Spm 1449 2006 Mathematics p2 Berjawapan

8/2/2019 Spm 1449 2006 Mathematics p2 Berjawapan

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Paper 2 j rro.r.,ime:2Instrrrction s: This question pq'per consistsof 2 sections.. ection A and,section B. Ansu.er all.uestions z section A-?!d t;-; q";;;;;ns in sectio. n. wri',r"your answers in the spaceprot,ided':,::;,i::',tJ"[,,!i,!5i wo"ki"j ;;";;;'"t be wrim",. 'i.""rv. Non-programmabtescientinc

Answer:(a)

Diagram 1 shows atriangle t/pe is the

Identify and calculate the angle

right prism. The base .PQRS s a horizontaluniform cross section ofihe p"i"r*

Section A(52 merks)Answer al l questions n this section.

ITJ:T j;*trn in the answer pace hows etsp, e andR such9: tl" diagrams n the answerspace, hade(a) the setp'n e,(b) the set(pue,)nB.

that the universal set.

[3 murks]

rectangle. The right angled

Diagram I


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3 Solve the quadratic equationAnswer:

4 Calculate the value of r and

Answer:

U&-"_)=x+,6:

of y that satisfr the following simultaneousr+2y=$3-,rc-!=-t

[4 marhs)

Iinear equations:

14marksl

o Diagram 2 showsa combinedsolid consistsof a right prism and a right pyr.amidwhich arejoinedat the planeEFGH.vis vertically abovene trasercg. Trapeziut iaep i"the uniform'cross ection f the prism., ,,,,,. .,;.i.,,;;;,i ,

-___-

_=_1


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fte^trliSft of the pyr.amid is g cm andFG = 14 cm.a) Calculate the volume, ir, "rnr. of1-il right pyramid.ft) It is given that the volume of the combined solid is 5g4 cma.alculate the length, in cm, of af. "'

Answer:(a)

[4 marks]

ft)

6 (a) bcompleTil|Xllff*"Jing statements ith thequantifier ,all,,or,,some,,sothat it will(i ) .......of the prime numbers are od.dnumbers.(ii) ............. pentagonshave five sides.

*' irt?::jt"?;:rx::""r of the following statement and hence determine whether its converse

(c) Complete the premise in the following argument:Premise 1 : IfsetKis a subsetofset L,


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Answer:(a) (i) .......of the prime numbers are odd numbers.

(ii) ............. pentagonshave five sides.(b)

(c) Premise2:

7 In a quiz contest, there ardrttrree categories of questions consisting of 5 questions on sport,3 questions on entertainment and 7 questions on general knowledge.Each question is placed inside,an envelope. All of the envelopes are similar and put insidea box.All the participants of the quiz contest are requested to pick at random two envelopes fromthe box.Find the probability that the first participant picks(a) the frrst envelope with a sport question and the second envelope with an entertainmentquestion,(b) two envelopes with questions of the same category. 15markslAnswer:(a)

(b)


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8,t$n:Dfiftgrtarn{BtOhIeN is a quadrant of a,circhrnri&aestme@dri}dffiffiah;aec,of,another circlewith eentre O.OMP and ORQ arc straight lines. i : i I ' t i r : . I*,

i

, , . . ] inM

Oiagii#litaOM = MP = 7 cm and ZPOQ = 60".Using = ?,caloulate:

. . i : . ; l

tr. :ri

, . t t ; : r : - r l ., : { l re{. . i ;: : ! : i , l l ! :!: ; r'

[6 marks](a) the perimeteq in cm, of the whole diagram,

' . - : , : j l , '; j , ;' ,,,,(h).!hp.area, n cm2,of the shaded egion.Answer:(a)

ft)


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Speed (m s-r)

Diagrhrn 4 shtrws the speed-time'graph'for.ths'rroveilrerit of .a:,partiele.for a period20 seconds.

Diagram 4(a) State the uniform speed, in m s-1,of the particle.(b) The distance travelled by the particle with uniform speed is g4 m.Calculate(i) the value of t,(ii) the average speed, n m s-1,of he particle for the period of 20 seconds.Answer:(a)ft ) (i)

[6 marks]

(ii)


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10 Diagram 5 shows a straight line PQ and a sraight line RS drawn on a Cartesian plane.PQ is parallel to RS.

Diagram 5Find(a) the equation of the straight line PQ,(b) the r-intercept of the straight line P@ ", 'Answer:(a)

: ,. ' :.i

it

. :11 2\ to r \11 (a) It is given thatl i - | is the inverbe matrix of { _"' ; I\ i , "1 \- - tFind the value of z.

$)

15marksl


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Answer:(a)

Section B(48 marks)Answer four questions from this sectioru.

12 (a) Transformation T is a translatio" (;3 ) and transformation p is an anticlockwise rotationof 90. about the centre (1, 0). \ 4 I3i'H,iffi:llinates of heimage fpoint 5,1)undereach f he ollowingransformations:(ii) T'r"tmr.tion T,(iii) Cs66itred transformations T2.oortffi..* 6 shows three quadrilaterals, ABCD, EFGH andJI{LM,drawn on

(b)

(b) [4 marhs)a Cartesian


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(i) JKLM is the image of ABCD under the combined transformations VU.Describe in full the transformations:(a) U(b) v.(ii) It is given that quadrilateral ABCD represents a region of area 18 m2.Calculate the area, in m2, of the region represented by the shaded region.[8 m.arks]

Answer:(a) (i)

(iii)

(b) (i) (a) U:

(b) v

13 (a) CompleteTable 1in the answer space or the equationy -24 bywritingdownthevaluesxof y when r = -3 and r = 1.5. [2 morks]

(ii)

(ii)


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Graph for Question lB


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(c) From your graph, find(i) the value of y when x = 2.9,(ii) the value of r when y = -13. [2 marksl(d) Draw a suitable straight line on your graph to find a value ofr which satisfies the equation2x2+5x=24for-4


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Answer:(a)

(c) Refer graph on page 16J.(d)

(b)

15 (a) Diagram8(i);fows,a solidright prism.The surface KLSRis ts uniform crosssection.TheaseJKLM is a rectangl" ot'-" hirizo-ntalp;;". ri.;;edg3sle JKReis an incrinedplanend the squareQBsrls"

rr*J""trl pd;. i;"'ffi, sz and TM arevertical.

Class interval--_ 27_25----_26_30Midpoint Frequency23 D

Table 2

Draw to full scale,


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the elevation of the solid on a verticalfrom X.Antswer:(a)

plane,paralLel to JK as viewed

(b) {lother solid right prism is joined to the solid in the Diagram g(i) at the vertical planeKLSG to form a combined solid as shown in Diagram g(ii). The trapezium ABCD is itsuniform cross section and AFKB is an inclined plane. The rectangl. Annp is a horizontalplane. The base JKBCLM is on a horizontal pLt". The edge oc is vertical.

* l/ l' l IIIIIII

r" l


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Draw to full scale,(i ) the elevation of the combined solid on a vertical plane parallel to JB as viewedfrom Y, 14marksl(ii) the plan of the combined solid. L5markslAnswer:(b) (i),

(ii)


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16 Diagram 9 shows four points, P, Q.,R and X, on the surface of the earth. P lies on longitude80"W. QR is the diameter of the parallel of latitude of 50'N. X lies 5 820 nautical miles duesouth of P. N

(a)(b)

(c)(d)

Diagrqrn 9Find the position of R.Calculate the shortest distance, in nautical miles, from @ tosurface of the earth.Find the latitude of X.An aeroplane took off from P and flew due west to R along thean average speed of 600 knots.Calculate the time, in hours, taken for the flight.

lJ marksl.R, measuredalong the[2 marksl

13markslparallel of latitude with

14markslAnswer:(a)

ft)

(c)

II

SPM 2006 EXAMINATION


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Paper 1lD 26D 7r tB t216C t7zLD 22264 2731A 3236C 37Paper 2Section AI (a)

PAPER Volume of 1pyramid = 5xArea of base x heightI= ;* 6 x 14 x 8= 224 cmsVolume of combined solid = bg4 cmg1224 + j(70 + I )(AF)Gt = 584

72 xAF = 360AF -5cm(i) Some(ii) AlrI fr>5,thenr>gFalseKvL*L-5-q _ 1575"14-2tO

174f+, .+) * /3,.2)*/7,. 6 ' l\15 '14l ' \ rs ^ Ml ' \ ts^ t+J=20*6,42270 270 270_68270=34

105LengthofarcN-t? rr&, ZxffxZ

= 8.6ZcmLengthfatcMR ff i , . ZxffxZ= 7.BB mLength or arc pO = fo 'o" ,60x2xfx74= 14.67 cm.' . Perimeter = 7(4) + 3.67 + 7.BB+ 74.67= b3.67cm

Area of ONft = 30 "860xTx7x7= 12.88cm2Area of PQRM160 22 \ 160 2' \[360T " u x ra)_(]#xf xzxz)= 102.67 25.67

(b)B 3C 4A 5BD 8A 9D IOBC 13B 14 C 15AC 18A 19D 20 AA 23 C 24 A 25 BD 28 B 29 D 30CB 33 D 34 B 35 CB 38D 39A 40 C

5 (a)

6 (a)

(b)(c ,

't ta)

8 (a)

(b)(b)

2 ZPRU/ZURPU9cmP

tang = 9130 =3442'3 Br(r-1) _Z = x+63x2-3x = 2x+723x2-5x-L2 = QGx+4)(x-3) - 0

4tr = -5 orx+2y-6

3_2"-Y =-' t3x-2y=-t4

3o/6\""" . ' " " \ : /

(b)

@"@

P'oQ

(PvQ')63

(ii) Total distance (a)


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lO (a)

11= *e3 + 14)(12) 84 + *tZl\+lz2= 222+84+t4= 320mTotal distanceAveragespeed = --Ttr"":

_ 32020=L6ms-1Gradient of .RS = a-2=-lEquation of PQy-5 = -2(x-3)

Y-5 = -2x+6t = -2x+L7Onr-axis,y=0O = -2x+tL2x = 1l -11, =t t1.' . t-rntercept = i

13

14 (a)

.b)

1l (a)

. . . u=_I,u=Section B12 (a) (i) (0, 4)(ii) (2, 3)(i i i) (-1, 5)

(c)(d)

1 2\1l,nt1 2\1l,n l1 2\1ltnt

4) =3t2\el =, /

1")= -s\u/ \21(" \ =(1 ?)/-5\\u/ \ i ; t \2t[ -5 + 4 \=\-**s/

/ -1 \=l1 l'2 '

{b)

1 12o=a \r1127\rI7lr\ t

. ' . , =12(3 -4\\-1 2l

4\ |3i= \

1,

(i ) 8(ii) 1.852x2+5x=24u*s =TI =2x+50 2

v o I. ' . x = 2.45

(b) (i ) (a) U is

tc -3 1.5v -8 16(b)

T

ti4\

:r r l:r j l o :}:.Y r 4;

\;4

4.4Fl t, t : ri aU ;' .$4

l t : l lx t : : i i ' ul:::;\ :

*li i . l\

Class interval Midpoint Frequency2I-25 23 526-30 28 631-35 33 836-40 38 104l-45 43 n

rtr4 6cm EISIL 1cm


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I \.' / ,d:; \;:\!

/ \ i . . l$11,1

t \ l

t l ''i t

tclFrequency

15.5(d)

15 (a)

6l DIC

10q

20.5 25.5 30.5 35.5 40.5 45.5 50.5 55.5Donations (RM)Modal class of the donations is 36 - 40.

RIQ 6cm SIT

KIJ 5 cm LIMAID

cm

R (50"N,135'E)Shortest distance from Q to,R = I x 60=80x60= 4800 n.m.Distance of PX = 5820 n.m.0x60 = 58200=97'Latitude of X is 47'S.Distance of PR = 145 x 60 cos 50"N= 5592.252 n.m.

DistanceIrme = _=-SPeed_ 5592.252

600= 9.32 hours

(a)(b)

16

87o

^

210

(c)

(d)

ft) (i )Qlr

6cm

FIEIcm

'_-_-_F--'

I

JIM BIC

Spm 1449 2006 Mathematics p2 Berjawapan

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