26
PROGRAM DIDIK CEMERLANG AKADEMIK SPM SPM PAPER 2 MODEL & SOLUTIONS ORGANISED BY: JABATAN PELAJARAN NEGERI PULAU PINANG ADDITIONAL MATHEMATICS FORM 5 MODULE 20 http://mathsmozac.blogspot.com http://sahatmozac.blogspot.com

# SPM - Penang Free School papers Add Maths... · program didik cemerlang akademik spm tttt spm paper 2 model & solutions organised by: jabatan pelajaran negeri pulau pinang additional

ngoanh

• View
221

0

Embed Size (px)

Citation preview

SPM

TTTT

SPM PAPER 2 MODEL & SOLUTIONS

ORGANISED BY:

JABATAN PELAJARAN NEGERI PULAU PINANG

MODULE 20

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

SULIT3472/2MatematikTambahanKertas 22 2

1 hours

JABATAN PELAJARAN PULAU PINANG

SPM EXAMINATION MODEL PAPER

Paper 2

Two hours and thirty minutes

DO NOT OPEN THE QUESTION PAPER UNTIL YOU ARE TOLD TO DO SO

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

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

from Section C.

3. Give only one answer/solution to each question.

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

6. The marks allocated for each question and sub-parts of a question are showns inbrackets.

7. A list of formulae is provided.

8. A booklet of four-figure mathematical tables is provided.

9. You allowed to use a non-programmable scientific calculator.

This question paper consists of 9 printed pages.

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

2

Section A

[ 40 marks ]

Answer all questions in this section.

1. Solve the simultaneous equations 3x + 2y = 5 and 2x 2 + xy = – 3 .

[5 marks]

2. The equation of a curve is y = 3x 2 + 5x + c where c is a constant. The straight line

y + 7x + 5 = 0 is a tangent to the curve at a point P. Find

(a) the coordinates of P, [4 marks]

(b) the value of c. [2 marks]

3. Figure 1 shows the first few triangles of a series of triangles of equal heights drawn ona straight line, AB.

The lengths of the bases of the triangles form a geometric progression. The length ofAB is 16.38 m.

(a) Calculate the area of 8th triangle, [3 marks]

(b) If n triangles are drawn on the line AB, find the maximum value of n. [3 marks]

A B

3 cm

Figure 1

0.4 cm 0.8 cm 1.6 cm

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

3

4. Table 1 shows the score distribution obtained by a group of contestants in a contest.

Table 1

(a) Calculate the varians of the distribution. [4 marks]

(b) If each score in the distribution is multiplied by 2 then subtracted by c, the mean

becomes 2.8. Calculate

(i) the value of c,

(ii) the new standard deviation. [3 marks]

5. (a) Given tan A =4

3, find the value of cos 2A. [3 marks]

(b) (i) Sketch the curve y = sin 2x for 0 x 2 .

(ii) Hence, by drawing a suitable straight line on the same axes, find the number

of solutions satisfying the equation sin x kos x =2

1

4

x

for

0 x 2 . [5 marks]

6. In figure 2, PQRS is a quadrilateral and PTR is straight line.

Given thatPQ = 6x,

QR = 24y,

SR = 15x + 12y dan PT =

3

1TR

(a) Express in terms of x dan y

(i)PR

(ii)QT . [3 marks]

(b) Show that PS is parallel to QT. [3 marks]

(c) If cm7x and o60TPQ , findPR [2 marks]

Score 1 2 3 4 5Number of contestants 4 6 12 5 3

S

T

R

P Q

Figure 2

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

4

Section B

[40 marks]

Answer four questions from this section.

7. Use the graph paper provided to answer this question.

Table 2 shows the values of two variables, x dan y, obtained from an experiment.x dan y are related by the equation y – q = p( x + 1 )( x – 1 ), where p and q areconstants.

x 1.0 2.0 3.0 4.0 4.5 5.5

y 12.0 16.0 24.2 34.3 40.5 55.6

Table 2

(a) Plot y against x 2 using a scale of 2 cm to 5 unit on both axes.Hence, draw the straight line of best fit. [4 marks]

(b) Use your graph from (a) to find the value of

(i) p

(ii) q

(iii) y when x = 5. [6 marks]

8. In figure 3, the straight line PQ is a tangent to the curve y = x 3 + 4 at the point

T(2 , 12 ).

Find

(a) the value of k, [3 marks]

(b) area of shaded region, [4 marks]

y

xO

P

Q(k , 0 )

T(2 , 12)

y = x 3 + 4

Figure 3

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

5

9. Solution by scaled drawing will not be accepted .

In figure 4, the straight line PR cuts y-axis at Q such that PQ : QR = 1 : 3. Theequation of PS is 2y = x + 3.

(a) Find

(i) the coordinates of R,

(ii) the equation of the straight line RS,

(iii) the area PRS.

[7 marks]

(b) A point T moves such that its locus is a circle which passes through the points

P, R dan S. Find the equation of the locus of T.

[3 marks]

P(- 3, 0 )

Q( 0, 4 )S

Ry

xO

Figure 4

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

6

10. Figure 5 shows a semicircle APD with AD as a diameter, and PBQD is a sector whose

centre is D. AB = BC = 1 cm dan AD = 10 cm.

Calculate

(a) the value in radian, [4 marks]

(b) the length the arc AP in cm, [2 marks]

(c) the area of shaded the shaded region in cm 2. [4 marks]

11. For this question, give your answer correct to 3 significant figures.

(a) A survey done on a group teachers shows that 32 teachers of 40 teachers in a

district posses laptop computers. If 6 teachers from that district are selected at

random, calculate the probability that

(i) exactly 3 teachers posses laptop computers,

(ii) more than 2 teachers posses laptop computers.

[4 marks]

(b) The heights of teenagers in a residential area were found to have normaldistribution whose mean is 150 cm. It was found that a teenager whose height is156 cm has a standard score of 1.5 . Find(i) the value the standard deviation, ,

(ii) the probability that a teenager selected at random from that area has aheight between 148 cm and 158 cm,

(iii) the value h if 70% of the teenagers in that area are less than h cm tall.

[6 marks]

B DC

θ

A

P

Q

Figure 5

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

7

Section C

[ 20 marks ]

Answer two questions from this section.

12. In Figure 6, BC is parallel to ED.

Find

(a) the length, in cm, of(i) EC,

(ii) EB .[6 marks]

(b) AEB,[2 marks]

(c) the area of ABE in cm2.[2 marks]

A

B C

DE

4 cm

7 cm

6.5 cm

3.5 cm

70 o

Figure 6

75 o

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

8

13. Table 3 shows price indices of sport shoes of four brands sold in a shop. Figure 7 is a pichart which represents the number of pairs of shoes by brands sold in 1997.

Brands of

sport shoes

Price indices in 1994

(1991=100)

Price indices in 1997

(1994=100)

P 112 130

Q 108 125

R 123 X

S Y 110

Table 3

(a) (i) If the price of the P brand shoes in 1991 was RM50, calculate the price in

1994.

(ii) If the price of the Q brand shoes in 1997 was RM60.75, calculate the

price in 1991.

[3 marks]

(b) (i) Given that the price of the R brand shoes increased by 18% in 1997

compared to the year 1994. Find value x.

(ii) The price index for the S brand shoes in 1997 based in on the year 1991 is

165. Find the value of y.

[4 marks]

(c) Calculate the composite index number for the shoes of the four brands for the year

1997 based on the year 1994.

[3 marks]

110o

30o

RQ

SP

Figure 7

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

9

14. Use the graph paper provided to answer this question.

The Mathematics Society of a school is selling x souvenirs of type A and ysouvenirs of type B in a charity project based on the following constraints :

I : The total number of souvenirs sold must not exceed 75.

II : The number of souvenirs of type A sold must not exceed twice the numberof souvenirs of type B sold.

III : The profit gained from the selling of a souvenir of type A is RM9 whilethe profit gained from the selling of a souvenir of type B is RM2. The totalprofit must not be less than RM200.

(a) Write down three inequilities other than x 0 dan y 0 which satisfy the aboveconstraints. [3 marks]

(b) Hence, by using a scale of 2 cm to 10 souvenirs on both axes, construct and shadethe region R which satisfies all the above constraints. [3 marks]

(c) By using your graph from (b), find

(i) the range of number of souvenirs of type A sold if 30 souvenirs of type B aresold.

(ii) the maksimum which may be gained. [4 marks]

15. An object, P, moves along a straight line which passes through a fixed point O.

Figure 8 shows the object passes the point O in its motion. t seconds after leaving the

point O , the velocity of P, v m s─1 is given by v = 3t2 – 18t + 24. The object P stops

momentarily for the first time at the point B.

P

O B

Figure 8(Assume right-is-positive)

Find:

(a) the velocity of P when its acceleration is 12 ms – 2 , [3 marks]

(b) the distance OB in meters, [4 marks]

(c) the total distance travelled during the first 5 seconds. [3 marks]

END OF QUESTION PAPER

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

3472/2

Paper 2

JABATAN PELAJARAN PULAU PINANG

Paper 2

Solutions and Marking Schemes

This document consists of 16 printed pages

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

2

1. 3x + 2y = 5 ……………………………. (1)2x 2 + xy = – 3 ……………………………. (2)

From (1) : y =2

x35 ………………... …… (3) [1]

Substitute (3) into (2) :

2x 2 + x

2

x35= – 3 [1]

2 : 4x 2 + 5x – 3x 2 = – 6x 2 + 5x + 6 = 0

(x + 2)(x + 3) = 0 [1]x = – 2, – 3 [1]

When x = – 2 , y =2

11

2

)2(35

When x = – 3 , y = 72

)3(35

[1]

2. (a) Equation of curve : y = 3x 2 + 5x + c

dx

dy= 6x + 5 [1]

The given equation of tangent at P is y + 7x + 5 = 0,Rearranging, y = – 7x – 5

Gradient of tangent at P = – 7 6x + 5 = – 7 [1]

x = – 2 [1]and y = – 7(– 2) – 5 = 9

coordinates of P = ( – 2 , 9 ) [1]

(b) Point of contact P( – 2 , 9 ) lies on the curve y = 3x 2 + 5x + c

9 = 3(– 2 ) 2 + 5(– 2) + c [1] c = 7 [1]

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

3

3. (a) Length of bases of triangles: 0.4 cm , 0.8 cm , 1.6 cm , ………..form a geometric progression with a = 0.4 and r = 2 [1]Length of base of 8th triangle = T 8

= ar 7

= 0.4 (2) 7 [1]= 51.2 cm

Height of triangle = 3 cm

area of 8th triangle =2

1 51.2 3 = 76.8 cm 2 [1]

(b) Length of AB = 16.38 m = 1638 cm and number of triangle drawn = nSum of lengths of bases of triangle 1638 cm

S n 1638

1r

)1r(a n

1638

12

)12(4.0 n

1638 [1]

2 n – 1 4.0

1638= 4095

2 n 4096 = 2 12

n 12 [1] the maximum value of n = 12. [1]

4. (a)Score, x Frequency, f Fx fx 2

1 4 4 42 6 12 243 12 36 1084 5 20 805 3 15 75

f = 30 fx = 87 fx 2 = 291[1] [1]

2 = 23087

30291

[1]

= 9.7 – 2.9 2 = 1.29 [1]

(b) (i) Old mean =30

87= 2.9

When every score is multiplied by 2 and then subtracted by c,new mean = 2.8 2(2.9) – c = 2.8 [1]

c = 3 [1]

(ii) new standard deviation = 2 ( 29.1 ) = 2.272 [1]

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

5. (a) Given tan A =4

3

cos A = 5

4[1]

cos 2A = 2 cos 2 A – 1

= 2 1254

[1]

=25

7[1]

(b) (i)

(ii) The given equati

2 :

The required st

From the grap

y

xO

1

- 1

y = sin 2x

y = 12

x

3

– 3 4– 4

5

5

http:/

http://sahatmozac.blogspot.com

4

on : sin xkosx

kxsin2

sin

raight line has th

x 0y – 1

h, the number of

21

/mathsmozac

4

x

xos

x2

e equati

2

xy

2 0

solutio

.blogs

2

1

12

x

12

x

on

1

n = 4.

23

pot.com

2

Lengkung [2]

Garis lurus [1]

[1]

[1]

5

6.

(a) (i)PR =

PQ +

QR

= 6x + 24 y [1]

(ii)QT =

QP +

PT

=QP +

QR

4

1

= –6x +4

1( 6x + 24y) [1]

= –2

9x + 6y [1]

(b)PS =

PR +

RS

= )y12x15()y24x6( [1]

= – y12x9

= 2 ( – yx 629 )

= 2QT [1]

PS is parallel to QT. [1]

(c) SPT = 90 O and PS is parallel to QT PTQ = 90 O

PQ

PTcos 60 o =

2

1

PT =2

1 ( 6 7) = 21 [1]

PR = 4 PT

= 4 (21) = 84 cm [1]

60 o

24 y

6 x

15 x + 12 y

S

T

R

P Q

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

6

7. (a)[1]

Refer to Appendix A

Correct axes and uniform scale [1]6 points marked correctly [1]Straight line of best fit [1]

(b) (i) y = p( x + 1 )( x – 1 ) + q= p ( x 2 – 1 ) + q= p x 2 – p + q [1]

=125.20

125.40

[1]

= 1.48 [1]

(ii) – p + q = y intercept

–1.48 + q = 10.5 [1]q = 11.98 [1]

(iii) When x = 5, x 2 = 25y = 47.5 [1]

8. (a) Equation of curve : y = x 3 + 4

2x3dx

dy [1]

Gradient of tangent at T( 2 , 12 ) = 3 ( 2 2 ) = 12

12k2

012

[1]

2 – k = 1k = 1 [1]

x 2 1 4 9 16 20.25 30.25y 12.0 16.0 24.2 34.3 40.5 55.6

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

7

(b)

Area under the curve RT = 2

0dxy

= dx4x2

0

3 [1]

=

2

0

4

44

x

x[1]

= 4 + 8 – (0) = 12

Area of QST = 612121

area shaded region = 12 – 6 [1]= 6 unit 2 [1]

(c) Volume generated = dyx21

4

2

= dy])4y([12

4

23

1

[1]

= dy)4y(12

4

3

2

=

35

)4(53

12

4

y [1]

= ]08[ 3

5

53

= 596 unit 3 [1]

y

xO

P

Q(k , 0 )

T(2 , 12)

y = x 3 + 4

R

12

12

S

y = 12

2

4

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

8

9. (a) (i) Let the coordinates of R be (h , k)

04

9h

and 4

4

0k

[1]

h = 9 k = 16 coordinates of R = (9 , 16) [1]

(ii) The given equation of PS is 2y = x + 3

Rearranging, y = 21 x + 2

3

m PS = 21

From the diagram, RS PS.

m RS m PS = – 1

m RS = – 2 [1]

The straight line RS passes through the point R(9 , 16)

the equation of RS is y – 16 = – 2 (x – 9 )

Simplifying and rearranging, y = – 2x + 34 [1]

(iii) Equation of RS: y = – 2x + 34 …………….. (1)

Equation of PS: 2y = x + 3 ……………… (2)

Substitute (1) into (2) :

2(– 2x + 34) = x + 3 [1]

– 4x + 68 = x + 3

65 = 5x

x = 13

and y = – 2(13) + 34 = 8

coordinates of S = (13,8)

area of PRS =08160

31393

2

1

= )242080(072482

1 [1]

= 80 unit 2 [1]

(b) Centre of circle = mid-point of PR = )2

160,

2

93(

= (3 , 8) [1]

Radius of circle = 21 PR = 22

21 )160()93( = 10 [1]

Therefore, equation of locus of T is (x – 3) 2 + (y – 8 ) 2 = 10 2

Simplifying and rearranging, x 2 + y 2 – 6x – 16y – 27 = 0 [1]

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

9

Other method:

Let coordinates of T be (x,y)

PR is a diameter. TP is perpendicular to TR.

,19

16

)3(

0

x

y

x

y

1)9)(3(

)16(

xx

yy

y(y – 16) = ─(x + 3)(x ─ 9)

y2 – 16y = ─(x2 – 9x +3x – 27)

= ─(x2 – 6x – 27)

= ─x2 + 6x + 27)

x2 + y2 – 6x – 16y – 27 = 0

10.

(a) Let O be the centre of semicircle APD.

Radius of semicircle APD = 21 10 = 5 cm

OP = 5 cm , CO = 5 – 1 – 1 = 3 cm [1]

kos POC =5

3[1]

PO = OD POD is an isosceles triangle

ODPOPDPOC

0.9273 = = 2

0.9273 2 = 0.4637 radian. [1]

B DC

θ

A

P

Q

O1 cm1 cm 5 cm3 cm

5 cm

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

10

(b) Length of arc AP = 5 0.9273 [1]

= 4.637 cm [1]

(c) Using Pythagoras’ theorem, PC 2 = 5 2 – 3 2 = 16

PC = 16 = 4 cm [1]

sinPD

PC

PD

4= sin 0.4637 [1]

PD = 8.943 cm

and PDQ = 2 = 0.9273

area of shaded region = 21 (8.493) 2 ( 0.9273 – sin 0.9273 ) [1]

= 4.591 cm 2 [1]

11. (a) (i) P(one teacher posseses laptop computers ) =40

32= 0.8

P(exactly 3 teachers posses laptop computers)

= 333

6 )2.0()8.0(C [1]

= 0.0819 (correct to 3 significant figures) [1]

(ii) P(more than 2 teachers posses laptop computers)

= P( X = 3, 4, 5, 6)

= 1 – P (X = 0, 1, 2)

= 1 – [ 0.2 6 + 16 C (0.8) 1(0.2) 5 + 2

6 C (0.8) 2 (0.2) 4 ] [1]

= 0.983 (correct to 3 significant figures) [1]

(b) (i) Given the standard score for 156 cm height = 1.5

5.1150156

6 = 1.5

=5.1

6= 4 [1]

(ii) P( 148 < X < 158 )

= P (4

150148 < Z <

4

150158 ) [1]

= P ( – 0.5 < Z < 2 )

= 1 – ( 0.3085 + 0.0228) [1]

= 0.669 (correct to 3 significant figures) [1]

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

11

(iii) Given P( X < h ) = 70%

P( Z < )4

150h = 0.7

P( Z > )4

150h = 1 – 0.7 = 0.3

From tables, P(Z > 0.524) = 0.3

524.04

150

h[1]

h = 4(0.524) + 150

h = 152 (correct to 3 significant figures) [1]

12. (a) (i) Using the cosine rule in CDE :EC 2 = 6.5 2 + 3.5 2 – 2 (6.5)(3.5) cos 70 o [1]

EC = 94.38 = 6.24 cm [1]

(ii) Using the sine rule inCDE :

24.6

70sin

5.3

DECsin o

[1]

sin DEC = 0.5271

DEC = 31.81 O

Given ED is parallel to BC

BCE = DEC = 31.81 O [1]

From EBC , EBC = 180 o – 75 o – 31. 81 o = 73.19 o [1]

Using the sine rule in EBC :

oo

EB

19.73sin

24.6

81.31sin

EB = 3.436 cm [1]

(b) Using the cosine rule in ABE :

cos AEB =)436.3()4(2

7436.34 222 [1]

= – 0.7710 AEB = 140.44 o [1]

(c) Area of ABE

=2

1( 4 ) ( 3.436 ) sin 140.44 o [1]

= 4.377 cm 2 [1]

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

12

13. (a) (i) Given the price of P brand shoes in 1991 = RM50 and

its price index in1994 based on the 1991 = 112

Its price in 1994 = 112 % RM50 = RM56 [1]

(ii) For the Q brand shoes,

price index in 1994 based on the year 1991 = 108

08.1100

108

H

H

91'

94' ……………………………….. (1)

price index in 1997 based on the year 1994 = 125

25.1100

125

H

H

94'

97' ………………………………. (2)

(1) (2) gives 91'

97'

H

H1.08 1.25 = 1.35 [1]

Given 97'H = RM60.75, 91'H =35.1

75.60= 45

price of Q brand shoes in 1991 = RM45 [1]

(b) (i) Given the price of R brand shoes increased by 18 % in 1997

compared to the price in 1994.

x = price index in 1997 based on the year 1994

= 100 + 18 = 118 [1]

(ii) Price index for S brand shoes in 1997 based on the year 1991

= 91'

97'

H

H100

= 100H

H

H

H

91'

94'

94'

97'

= 100100

y

100

110 = 1.1 y [1]

Given the price index = 165 1.1y = 165 [1]

y = 150 [1]

(d) Angle of sector for S brand shoes= 360 o – (30 o + 110 o + 90 o ) = 130 o [1]

composite price index in 1997 based on the year 1994

=360

)130(110)90(118)110(125)30(130 [1]

= 118.25 [1]

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

13

14. (a) The three inequalities are

x + y 75 , x 2y dan 9x + 2y 200 [3]

(b) Refer to Appendix B :

All lines drawn correctly [2]

(c) (i) 16 number of A type souvenirs sold 45 [2]

(ii) Maximum profit

= RM [ 9(50) + 2(25) ] [1]

= RM500. [1]

15. (a) Velocity, v = 3t 2 – 18 t + 24

Acceleration, a =dt

dv= 6t – 18 [1]

When a = 12 ms – 2 ,

6t – 18 = 12

6t = 30

t = 5 [1]

v = 3( 5 2 ) – 18 (5) + 24 = 9 ms – 1 [1]

(b) When P stops momentarily, v = 0

3 t 2 – 18t + 24 = 0

:3 t 2 – 6t + 8 = 0

( t – 2 )( t – 4 ) = 0 [1]

t = 2 , 4

P stops momentrarily for the first time when t = 2 seconds [1]

v = 3 t2 – 18t + 24

Displacement, s = ( 3 t 2 – 18t + 24) dt

= t 3 – 9 t 2 + 24t + c [1]

When t = 0 , s = 0.

c = 0

s = t 3 – 9 t 2 + 24 t

When t = 2, s = 2 3 – 9 (2 2 ) + 24 (2 ) = 20

OB = 20 m [1]

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

14

(c) When t = 5 , s = 5 3 – 9 ( 5 2 ) + 24( 5 ) = 20

When t = 4 , s = 4 3 – 9 ( 4 2 ) + 24( 4 ) = 16 [1]

P t=4 t=2, 5

O B

16 m 4 m

From the diagram, total distance travelled during the first 5 seconds

= 20 + 4 + 4 [1]

= 28 m [1]

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

15

7 (a)

Appendix A

y

0x 2

10

10

20 30

20

30

40

50

x

x

x

x

x

x

10.5( 1 , 12 )

( 20.25 , 40.5 )

47.5

25

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com

16

14 (b)

Appendix B

100

40

90

y

80

70

60

50

30

20

10

O 10 20 30 40 50 60x

70 80

9x + 2y = 200

x = 2y

x + y = 75

y = 30

( 50 , 25 )

R

16 45

http://mathsmozac.blogspot.com

http://sahatmozac.blogspot.com