is parallel to OS.

Tunjuklmn bahawa PQ adalah selari dengan OS.

[ 3 marks ]

13 markahl

12 marksl

12 markahl

[Lihat halaman sebelah

13 marksl

13 markahl

(c) Given that lal = , ,

l1l = 5 , find the area of trapezium OPQR.

Dibert bahawa l1l

= t , ll = 5 , cari tuas trapezium }PQR.









l1

Solution by scale drawing will not be accepted.

Penyelesaian secara lukisan berslwla tidak akan diterima.

ln Diagram 8, AOC is a straight line and point B lies on the x-axis.Dalam Rajah 8, AOC ialah suatu garis lurus dan titik B terletak pada palai-x.

347212

8.

Given the area of AOAB is 10 unit2, white the area of AOBC is 20 unif,find the coordinates of B. Hence, show that the coordinates of C is (6, 8).

[ 5 marks ]Diberi luas AOAB ialah l0 unif , manalmla luas AOBC ialah 20 unif , carikoordinat titik B. Seterusnya tunjukkan bahawa koordinat titk C ialah (6, 8).

| 5 marknh I

A point P moves such that its distance from C is always twice the distancefrom A.Satu titik P bergerak dengan keadaan jaralotya dari C adalah sentiasa dua kalijarahtya dari A.

(i) Find the equation of the locus of P,Cari persamaan lolats bagi P,

(ii) Hence, determine the coordinates of the points where this locuscuts the x-axis.Seterusnya, tentukan koordinat titik-titik padamana lola$ ini memotongpalai-x.

[5 marks ]15 markahl

(a)

(b)

@ 2011 Hak Cipta MPSM Pahang [Lihat halaman sebelah








t2

Rajah 9Diagram 9

3472t2

12 marksl| 2 markah )

| 4 marks I| 4 markah l

| 4 marks I| 4 marlwhl

9. Diagram 9 shows a sector OPR of a circle, centre O and radius 10 cm. pSRf isa circle, centre Q. oP and oR are tangents to the circle.

Raiah 9 menuniukkan sector bulatan OPR berpusat O dan berjejari 10 cm p^ffif iahhsebuah bulatan berpusat Q. OP dan OR adalah tangen kepada bulatan itu.

[UselGuna n =3.1421

(a) Show that the radius QR = 6.84 cm.Tunjukkan bahawa jejari QR = 6.84 cm

(b) FindCari

(i)

(ii)

the length, in cm, of the major arc pSR,panjang, dalam cm, lengkok major P^tR,

the area, in cm2, of the shaded region.luas , dalam cfrZ , kawasan yang berhrek.









347212

10 (a) The probability that rain will fall on a certain day in the month of

December is ] . Find the probability that in a certain week in December,4

rain will fall in

Kebarangkalian bahawa hujan alwn turun pada sesuatu hari dalam bulan

Disember nhn ] . Cari kebarangkalian bahawa pada satu minggu tertentu dalam4

bulan Disember, hujan alcan turun

(r) exactly 3 days,tepat 3 hari,

(ii) not more than 5 days.tidak melebihi 5 hari.

[ 5 marks ][ 5 marlwh ]

(b) The height, in cm, of a group of students is found to be normally distributedwith mean 163 cm and standard deviation 16 cm.Tinggi sehtmpulan murid didapati bertaburan secara normal dengan min 163 cmdan sisihan piawai 16 cm

FindCari

(i) the standard score for a height of 175 cm,skor piawai bagi ketinggian 175 crc+

(ii) the height of a student which corresponds to a standard score of -0.6,tinggi seorang murid yang mempunyai skor piawai -0.6,

(iii) the probability that the height of a student chosen at random from thegroup is between 14T cm and 125 cm.ltebaranglwlian bahawa tinggi seorang murid yang dipilih secara rawakdaripada kumpulan itu adalah antara r47 cm dan-l7i cm

[5 marlrs J

[ 5 markah ]

13









11.

l4 347212

Diagram 1 1 shows part of the graph of the function y = flx) which touches thex-axis at point P and cut the y-axis at point Q. The straight line QR which isparallel to the x-axis is the tangent to the curve at point Q.

Rajah 1l menunjuklwn sebahagian daripada graf fungsi y : f(x) yang menyentuhpalrsi-x pada titik P dan memotong paksi-y di titik Q. Garis lurus QR yang selaridengan palai-x ialah tangen kepada lengkung itu pada titik Q,

Diagram 11

Rajah tl

Given that f '(*) : 3x2 - 12 x.Diberi bahawa f '(x) = 3x2 - 12 x.

(a) FindCari(i) tn:;;,:inates of P,

(ii) (x).

(b) Find ,nli."roinates of R and hence, find the areaof the shaded region.Cari koordinat R dan setentsnya, cari luas rantau berlorek.

[2 marksl12 markah l

[ 3 marks ]| 3 markah j

[ 5 marks ][ 5 markah ]









12.

l5

Section GBahagian C

[ 20 Marks ]120 Markah l

Answer any two questions from this section.

Jawdb mana-mana dua soalan daripada bahagian ini.

(a) the initial velocity of the particle,halaju awal zarah itu,

(b) the minimum velocity of the particle,halaju minimum zarah itu,

(c) the acceleration of the pailicle at point P,pecutan zarah itu pada titik P,

(d) the range of values of f during which the particle moves to the left.

julat nilai t ketika zarah itu bergerak ke kiri.

A particle moves along a straight line such that its displacement, s fft, is given

by s = f -A( - 15f, where tis thetime, in seconds, after it passes through a

fixed point O. The particle is instantaneously at rest at point P.

[Assume motion to the right is positive]

Satu zarah bergerak di sepanjang suatu garis lurus dengan keadaan sesarannya, .s rn,

diberi oleh s : f - 6/ - I5t, dengan keadaan t ialah masa, dalam saat, selepas melaluisatu titik tetap O. Zarah itu berhenti seketika di titik P.

lAnggapkan gerakan ke arah kanan sebagai positiJl

3472t2

12 marksl12 marknh l

[ 3 marks ]13 marknh l

[ 3 marks ]| 3 markah l

12 marksl12 marlwhl

FindCari

@ 2011 Hak Cipta MPSM Pahang [Lihat halaman sehelah








13.

t6

Solution by scale drawing will not be accepted.

Penyelesaian secara lukisan berslwla tidak alun diterima.

12 cm

Diagram 13Rajah 13

ln Diagram 13 , AB is parallel to DC such that 3AB = 5CD andI ACB is obtuseDalam Rajah 13, AB adalah selari dengan DC dengan keadaan 3AB : 5CD dan

I ACB adalah cakah.

347212

[ 3 marks ][3 markah I

12 marksl

[2 markahl

[ 5 marks ]| 5 markah l

Find the length, in cm, of CD and AC.Cari panjang, dalam cm, bagi AC,(i) €, fr$(ii) Ac.

Find /. ACB .

Cari I ACB .

Find the area, in cm2, of MBC. Hence, findCari luas, dalam cnf , bagi MBC. Seterusnya, cari

(i) the area, in cm2, of MDC,Iuas, dalam crrf , bagi AADC,

(ii) distance between the two parallel lines AB and DC.jarak di antara dua garis selari AB dan DC.

(a)

(b)

(c)









347212

14. ln an effort to strengthen Bahasa Melayu and to improve the standard of Bahasalnggeris, a training centre offers two courses, BM and Bl. The number ofparticipants for the BM course is x and the number of participants for Bl course isy. The intake of pafticipants is based on the following constraints :

Dalam usaha memperlcasakan Bahasa Melayu dan memperlwkuhlun Bahasa Inggeris,sebuah pusat latihan menawarlcan dua lanrsus, BM dan BI. Bitangan peserta tatrsus nUialah x dan bilangan peserta kursus BI ialah y. Pengambilan peserta adalah berdasarpankekangan berikui

The number of BM participants is more than 20,Bilangan peserta BM adalah melebihi 20,

The number of Bl participants is at least 10,Bilangan peserta BI adalah sehtrang-larangnya 10,

The maximum number of participants is 80,Jumlah maksimam bilangan peserta ialah 80,

The ratio of the number of BM participants to the number of Blparticipants is not more than 4 : 1.Nisbah bilangan peserta BM kepada bilangan peserta BI adalah tidakmelebihi 4 : l.

17

il

ilt

IV

(a) write down four inequalities, other than x >0, y >0, which satisfyall the above constraints.Tulis empat ketaksam),aan, selain x 20, y 20, yang memenuhisetnua kekangan di atas.

| 4 marksl

| 4 markahl

(b) Using a scale of 2 crn to 10 participants on both axes, constructand shadethe region which satisfy all the constraints. [ 3 marks ]Menggunakan skala 2 cm kepada l0 peserta pada kedua-dua palai, bina dan lorekrantau Ryang memenuhi semua kekangan di atas. ll markahl

(c) Using the graph constructed in 14(b), findDengan menggunakan graf yang dibina di L4(b), cari

(i) the range of the number of BM participants if the number of Blparticipants is 20,julat bilangan peserta BM jika bilangan peserta BI ialah 20,

(ii) the maximum total fees that can be collected if the fees for BM and Blcourses are RM 200 and RM 900 respectively.Jumlah maksimum lwtipan yuran yang diperoleh jilw yuran bagi latrsus BMdan BI ialah RM 200 danRNl3}} masing-masing

[ 3 marks ][3 markah]









15.

18 3472t2

Table 15 shows the price indices of four food items P, Q, R and S in the year

2009, based on the year 2008 and the changes of the price indices from the

year 2009 to the year 2010. The pie chart in Diagram 15 reflects the proportion

of weightage of the 4 items.

Jadual 15 menunjukkan indeks harga bagi empat bahan malwnan P, Q, R, dan S dalamtahun 2009, berasaskan tahun 2008 dan perubahan indeks harga dari tahun 2009 ketahun 2010. Carta pai dalam Rajah 15 menggambarkan kndar pemberat empat bahantersebut.

ItemBahan

Price index inthe year 2009based on the

year 2008

Indeks hargadalam tahun 2009berasaslmn tahun

2008

Changes of priceindex from the year

2009 to the year 2010

Perubahan indeksharga dart tuhun 2009

ke tahun 2010

P 110 lncreased by 10%

Bertambah l0%

o 125 lncreased by 20o/o

Bertambah 20%

R 120 Unchanged

Tidak berubah

s 110 Decreased by 5%

Berhtrang 5o/o

DIAGRAM 15Rajah 15

TABLE 15Jadual 15

(a) Given that the price of item P in the year 2008 is RM 5,find its price in the year 2009.

' Diberi bahawa harga bahan P pada tahun 2008 ialah RM 5,

cari harga bahan P pada tahun 2009.

12 marksl

12 markahl

1500

0









19 347212

on the year

12 marksl

(b) The price of item Q in the year 2009 is RM 7.50. Find its correspondingprice(i) in the year 2008,(ii) in the year 2010.

12 marksl

Harga bahan Q pada tahun 2009 ialah RM 7.50. Cari harga yang sepadan

bagi bahan itu

(i) pada tahun 2008,

(ii) Pada tahun 2o1o' t 2 marrrahf

(c) Find the price index ol each item for the year 2010 based2008.

Cari indeks harga bast setiap bahan pada tahun 2010 berasaslmn tahun 2008.

12 markahJ

Calculate the composite index for the four items for the year 2010 basedon the year 2008.

| 4 marks I

Hitunglwn indela gubahan bast empat bahan itu pada tahun 2010 berasasknntahun 2008.

| 4 markah)

END OF QUESTION PAPERKERTAS SOAIAN TAMAT

(d)









3472/2 Trial SPM (PP)

1 3472/2

PEPERIKSAAN PERCUBAAN SPM 2011

ADDITIONAL MATHEMATICSTingkatan 5

KERTAS 2

PERATURAN PEMARKAHAN

UNTUK KEGUNAAN PEMERIKSA SAHAJA

3472/2(PP)Tingkatan LimaAdditional MathematicsKertas 2PeraturanPemarkahanSeptember2011








user

Text Box

user

Text Box


ADDITIONAL MATHEMATICS PAPER 2 TRIAL SPM 2011 (MARK SCHEME)

No. PERATURAN PEMARKAHAN MARKAH

MARKAH

PENUH

1 x – 2y = 8 ………… (1)

8 12

8x y

− = ………….(2) (any two equations : one linear, one non-linear) P1

x = 2y + 8 or y = 8

2

x − ( x in terms of y or y in terms of x) P1

Substitute x or y into non-linear equation …

8 128

2 8y y− =

+ or

8 128

( 8)2

xx− =−

K1

y2 + 5y +6 = 0 or x2 – 6x + 8 = 0 * (y +2) (y+3) = 0 or (x – 2) (x – 4) = 0 [factorisation or formula] K1

y = -2 , y = -3 ; x = 2, x = 4 N1

x = 4 , x = 2 ; y = -3 , y = -2

[ If answers given without factorisation or using formula : K1 N1 N1 OW-1]

N1

6


MARKAH

PENUH

2(a) Listing : 6x, 6x + 6, 6x + 12, … or a = 6x

d = 6

[ ]10

102(6 ) 9(6) 750

2S x= + = [Equating Sn = 750]

x = 8

P1

N1

K1

N1

(b) Area : 2(x)2, 2(x + 1)2 , 2(x + 1)2, … => 2(8)2, 2(9)2 , 2(10)2, …AP : x , x + 1, x + 2, … => 8, 9, 10, …T10 = *8 + (10 – 1)1

= 17Area of largest rectangle = 2[*8 + (10 – 1)1]2

= 578 cm2

ORPerimeter : 6(8), 6(8) + 6, 6(8) + 12, …T10 = 6(8) + (10 – 1)6 = 102 = 6y [any letter except x] K1

2y y = 17

y y Area of largest rectangle = y 2y = 17 2(17) K1

2y = 578 cm2 N1

K1

K1N1

7

2 3472/2








user

Text Box

user

Text Box

user

Text Box



MARKAH

PENUH

3 Shape of graph of y = sin x , 0 ≤ x ≤ 2π [must begin from origin, O(0, 0)] N1

Shape of graph of y = sin 2x , 0 ≤ x ≤ 2π [exactly 2 complete cycles] N1

Amplitude of graph = 3 N1

Reflection of graph y = 3 sin 2x at x – axis, 0 ≤ x ≤ 2π [all correct] N1

x

yπ

= [Equation of straight line]

Draw any straight line with positive gradient passing through (0, 0)Number of solutions = 5

P1

K1N1

7

4 (a) k = (-6) [4 – 3(2)] = 12 P1

(b) )3()34)(1(12* 2 −−−= −xdx

dyK1

[ ] )3()2(34)1(12* 2 −−−= −

dx

dy[substituting x = 2 in his

dx

dy]

K1

= 9 N1

(c) y – (–6) = 9(x – 2) or equivalent [finding equation of tangent using his dy/dx ]

y = 9x – 24

K1N1

6

3 3472/2

x

y

O 2ππ½ π 3/2 π

+3

-3

2

y = - 3 sin 2x








user

Text Box

user

Text Box



MARKAH

PENUH

5 (a)Marks

(Markah) 0-19 20-39 40-59 60-79 80-99

Number of students(Bilangan murid) 3 10 14 8 5 N1

(b) Percentage of students scoring A+ = 10040

5* ×

= 12.5%

K1

N1

(c) L = 39.5

[ ]( )

140 13

239.5 2014

m−

= +

= 49.5

P1

K1

N1

6

6 (a) ~ ~

6 4PR x y= − +uuur

N1

~ ~

3

49

32

PS PR

x y

=

= − +

uuur uuuvK1

N1

(b)

~ ~

33

2

OS OP PS or OR RS

x y

= + +

= +

uuur uuur uuur uuur uuur

N1

~~42 yxPQ += N1

+=

~~4

2

3

3

4yxPQ and PQ is parallel to OS

or

+=~~

424

3yxOS and OS is parallel to PQ .

N1

(c) OP = 12, OR = 20 and QR = 16 [Using

~x = 2,

~y = 5 to find length]

K1

Area of OPQR = 280 unit2 N18

4 3472/2










MARKAH

PENUH

7

x2 1.00 4.00 9.00 12.25 16.00 30.25

xy 4.50 8.90 16.20 21.53 27.00 48.40

N1

N1

(a) Graph of xy against x2 : Please refer to Appendix 1 (End of Mark Scheme)Correct axes with uniform scale and a pointAll points correctly plottedLine of best fit

K1N1N1

(b) 22k

xy hxh

= + P1

(i) From Graph, gradient, m = 2h ≈ 1.50 h ≈ 0.75 [0.70 – 0.80] N1

(ii) xy-intercept, c = 3.0

3.00.75

2.25 [ 2.1 2.4]

k

k accept

≈

≈ − N1(iii) xy = 30

From graph, x2 = 184.24x∴ ≈ N1

108

(a) 10)0)4(0()000(2

1 =+−+−++ OB or 1

4 102

OB × = K1

OB = 5B(5, 0) N1

Let C be (h, k),From area ratio, OR Equation of OC : OR Area of ΔOBC

AO:OC = 10 : 20 xy3

4= = 20)5(*2

1 =×k K1

021

)3(2)(1 =+

−+h 20)000()05*0(

2

1 =++−++ k OR hk3

4=

or 021

)4(2)(1 =+

−+kK1

C(6, 8) N1

(b) PC = 2 PA2222 )]4([)]3([2)8*()6*( −−+−−=−+− yxyx K1

(x – 6)2 + (y – 8)2 = 4 [(x + 3)2 + (y + 4)2] N1x2 + y2 + 12x + 16y = 0 N1

At x-axis, y = 0 : x2 + 12 x = 0 K1x (x + 12) = 0x = 0, -12(0, 0) , (-12, 0) N1

10

9 (a) 0.6 radian = 34.370

5 3472/2

K1Equating 2h with his gradient ORequating k/

h with his xy-intercept OR

construction of the line xy = 30 on his graph.









tan 0.610

QRrad = or 0tan 34.37

10

QR= K1

QR ≈ 6.84 cm N1

(b) (i) °=∠ 55.63or rad 971.0RQO OR °=∠ 111.26or rad 942.1PQR

OR )90(34.37or rad )2

6.0(2

1major °+°+=∠ π

PQR

°=∠ 248.74or rad 342.4majorPQR

Arc PSR = 6.84 4.342 or 84.62360

74.248 ××°

° π

= 29.70 cm

P1

N1

K1

N1

(ii) Area of OPQR = 284.6102

1 ××× K1

Area of sector OPR = 2.1102

1 2 ×× or 210360

74.68 ××°° π K1

Area of shaded region = 68.4 – 60 or 68.4 – 59.99 K1

= 8.4 cm2 or 8.41 cm2 N110

10 (a) p = 0.75 , q = 0.25

(i) P(X = 3) = 7 3 43 (0.75) (0.25)C or

43

37

4

1

4

3

C K1

= 0.05768 or 16384

945N1

(ii) P(X ≤ 5) = 1 – [P(X = 6) + P(X = 7)] K1= [ ]7 6 1 7

61 (0.75) (0.25) (0.75)C− + K1= 0.5551 N1

(b) (i)175 162

16z

−=

= 0.75 N1

(ii)163

0.616

x − = −

x = 153.4 cm N1

(iii) P(147 < X < 175) = P (–1 ≤ Z ≤ 0.75) K1 = 1 – P(Z> 1) – P(Z > 0.75) K1 = 1 – 0.1587 – 0.2266 = 0.6147

N110

11 (a) (i) At P, 2'( ) 3 12 0f x x x= − = K1 3x (x – 4) = 0

6 3472/2









x = 0 , x = 4P(4, 0) N1

(ii) 2

3 2

( ) (3 12 )6

f x x x dxx x c

= −= − +

∫K1

c = 32 N1f(x) = x3 – 6x2 + 32 N1

(b) R(6, 32) P1

Area of = 6 x 32 OR *32 – *(x3 – 6x2 + 32)= 192 …………….… (I) = 6x2 – x3

K1

Area under curve from x = 0 to x = 6 Area of shaded region

( )∫ +−=6

0

23 326* dxxx ( )∫ −=6

0

326 dxxx K1

= 64

3

0

2 324

xx x

− +

6

0

43

43

6

−= xx

K1

= 84………………………………. (II)

Area of shaded region = (I) – (II) = 192 – 84 04

)6(

3

)6(6 43

−

−=

= 108 cm2 N110

SECTION C (20 marks)12 S = t 3 – 6t 2 – 15t

(a) 23 12 15ds

v t tdt

= = − − K1

t = 0, v = –15 ms-1 N1

(b) For minimum velocity, 0126 =−= tdt

dvK1

t = 2 N1vmin = 3(2)2 – 12(2) – 15

= – 27 ms-1 N1

(c) At P, v = 0 : 3(t + 1) (t – 5) = 0 K1 t = 5 N1

a = 6 (5) – 12 = 18 ms-2 N1

(d) v < 0 3 (t + 1) (t – 5) < 0 K1 0 < t < 5 N1

10

13 (a) (i) AB = 20 cm

(ii) 2 2 2 020 8.9 2(20)(8.9) cos 54AC = + −

P1

K1AC = 16.43 cm N1

7 3472/2









(b) ACB∠−+= cos)43.16)(*9.8(243.16*9.820 222 K11739.0cos −=∠ACB

∠ ACB = 100 o [accept 100.01o ] N1OR

Using sine rule,

43.16*

54sin

9.8

sin °=∠CABOR

43.16*

54sin

20

sin °=∠ACB K1

°=∠ 99.25CAB°−°−°=∠ 99.2554180ACB

= 100.01o °=∠ 100ACB N1

(c) Area of ΔABC = ½ (8.9) (20) sin 540 = 72.00 cm2

K1N1

(i) Area of ΔADC = ×5

3 Area of ΔABC OR °99.25sin)43.16)(12(

2

1

= 43.2 cm2 = 43.20 cm2

OR 2.772202

1 ==>=×× hh

Area of ΔADC = 2.7122

1 ×× K1

= 43.2 cm2 N1

K1N1

(ii) Let h cm be the distance between AB and DC

72202

1 =×× h

h = 7.2 cm N110

14 (a) x > 20 y ≥ 10x + y ≤ 80x ≤ 4y (or y ≥ ¼ x )

(b) Axes correct and one *straight line correct

Draw correctly all 4 straight linesRegion R is correctly shaded and labelled.(Please refer to appendix 2)

(c) (i) 20 < x ≤ 60

(ii) 200 (21) + 300 (59)

The maximum fees collected is RM 21 900

N1N1N1N1

K1N1N1

N1

K1

N1

10

15(a) Item P : 110100

509 =×

PK1

8 3472/2









P09 = RM 5.50 N1

(b) Item Q : (i) Q2008 = RM 6.00 N1

(ii) Q2010 = RM 9.00 N1

(c)Item 2008I2010

P 121Q 150R 120S 104.5

N1

N1

(d) Item 2008I2010 W WI

P 121 80 9680Q 150 150 22500R 120 90 10800S 104.5 40 4180

∑W = 360 ∑WI = 47160

Weightage for item S = 40o or 4 [as seen in his formula ]

ii IW∑ = 80(121) + 150(150) + 90(120) + 40(104.5)or 8(121) + 15(150) + 9(120) + 4(104.5)

360

)5.104(40)120(90)150(150)121(80 +++=I

or 36

)5.104(4)120(9)150(15)121(8 +++

= 131

P1

K1

K1

N110

Skema Pemarkahan Tamat

Appendix 1Graph for Question 7

9 3472/2

One mark each for every TWO correct answers.








10 200 30 40 50 60 70 80

80

50

40

30

70

60

20

10

xy

R

x

100

90

(21, 59)

x+ y = 80

X

x = 4y

x = 20

y = 10


Appendix 2Graph for Question 14 [LINEAR PROGRAMMING]

10 3472/2

5 100 15 20 25 30 35 40

40

25

20

15

35

30

10

5

xy

X

X

x

x

x2

50

45

(0, 3.0)

x

18

(28, 45)x

Gradient ,

45 3

28 0

1.50

m−=−

=

x









END OF MARKING SCHEMESKEMA PEMARKAHAN TAMAT

11 3472/2

y








Documents

add. maths pahang 2011 - tutormansor.files.wordpress.com · The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used. Rumus-rumus