penang trial spm 2013 maths

7/27/2019 Penang Trial SPM 2013 Maths

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SULIT1449t1MATHEMATICSKERTAS 1I1^ JAM4

1449t1

MAJLIS PENGETUA SEKOLAH MALAYSIA(CAWANGAN PULAU PTNANG)MODUL LATIHAN BERFOKUS SPM 2013

MATHEMATICSKERTAS 1lJAM 15 MINIT


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t149fiM,4THEIVTAT ICAL FORMULAERUMUS MATEMATIK

The following formulae may be helpful in answering the questions. The symbolsgiven are the ones commonly used.Rumus-runlus berikut boleh membanlu anda menjawab soalan. Simbol-simbol yangdiberi adalah yang biasa digunakan.

R.ELATIONSPERKAITAN

I a**an:a**n2 a"' * a" : om+n3 (o* )n - ot?,n4 A-t--l-( d -b)r 1' -o4-6o[-, ,)5 Distance I Jarak

t-- f,(rz - r)2 + (yz - y)2

10 Pythagoras' TheoremTeorem Pithagoras,2 -12 +b2

lt P(A\-n(A)n(,s)12 P(A'):1-P(A',)13 *_lz-Ytx2-xr


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')J

SHAPES AI.JD SPACEBENTT]K DAN RUANGI Area of trapezium: 1* ,r* of parallel sides x height2

Luas trapeziuu, : :.x ltasil tambah sisi selari x tinggi2 Circumference of circle : nd : 2nr

Lilitan bulatan : nd : 2nj3 Area of circle : nr2Luas bulatan : fi.i2

4 Curved surface area of cylinder : 2nrltLuas perruukaan ntelengkung silinder : Zn.jt

5 Surface area of sphere : 4 x 12Luas permukaan sfera : 4n j26 Volume of right prism : cross sectional area x length

Isi padu prisma tegak : luas keratan rentas x panjang7 Volume of cylind er - xr2 h

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1., arc length _ angle subtended at centrecircurnference of circle 360"panjanglengkok _ sudut pusatlilitan bulatan 360o

I - area of sectoratIJ angle subtended at centrearea of circle 360"luas sehor _ sudut pusatl"^ brt"t"r- 360"

14 Scale factor. k - PA''PAFaktorskala. k-PA'PA

15 Area of image : P x area of objectLuas irnej : lP x luas objek


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Find the value of ( 6-0.418) x 0.005, and round off the answer correct to twosignificant figures.Cari nilai (6 - 0.418) x 0.005 , dan bundarkan jawapan itu behi kepada duaangka bererti.A 0.027B 0.028c 0.02D 0.03

Express 0.000 021 9 in standard fonn.Ungkapkaz 0.000 021 9 dalam bentuk piawai.A 2.19 x 10-7B 2.19 x 10-5C 2.19 x 10s


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0.0025 +7 xl0-3 :A 9.5 x 103B 3.2 x 103C 3.2 x 10-3D 9.5 x l0-3

Given 2 x 53 + 4 x 52 + 5_t,:2430s, find the value of y.Diberi 2 x 53 + 4 x 52 + 5y : 2430s, cari nilai y.AOB1C3D6


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t449lt

In Diagram I , PQRST is a regular pentagon. STU is a straight line.Dalam Ra.jah 1, PQRST ialah sebuah pentagon sekata. STU ialah garis lurus.

Diagram IRajah I

Find the value of x.Cari nilai x.


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In Diagr am:,, PQSis a tangent to the circlewith centre o at point B'

Daram Rajah 2, pes iarah tangen kepada buratan berpusat o pada titik Q-

ODragram2Raiah 2


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IO 1449 I 1

In Diagratn 3, PQRS is a parallelogram and RSI is a straight line.Dalam Raiah 3, PQRS ialah sebuah segi empat selari dan RST ialah garislurus.

Dragram 3Rajah 3

Calculate the value of x + y.Hitungnilaix+!.

T I l0o


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ll 1149il

EIolotEl(t)lLIEl-clololol-clolaDl>lEI;l=t=IEIot9tLI(t)lLI'tr1EIolololslolal

l0 Diagram 4 shows two triangles, PQR and SQT, drawn on square grids.Triangle PQR is the image of triangle SQT under an enlargement.Raiah 4 menuniukkan dua segi tigct, PQR dan SQT, dilukis pacla gr"i4 segientpat sama. Segi tiga PQR ialah imej bagi segi tiga SQT di bau,ah syatuperubesaran.

P\

\\

.s' \\ \

\ \O T R


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t2 t449lt

1t ln Diagram 5, trapezium LDalam Raiah 5, trapeziuntputaran 90" ikut arah iam.

is the image of K under a clockwise rotation of 90".L ialah intej bagi trapezium K di bawah suatu

Diagram 5Raiah 5which of the points, A, B, C or D, is the centre of rotation?Antara titik A,B, C danD,yang manakah pusat putaran itu?

Y LAft 1-It


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t3 1449/t

13 In DiagrantT,JKLMis a rectangle. KNLis a srraighr line a,d Kil :1or.Dalam Raiah 7,JKLM ialah 'ebtnh segi empat tepat. Klvl- ialah goris lurusdartKN-3 KL.5

6 crn

Diagram 7Rajah 7

The value of tan ZMNK isNilai bagi tan /"Ml{K iuluh

M20 cm


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t4 1449/t14 ij:ff[i;]r".:. a right-angred rriangurar prisnr with refia,gre peRsas rhe

Rajah B ruertunjukkan sebuahpeRs. .'-""'t'4't"L't" ovuuutt pr*ma segi figa tegak dengan tapak rutertgtlfirk

Diagram 8Rajah 8

Name the angre between the pran e eTUand the prane RsTu.Nqruakan sudut di antara satah eTtl dengart satah RsTu.


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t5 t44qt1

[5 In Diagram 9, JK is a vertical pole. KL is horizontal.Dalarn Rajah 9, JK ialah sebatang tiang tegak. KL adalah mengufuk.

2JmDiagram 9

Ra.iah 9

The height of the vertical p


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l6 1449n

r6 ln Diagram I 0, PQR and SZ are two vertical poles on a horizontal plane. Q ts apoint on PR such that QR: S7.Dalam Rajah 10, POR dan ST ialah dua batang tiang tegak pada satahmengduk. I adalah satu titik di atas PR dengan keadaan QR: SZ.

Diagram l0Rajah L0


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t1 t449ll

t7 Diagram l1 shows three points, P, Q and R on a horizontal plane. R lies duenorth of Q.Rajah ll menunjukkan tiga titik, P, I dan R yang terletak pada suatu satahmengufuk. R berada ke utara Q.

Diagram 11Rajah ll

Find the bearing of P from Q.Cari bearing P dari Q.


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Itt 1449 n

t8 ln Diagram l2, Z ts a point on the eafih..fu is thc NorthPole and //OS is the axis of the eafih.Dalarn Rajah 12, Z ialah titik di stas permukaan bumi..ialah Kutub Selatan dan UOS ialah paksi bumi.

Greenwich MeridianMeridian Greenv,ich ,s

Pole, S is the South

U ialah Kutub [Jtaro, S

EquatorKhatulistiwa

N/U

50" \Zt

I- - 'I- -


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l9 1419t1

zo Express 2+k-2h-l3k (th as a single fraction in its simplest form.(Jngkapkan +! -?!| ,sebagai satu pecahan tunggal daiam bentttk3k (th utermudah.

l9 (m - n)(m + n) -2n(,n -3n) _A nf + 5n2 -2mnB m2 + 5n2 + 2rnnC m2 - 7n2 - 2mnD m2 + 7n2 - 2mn

4h+k6hk

4h-k6hk

4+3k-2hk6hk

AB

C


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20 t449lr

22 Given that -: * 4y - 2(l- y) , calgtrlate the value of 1'.3Diberi bahav'a - I * 4y - 2(l - 1t) , hitttng nilai y-3

6A 1llll8_ t8111l6

A223 I! - can be written as\lm-,l2/.n,.' n = boleh ditulis sebagai\lm'


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24 SirnplifyRingkaskanJ!:!'t'_-

I(h")'koA h-o k'B h-'k-'c h-2 k2D h2k2

25 Diagram 13 represents two simultaneous linear inequalities on a number line.Rajah 13 mewakili duu ketak.sumaan linear pada garis nombor.


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22

List all the intcgcrs of /r which satisfy the inequality 5 - 2k < 7 .Senaraikan semua integer k vang memua.skan ketalcsctmaan linear 5 - 2kA -3,-2.-l ,...-2,-1,0,...-1,0n1,...

0, I ,2,...

The pie chart in Diagram 14 shows hobbies of a group of teenagers.Carta pai dalant Rajah 14 menunjukkan hobi sekumpulan remaja.

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2t| I)iagrarn l5 is a piclograph which shorvs the nunrbcr of houscs inl'arnan Pelangi. Thc nurnber of single storey serni-D houses is not slrown.lla.jah 15 ialah piktogra/'yang menunjukkan bilangan rumah diTaman Pelangi. Bilungon rurnah kembar satu lingkat tidak diturtjukkan.

represents 5 houscsmewakili 5 huuh nrmuh

Double storey terraceTeres dua tingkat ooooo,aoooooa

oao00Single storey serni-DKembar satu tingkal

Double storey serni-I)Kembar dua tingkut


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24

29 Table I shows the weight, in kg, of a group of pupils in a certain school.Jadual L menunjukkan berat. dalam kg, bagi sekumpulan murid disebuah sekolah tertentu.

Weight I Berat ( ke ) 30 -34 35 -39 40-44 45 49Frequency I Frekuensi 5 6 l0 4Table I

Jadual IFind the mean weight of the pupils.Cari min berat murid-murid itu.

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ABCD

36.939.6369396


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3t 'l'lrc Vcrln diagram in Diagram I7 shows the number of elements of setA,sct 1J and set C.(iumbar raiah Venn dalant Rajah 17 menunjukkan bilangan unsur dalam set A,set B dan set C.

Diagram 17Rajah 17

It is giventhatthe universal sct ( - A \-t B w C and n(B c\ C) - n(C') .Find the value of x.


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26

Whicir of thc follorving graphs represents 1: - -1 trYang manakah rJi antara graf berikut nrcv,okili )' -

t449ll

32 _!,x

A

ts


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1-/-I2

In Diagram 18, JKL isDalam Rajah 18, JKl,

2l

an ist'rsceles triangle.iulalt ,segi ligo sama kaki.

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1,1 5Ilrc srirdicnt ol'thc straight line 5r + hy -12 is - 1. Find the value of h.25Ke(crttnun garis lurus 5x + hy - 12 ialah -: Cori nilai h.2

1

22

AB

CD

34


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28 1419n

35 It is given that set R : t\21


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29 l449lt

l7 lllrlc 2 shows some values of the variables m and ru such thatm varies inverselyrrs llrc scluarc root of z..ltrtlrrrtl 2 menunjukkan beberapa nilai bagi penfioleh ubah m dan n denganAcrrtluurt tn berubah secot"o songsang dengan punca kuasa dua n.

Table 2Jadual 2

f iirrd the relation betrveen m and n.('ut'i hubungan antara m dan n.

m n

l0 4

4 25


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30

Tablc 3 shows two sets of rralues of the variables 1', V and W.Jadral 3 di bau,ah menunjukkan duu set nilai bagi pemboleh ubah Y, V dan W.

Table 3.Iadual 3

It is given that IZvaries directly as the square of V and inversely as W.

1449fi

38

Y V W

1J5

aJ t2

m 5 18


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3r 1449fi(t _2) ^(t 3) (t _8.3e ciu"" Ii -; )-'[; ;]: [i -i,] n,d trre varuc or/'l(t _2) ^(r 3) (t _s)niL,", i li ; )- ri; ;,j : l; - i). c,,ri ,,itai k

A-1B3C4D8

41' t; ili -:):A t: )


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SULIT144912MATEMATIKKERTAS 2SEPTEMBER2 Yz J^LM

MAJLIS PENGETUA SEKOLAH MALAYSIA(CAWANGAN PULAU PINANG)MODUL LATTHAN BERFOKUS SPM 2OI3

MATEMATIKKERTAS 2

2 JAM 30 MINIT

JANGAN BUKA KERTAS SOALAN INIUntuk Kegunaan PemeriksaKod PenLeriksa : Markah Markah


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SULTT2

MATHEMATICAL FORMULAERUMAS MATEMATIK

rr r(a) =_M

t44912

The following formulae may be helpful in answering the questions. The symbols given arethe ones commonly used.Rumus-rumus berikut boteh membantu anda menjawab soalan. Simbol-simbol yang diberiadalah yang biasa digunakan.

RELATIONSPERKAITAN10 Pythagoras TheoremTeorem Pithagorasc2=a2+b2

1 afr xo" = 11mtn2 a* n an -- gil-n3 (a*)n = dmn , (s)\.) /

-l t Ia, ul4 A'='((t -blad-Dc\-c a) LZ p(A,) :1-f(ri

5 Distance I Jarak ^ v, - lt= 13 m=ffi


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SULIT 144912

SHAPES AND SPACEBENTUK DAN RAANGI

ILuas trapezium : Z * hasil tambqh dua sisi selari x tinggi2 Circumference of circle = xd = 2nrLilitan bulatan : nd : Znj3 Area of circle = rcP' Luas bulqtan : fijz4 Curved surface area of cylinder = ZnrhLuas permukaan melengkung silinder = 2ttit5 Surfacc arca of sphere - 4 n P{.uas permukaan- sfera :- 4njz6 Volume of right prism : cross sectional area x lengthIsipadu prisma tegak = luas keratan rentas x paniang7 Volume of cylinder n Ph


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SULIT 1449t2

12.

lnI J area of circle

arc lenglh angle subtended at centrengle subtended at centreedcircumference of circle 3600panjan_g l3ngfu!( sudut pusat

lilitan bulatan 360oarea of sector - angle subtended at centre

3600luas seHor sudut pusat

luas bulatan 3600L4 Scale factor, k = #

Faktorskala,k=#


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5SULIT1 On thc graph in the answer space, shade the region which satisfy the three inequalites2y *x )B,x B,x


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6SULIT2 Solve the following quadratic equation using factorization method.

Sel es a i kan pers amaan kuadra t ik b erikut menggunakan kaedah pentfaktoran.

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[ 4 rnarks][4 ruarkah]Answer/Jawapan

3 Calculate the value of x and ofy that satisfy the following sirnultanc()us linear equation:Hitungkan nilai x dan nilai y yang memuaskan persamaan linacrr scrt:nltrk berikut:


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SULIT4 Diagram 4 shows a cuboid ABCDEFGH u,ith a horizontal square base EFGH.Rajah 4 menunjukkan sebualt kuboid dengan tapak segi empat set?to EFGH cLi atassotah mengufuk.

Diagram 4Rajah 4

/ is tlre midpointof AD. BC: 14 cmandCG:5 cm.I udulah titiktengah AD. BC: 14 cm danCG:5 cm.

1449t2ForExaminer'sUse


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8SULIT5 (a) State whether the following compound statement is true or false. 144912Nyatakan somo ada pernyataan majmuk berikut benar alau palsu.

0 x 12: 0 and 12":00x12:0danl\i:0

(b) Complete the following statement with the quantifier "all" or "some"so that it will become a true statement.Lengkapkan pernyataan berikut dengan pengkuontiti "semua" otuu"sebilangan" supaya menjodi suatu pernyataan benar(i) : ,;:i:;,';ff"Y;;"fr:":';;i";';;,:X;:i)ll]r'ffo,

(c) Write down Premise I to complete the fotlowing argument:Tuliskan Premis I untuk melengkapkan hujah berikut:Premise 2: n is not an even number.Premis 2: n bukan nombor genop.Conclusion: ( -3)' is not a positive number.Kesimpulan: (-3)' bukan nombor positif.

(d) It is given that the gradient of a straight line can be defined by


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9STJ LIT(r In Diagraln 6, thc straight line AB is parallel to ttre straight linc C.D. Point O is thc

origin. Given that the equatiou of the straight hne CDis J,- -f t-+2Dalam Rajah 6, garis lurtts AB odtrlalt .selari dengan garis lttnt,s CD. Titik O ialoltasalon. Diberi persamaan garis lurtrs CD iatah l, - + x - 4 .-2

[)iagrarn 6lluiult 6

t449t2

c (2,-3)

ForEraminer's

Usc


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ForExaminer's

Use

l0SULITAnswerlJawapon:

r449t2

(a)

(b)


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11STJI,IT

1 Diagrarn 7 shows a composite solid, formed by a combination of a cylinder and twoidentical cones. Given the diameter of the cone is 7 cm and its height is 9 crn. Thevolume of the composite solid is 1078 .*t.Raiah 7 menunjukkan gabungan pepe.lal yang terbentuk daripada cantuntan satusilinder dan dua kon yang sama. Diberi diameter kon ialah 7 cm dan tingginyq 9 cm.Isipadu gabungan pepejal ialah I078.rrr'.Using , calculate the height, in cm, of the cylinder.Menggunakan hitung tinS;gi, dalam cm, bagi silinder itu.

t449 t2

227 22

7

Diagram 7Rajah 7

[4 marks][4 markah]

AnswerlJawapan:

ForExaminer'sUse


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t2$ULIT l44gt2' -3 rz) (l -z\g (a) It is givcn that +l : -' ll, the inverse matrix of I | . pina rhe value of 3(-8 7 )- - [a 4) -"'p and of m. (t -2\Diberi bahawa +(-: "-'\,otot, matriks songsutlg basi | ' - l. Cari nilai p3[-s 7) o-----.-.[8 4)

dan nilai m.(b) Write the following simultaneous linear equations as a matrix cquations.Tulis persamaan linear serentak berikut dalam bentuk persamoan mcttriks.

7 x-2Y:58x - 3Y:10Hence, by using matrix method, calculate the value ol'x and of -y.Seterusnya, dengan menggunakan kaedah matriks, hitung nilai x dannilai y.

[6 marks]16 markah)

AnswerlJowapon:(a)

F'or[:xaminer's

{Jse


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t3ST]LIT() In Diagram9, OQS is a quadrant of a circle with centre O. OPT is another quadrant alsowith centre O. OUR is a straight line.Dalam Rajah 9, OQS ialah sukuan bulatan berpusat O. OPT ialah satu lagi sukuanbulatan jugo berpusqt O" OUR ialah garis lurus.

Diagram 9Rajah 9

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OT : TS : J cm and /- QOR : (r0"))Usins lr - ::. calculatee7OT : TS : 7 cm dan /- 60o

IiorExaminer'',s[-J,se


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t4SULITAnswer I Jawapan:(a)

(b)

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l5SULITl0 Diagram l0 shows two spinning discs. Disc K is divided into four equal sectors anddisc L is divided into three equal sectors. The pointers at the centr. of tlr. discs arespun simultaneously so that they will stop at random at one of the sectors.Rajah l0 menunjukkan dua buah cakera berputar. Cakera K dibohagikan kepadaempat sektor yang sama besar dan cakera L dibahagikan kepada tiga sektor yang san?ctbesar. Petunjuk -petunjuk pada pusat cakera itu diputorkon serentak sehingga merekaakan berhenti secara serentak pada mono-mena satu sektor. www.myschoolchildren.com

t 449 t2

Disc KCakera K Disc LCakera L

Diagram l0Rajah 10

ForExaminer's(Jse


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U.se

l6SUI-[TAnswerlJa\uapan .(a)

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l7STJ I,I TI I Diagram l l shows the speed-tirne graph

I 8 seconds .Rajah I I meruujtrkkan gruf'laju-n'tasasaat' Speed (*, ')Lqju (*t ')

for the ffrovement of a particle for a period ofbagi pergerakan sattt zarah dalam tempoh l8

Time (s)Masa (s)

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Diagram l1Rajah ll

ForExaminer's

Use


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I8SULITAnswer I Jawapan;(a)

t44912

(b)

(c)


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SULIT t9Section B

Bahagian B148 marksl

f48 markah)Answer any four questions from this section.Jawab nlana-mana empat soalan daripada bahagian ini.

12 (a) Complete Table 12 in thc answer space for the equation !: -x3 +2x+8 bywriting down the values o{'.y when x : -Z and x - 3. [2 marks]Lengkapkan Jadual 12 di ruang jawapan bagi persamaan y:-x3 +2x+8dengan menulis nilai y apabila x - -2 dan x * 3. [2 markah)

(b) For this part of the qucstion, use the graph paper provided on page 21. You mayuse a flexible curve rulc.By using a scale of 2 crn to I unit on the x-axis and 2 cm to l0 units on they -axis, draw the graph f y: _x' +2x+8 for -4 < x < 4 and

[4 marks]48


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Use

20SULIT

(a)

Answer/awapan :

www.myschoolchildren.com U4glz

Table l2Jadual 12

(a) Reftr graph on page 21.Rujuk graf di halaman 21.(c ) (i) y: . .........

x -4 -J -2 -t 0 I 2 3 3.5 4v 64 29 7 8 9 4 -27 .9 -48


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22

13 (u) Transformation R is a reflection in the straight line y : 5.( z\Transformation T is a translation I /- I .[- 3,]State the coordinates of the image of point (1, 3) under the followingtransformations:PenjelmaanR ialah satu pantulan pada garis lurus y : 5.( z\PenjelmaanT ialah satu translasi I - l.

[- 3/Nyatakan koordinat imej bagi titik (1,3) di bawah penjelmaan berikut:(i) r(ii) rR

SULIT

AnswerlJawapan:(a) (i)

(ii)

1449/2

[3 marks][3 ntarkah)


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23SULITl3 (b) Diagram 13 shows threc quadrilaterals, ABCN, KLMN and PQRS, drawn on aCartesian plane.Rajah 13 menunjukkan tigu sisi emltat, ABCN, KLMN dan PQRS, dilukis padasatah Cartesan.

I)iagram 13llujah 13

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SULIT14 Table 14.1

Jadual 14.ldari laut.

24144912

shows the frequency distribution of the lengths of 84 fish caught fi orn the sea.menunjtrkkan taburon kekerapcrn bagi panjang 84 ekor ikan.t'ctng tlitcrngkap

Length (cm)Panjang km)FrequencyKekerapan

1l 15 516-20 82t -25 l426-30 243l - 3s 2l36-40 94t -45 3Table I4.lJadual 14.l

(a) Calculate the estimated mean of the length ol the fish.Hitung min anggaran bagi panjang ikan. [3 marks]13 markahl

(b) Based on table 14.1, complete Table 14.2 tn the answer space by writing down thevalues of the upper boundary and the cunrulative frequency. [3 marks]Berdasarkan jadual 14.1, lengkapkan.ltulrtul 14.2 di ruang jawapan dengan


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St ]l,ll'Answcr /Jawapan:(a)

(b)

25 r449t2

Upper boundarySempadan atas

Cumulative frequencyKekerapan Longpokan

10.5 015.5

45.5 84

ForExaminer'sUse


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27l5 You urc not allorved to use graph paper to answer this questiou.

Attdu tidak dibenarkan menggunokun kertqs gra/'untuk menjavt'ob soulun ini.(a) Diagrarn 15.1 shows a righl prism with rectangular base ABCD on a hori zontalplane. The surface AEHII-D is tlrc unifbnn cross-section. The reclangle IJKL is aninclined plane. The rectanglc Ef'-Gt"l is a horizontal plane. The edges AE , BF, CK,DL, GJ and HI are vertical cdges. Given EH: FG:2 cn. www.myschoolchildren,com

Raiah 15.l menunjukkutt ytri,srrtu tagak dengan dengan tapak segiempat tepatABCD terletak di atas :;uluh murgtfirk. Permukaan AEHILD ialah keratan rentasseragam prisma itu. Scgi urtltul Ittt'GH adalah satah menguJirk. Tepi AE, BF, CK,DL, GJ dan HI adalah si.si tt,gul;. Diberi EH : FG : 2 cm.

s U LI't' 1449t2For

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28SULTTAnswer / Jawapan :(a)

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SULIT(b)

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A solid right prism is joined to the solid in Diagram 15.1. The combined solid isshou'n in Diagram 15.2. Triangle CMN is the unifonn cross section of the prismand NMRQ is an inclined plane. The base ABSRMCD is a horizontal plane andNK:2 cm.Sabuah pepejal prisma dicantumkan kepada pepejal dalam Rajah 15.1.Guhungan pepejal adalah seperti dalam Rajah 15.2. SegitigacMlv adalah keratan rentas seragam kepada prisma itu danNMRO adalah satah condonE;. T'apctk ABSRMCD adalah satah mengq/uk danNK :2 cm.

I.l,I III

ForExctntiner'sUse


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For'xaminer's

{Jse

30SULITAnswer I Jawapan :(b)(i)

(ii)

1449t2


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3lSULITI6 V((r5"N. 40"W) and R are two points on the surface of the earth such that VR is thediantctr:r' o1- il parallel of latitude.V(r5 tl' 4O tl) clan R ialah riua titik padct permukaan bunti dengan keadsan YR iatahdi ttrrrcIt, t' lru,gi selarian latitud.

Diagrarn l6lRajah 16

1449t2

[2 marks]12 markahl

[2marks]12 markah)

(a)

(b)

ljirxl tlrrr location of R.('(rrt lt',ltulukan R.( ir vq:rr that VZ is a diameter of thc carth. State the locat ion of Z.l)ilu,r i \//. ialah diameter bumi. Nvotakan kedudukan Z.

N/U

S/.S

ForExaminer's(Jse


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ForExqminer's

Use

32SULTTAnswer I Jawapan :

(a)

(b)

(c ) (i)

1449t2


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MAJLIS PENGETUA SEKOLAH MATAYSIA(CAWANGAN PULAU PINANG)MODUL LATIHAN BERFOKUS SPM 2013MATHEMATICSKERTAS 1

AnswercSkema Jawapan

QuestionNumber Answer QuestionNumber Answer QuestionNumber Answer QuestionNumber AnswerI B u D 2t D 3l C


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SULIT

144e/z(PP)PeraturanPemarkahanMatematikKertas 2September2013

MAJLIS PENGETUA SEKOLAII MALAYSIA(cAwAI\tGAlt PULAU PINAI\IG)

MODT'L LATIHAN BERTOKUS SPM 2OI3

MATEMATIKKertas 2


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Solution and Mark Scheme

AAAA>(w-o

t

Line y: x drawnRegion shaded correctlyNote: Award lmark for the shade region withoutdashed line.2x2 -7x -15:e(2x +3Xx -S):03x=-|,x=g


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SULIT 1449/2Question Solution and Mark Scheme Marks

(c ) If (-3)' is a positive number, then n is an evennumber. ll Jika (-3f iahh nombor positi{ makan ialah nombor genap.(d) :-* (-;)

KI

KIN1

5

(a) Y--5-l' or s:1 r2)+corequivalentx-Z 2 2 \' "- IIy =zx+4 or equivalent-2

o)(c)

k:4

KINIPI


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SULIT tu9t2Solution and Mark Scheme

(b)

t; _i[) =[;)[,)- 1 (-3[.rJ- n-3H-2x8)[-gx=-l!=-6Note 1.

/x\ /-r\l.r] = t- tJas a finar answer, award NlZ.Do not accept answer not using matrixmethod

P1

KIN1NI

a) ffx z"+xt4ft x 2 "+x 14 +t4+14s0

KIKINI


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SULIT 1449/2Question Solution and Mark Scheme Marksl1 (a)13

(b) ( 18- l0) (13)or ( 8 ) ( 13 )104 m

1i(r+ l3Xl0) +(5X13)(c)

-=l4or

any equivalentv- 16Note, 1(r+13)(rO) ," 5 (13) award KI2'

PIKINIK2

N1 6

12 (a) y: 12 'i!='13(b) Refer graph.Axis correctly drawn wittt correct direction

and trniform scale in the range 4 3 x< 4 and-48 s y s64points points* plotted in the

P1PI

PI

TO


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SULIT 1449t2on Solution and Mark Scheme Marks

Graph for question no 12 lGraf untuk Soalan 12


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itL!

r3)C=i=ts.'..i--

-ginwOJ

SULIT 1449t2Question Solution and Mark Scheme MarksI3 (a) (i) (3, 0)

(a) (ii) (3,4)Note:(l, 7) award Pl

(b) (i) E : Rotation of 90" anticlockwise at point(4,4).Note:L Rotation only. Award Pl2. Rotation at point (4, 4) orRotation of 90o anticlockwise. Award P2

O) (il) F : Enlargement of scale factor 2 about pointN or (5, 5).Note:1. Enlargement only. Award Pl2. Enlargement of scale factor 2 orEnlargement about point N or (5, 5). Award P256 or

PIP2

P3

P3

K2


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Solution and Mark Scheme

Upper boundarySempadan atas CumulativefrequencyKekeraponLonggokan

r 0.5 015.5 524.5 1325.5 2730.5 5l35.s 7240.5 8l45,5 84tUpper Boundary

Cumulative frequencyOgwe (Refe, graph on page )Axes drawn in the correct dirpction , uniform scalefor to.ss.r


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SULIT 1449t2on Solution and Mark Scheme Marks

Graph for Question t4/ Graf untuk soalan 14

10


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Question Solution and Mark Scheme Marksl5 (a)

Correct shape wittr hexagon ADLII{E.AII solid lines.AD>IL>IH>LD>AE=EHMeasurements comect to * 0.2 cm (one way) and all

KIKI

SULIT 144912

1.1


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SULIT M49nQuestion Solution and Mark Scheme Marksl5 (bxi)

72

SULIT 1449t2


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Question Solution and Mark Scheme MarksI5 (bxri)

Correct shape wittr rectangles fIKL, LKCD and tiangleCMN.All solid lines.(Igrtore HG)joined

KI

13

SULIT 1449t2


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free 'f, piipsrs. free skena at : vlwu;.myschooichiiCren coi'rt

Question Solution and Mark Scheme Markst6 (a) (651\ , l4ooE)

(b) (6505, l4ooE)(c ) (i) Distance VE = 375 x25: 937 .5 n.m.

(ii) g x 50 x cos 650 : 937 .5Note : cos 650 used correctly, award Kl0:37oLongitude of E : 30 W(d)Shortest distance EN : (90 - 65) x 60: 1500 n.m. r

PI PIPI PIKIN1K2NINIK1N1 t2

MODUL BERFOKUS 1VIATEMATIK SPM 2013EXAMINATION SPECIFICATION TABLE (EST)


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SPM MATHEMATICS . PAPER 2

Content Evaluation

Lines and Planes in 3-DimensionsGraph ofFunctions II

Gradient and Area under a

Plans'iurd Elevatiohs

TotalQuestion

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