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SULIT 1449/1

1449/1 [Lihat sebelahSULIT

SULIT

1449/2MATHEMATICS

PAPER 2

AUGUST 20082 HOURS 30 MINUTES

JABATAN PELAJARAN NEGERI SABAH

SIJIL PELAJARAN MALAYSIA TAHUN 2008

EXCEL 2

________________________________________________________________________

MATHEMATICS (MATEMATIK)

PAPER 2 (KERTAS 2)

TWO HOURS THIRTY MINUTES (DUA JAM LIMA TIGA PULUH MINIT)

________________________________________________________________________

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

1. Tulis nombor kad pengenalan dan angka giliran anda pada ruangan yang disediakan.2. Kertas soalan ini adalah Bahasa Ingerís.3. Calon dikehendaki membaca maklumat di halaman 2.

________________________________________________________________________This question paper consists of 27 printed pages.

(Kertas soalan ini terdiri daripada 27 halaman bercetak.)

NAMA : _____________________KELAS : _____________________NO K.P : _____________________A. GILIRAN : _____________________

MOZ@C

SMS MUZAFFAR SYAH , MELAKA

http://tutormansor.wordpress.com/

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2

INFORMATION FOR CANDIDATES

1. This question paper consists of two sections : Section A and Section B..2. Answer all questions in Section A and four questions from Section B.

3. Write your answers clearly in the space provided in the question paper.

4. Show your working. It may help you to get marks.

5. If you wish to change your answer, neatly cross out the answer that you have done. Thenwrite down the new answer.

.6. The diagram in the questions provided are not drawn to scale unless stated.

7. The marks allocated for each question and sub-part of a question are shown in brackets.

8. A list of formulae is provided on pages 3 to 4.

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

10. You may use a non-programmable scientific calculator.

11. This question paper must be handed in at the end of the examination.

The following formulae may be helpful in answering the questions. The symbols given are theones commonly used.

RELATIONS

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SULIT 1449/2


3

1 nmnm aaa

2 nmnm aaa

3 mnnm aa )(

4

ac

bd

bcadA

11

5)(

)()(

Sn

AnAP

6 )(1)'( APAP

7 Distance = 221

221 )()( yyxx

8 Midpoint,

2,

2),( 2121 yyxx

yx

9 Average speed =distance travelled

time taken

10 Mean =sum of data

number of data

11 Mean =sum of (class mark frequency)

sum of frequencies

12 Pythagoras Theorem222 bac

1312

12

xx

yym

14 m =intercept

intercept

y

x

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4

SHAPES AND SPACE

1 Area of trapezium =1

sum of parallel sides height2

2 Circumference of circle = 2d r

3 Area of circle = 2r

4 Curved surface area of cylinder = 2 rh

5 Surface area of sphere = 24 r

6 Volume of right prism = cross sectional area length

7 Volume of cylinder = 2r h

8 Volume of cone = 21

3r h

9 Volume of sphere = 34

3r

10 Volume of right pyramid = 3

1base area height

11 Sum of interior angles of a polygon = 180)2( n

120

arc length angle subtended at centre

circumference of circle 360

130

area of sector angle subtended at centre

area of circle 360

14 Scale factor,PA

PAk

'

15 Area of image = 2k area of object

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5

SECTION A(52 MARKS)

Answer all questions in this section.

1. The Venn diagram in the answer space shows sets X, Y and Z such that the universal set,.X Y Z

On the diagrams in the answer space, shade(a) the set Y Z (b) the set ( ) .X Y Z

[3 marks]

Answer:(a)

(b)

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2. By using factorization, solve the quadratic equation5 2

3n

nn

.

[4 marks]Answer:

3. Calculate the value of x and the value of y which satisfy the following simultaneousequations.

x – 2 y = 141

23

x y

[4 marks]

Answer:

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4. Diagram 1 shows a right prism with a horizontal rectangular base PQRS. PQVU is auniform cross-section of the prism.

Diagram 1

Calculate the angle between the plane TQR and the plane PQRS.[3 marks]

Answer:

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5. In diagram 2, PQ and SR are parallel lines. The equation of the straight line QRis 3y = x + 9.

Diagram 2Find(a) the value of t,(b) the equation of the straight line RS.(c) the y-intercept of RS.

[5 marks]

Answer:(a)

(b)

Q(-3,2)

R(6, t)

x

y

P

-4 S

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9

(c)

6. Diagram 3 shows two sectors OTS and ORQ with centre O. It is given that OQ = 21 cm andOT = 14 cm.

Diagram 3

By using22

7 , calculate

(a) the perimeter of the whole diagram,(b) the area, in cm2, of the shaded region.

[6 marks]

Answer :

(a)

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(b)

7. (a) Combine the following statements with the word “and” or “or” to form a truestatement.

I : 3 is an odd number.II : 14 ( 2) 7

(b) Complete the conclusion in the following argument.Premise 1 : If is less than 180o , then sin is positive.

Premise 2 : is 135o.

Conclusion : ………………………………………………………..

(c) Identify the antecedent and the consequent of the implication below:

“If 090 , then andA B A B are complementary angles.”

[5 marks]

Answer:

(a) …………………………………………………………………………................

……………………………………………………………………………………

(b) Conclusion : ……. ………………………………………………………..

………………………………………………………………

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11

(c) Antecedent : ………………………………………………………………

Consequent : ………………………………………………………………

8. The matrix G =1 3

.2 4

(a) Given that the inverse matrix of G is41

2 12

n

mn

, find the value of m and of n.

(b) Hence, using matrices, calculate the value of p and of q that satisfy the followingequation.

G6

2

p

q

[7 marks]

Answer:

(a)

(b)

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12

9. Diagram 4 shows 10 labeled cards in two boxes.

Box P

Box Q

Diagram 4

A card is picked at random from each of the boxes.By listing the outcomes, find the probability that(a) both cards are labeled with a number,(b) one card is labeled with a number and the other card is labeled with a consonant.

[5 marks]

Answer:

(a)

A 4 B F

G 8 I 9 J K

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13

(b)

10. Diagram 5 shows the speed-time graph of a particle for a period of t seconds.

Speed (ms-1)

22

10

0 4 16 t Time (s)

Diagram 5

(a) State the length of time, in s, when the particle moves with uniform speed.(b) Calculate the rate of change of speed, in ms-2, in the first 4 seconds.(c) Calculate the value of t , if the total distance travelled in t seconds is 306 m.

[6 marks]Answer:(a)

(b)

3

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14

(c)

11. Diagram 6 shows a solid cylinder with hemisphere PQR removed form one end. Both thecylinder and the hemisphere have the same radius of 3 cm.

Diagram 6

Using22

7 , calculate the length, t cm, of the cylinder if the volume of the

remaining solid is6

2827

cm3.

[4 marks]

Answer:

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1449/2

15

Section B[48 marks]

Answer any four questions from this section.

12 Diagram 7 shows quadrilateral JIGH, EFGH and ABCD drawn on aCartesian plane.

12 (a) M is a translation

2

3

coordinates of the image

(i) MN(ii) NM

(b) EFGH is the image of J

2

4

6

8

10

y

x0

-2

2 4 6 8-2-4-6

A B

C

D

E F

G

H J

I

MOZ@C

SMS MUZ

http://tutorm
[Lihat sebelahSULIT
. N is a reflection in the line x = 3. State the

of point H under each transformation below:

[ 4 marks]IGH under a transformation U. ABCD is the

Diagram 7

AFFAR SYAH , MELAKA

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SULIT 1449/2


16

image of EFGH under a transformation V. Describe in full

(i) the transformation U(ii) the transformation V

[ 5 marks]

(c) Given that the area of ABCD is 336 unit2, calculate the area, in unit2.of JIGH.

Answer :(a) (i)

(ii)

(b) (i)

(ii)

U :

V :

(c)

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17

13 (a) Complete Table 1 in the answer space for the equation

754 2 xxy by writing down the values of y when x = –1 and

x = 2.5.

[ 2 marks](b) For this part of the question, use the graph paper provided on page

18. You may use a flexible curve rule.By using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on

the y-axis, draw the graph of 754 2 xxy for .35.2 x

[ 4 marks]

(c) From your graph, find(i) the value of y when x = -1.5,(ii) the value of x when y = 22.

[ 2 marks]

(d) Draw a suitable straight line on your graph to find a value of x which

satisfies the equation 0224 2 xx for .35.2 xState these values of x.

[ 4 marks]Answer :

(a) x –2.5 –2 –1 0 1 2 2.5 3

y 30.5 19 -7 -8 -1 14

TABLE 1

(b) Refer graph on page 18.

(c) (i) y = ……………………………………..

(ii) x = …………………………………….

(d) x = …………………………………………….

Graph for Question 13


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18MOZ@C



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19

14

(a) The table 2 shows the masses, in kilogram, of waste paper collected by40 students.

5 12 7 3 32 17 22 29 35 34

29 4 9 14 20 18 15 23 21 8

26 38 25 13 19 24 27 1 30 21

10 25 27 24 16 11 4 26 23 17

TABLE 2

(a) Based on the data above and by using a size class of 5, complete the tableon page 20.

[ 4 marks]

(b) Based on the table 2 in (a), find the estimated mean for the mass of wastepaper collected by each student.

[ 3 marks]

(c) For this part of the question, use the graph paper provided on page 21.

By using a scale of 2 cm to 5 kg on the x-axis and 2 cm to 1 student onthe y-axis, draw a histogram for the data.

[ 4 marks](d) Based on the histogram in (c), state one information related to the mass of

waste paper collected by the students.[ 1 mark]

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20

Answer :(a)

Mass (g) Frequency Midpoint

1 - 5 5 3

(b)

(c) Refer graph on page 21.

(d)

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21


Graph for Question 14MOZ@C



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22

15 You are not allowed to use graph paper to answer this question.

(a) The diagram shows two prisms joined together at the vertical plane BMQG. Thebase AFGNKB is a rectangle lying on a horizontal table. ABCDE and BKLM arethe uniform cross sections of the respective prisms. AE and BMC are verticaledges of the same length. EJID and IHCD are inclined planes. MLPQ is ahorizontal plane. ED = CD and DI is 6cm vertically above the base of the solid.

Draw to full scale, the plan of the solid. [ 3 marks]

Answer :

B

A

N

P

K

C

D

G

Q

E

I

F

L

4cm

6cm

7cm

3cm

5cm

M

H

J

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23

(a)

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24

(b) A triangular prism is joined to the solid in (a) at the horizontal plane PQML. Thecombined solid is as shown in the diagram below. LMR is the uniform crosssection of the triangular prism. RM = RL and RS is 2 cm vertically above theplane PQML.

Draw to full scale,

(i) The elevation on a vertical plane parallel to ABK as viewed from X. [4 marks]

(ii) The elevation on a vertical plane parallel to KN as viewed from Y. [5 marks]

B

A

N

P

K

C

D

G

Q

E

I

F

L

4cm

6cm

7cm

3cm

5cm

M

Y

X

R

S

H

J

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25

Answer :

(b)(i)

(ii)

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26

16 A (54oS, 5oW), B (54oS, 25oE), C and D are four points on the earth’s surface. AC is adiameter of a parallel of latitude and D is due north of A.

(a) Find the longitude of C.[ 2 marks]

(b) Calculate the shortest distance, in nautical miles, from A to C.[ 3 marks]

(c) Calculate the distance of AB, in nautical miles, measured along the commonparallel of latitude.

[ 3 marks]

(d) A plane took off from A due north at a speed of 550 knots and took 6 hours toreach D. Find the latitude of D.

[ 4 marks]Answer :

(a)

(b)

(c)

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27

(d)

END OF QUESTION PAPER

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1

MARKING SCHEME PAPER 2 – EXCEL 2Question Solution and Mark Scheme Marks

1 (a)

(b)

P1

P2 3

2

23 5 2 0

3 1 ( 2) 0 or equivalent.

2

1or 0.3333

3

n n

n n

n

n

Note:1. Accept without “= 0”2. Do not accept solutions solved not using factorization.

K1

K1

N1

N1 4

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2

3 13 6 or 7 or equivalent

2x y x y

55 20 or 5 or equivalent

6y x

or

1414 2 or or equivalent

2

55 20 or 5 or equivalent

6

xx y y

y x

(K1)

(K1)or

3 2 141

1 1 61(3) ( 2)(1)

6

4

x

y

x

y

(K2)

Note:

3 211.

1 11(3) ( 2)(1)

62. as final answer, award N1

4

x

y

K1

K1

N1N1 4

4 Identify orTRS SRT

0' 13.53853

6

8tan

or

equivalentor

o

P1

K1

N1

3

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3

5(a)

2 1

6 ( 3) 3

t

t = 5

(b)2 0

3 ( 4)RSm

y = 2 x – 7

(c) y-intercept is – 7 or c = - 7.

K1

N1

K1

N1

N1 5

6 (a)135 22 45 22

2 14 or 2 21360 7 360 7

135 22 45 222 14 + 2 21 7 21 14

360 7 360 7

or equivalent

1 18391 or 91.5 or

2 2

(b)

2

2

45 22 121 or 14 14

360 7 2

45 22 121 14 14

360 7 2

1 30175 or 75.25 or

4 4

Note:1. Accept for K mark.2. Correct answer from incomplete working, award KK2.

K1

K1

N1

K1

K1

N1 6

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4

7 (a) 3 is an odd number or 14 ( 2) 7 .

(b) sin 135o is positive.

(c) A + B = 90o

A and B are complementary angles

P2

K1

P1P1 5

8 (a) n = 3

1 (4) – ( - 3)(2) = m(3) – 2

m = 4(b)

4 3 61

2 1 21(4) ( 3)(2)

3

1

3

1

p

q

p

q

p

q

Note:

1.

*p 6

q 2

inverse

matrix

or

4 31seen, award K1.

2 11(4) ( 3)(2)

2. Do not accept

* *1 3 1 0

or .2 4 0 1

inverse inverse

matrix matrix

3.3

as final answer, award N11

p

q

4. Do not accept any solutions solve not using matrices.

P1

K1

N1

K2

N1

N17

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5

9 (a) (4,8), (4,9)

2 1or

24 12

(b) (4, G), (4, J), (4, K)(B, 8), (B, 9), (F, 8), (F, 9)

Note: Any two groupings correct, award K1.

7

24

K1

N1

K2

N1 5

10 (a) 12s

(b)

2 2

10 3

4

7or 1.75

4ms ms

(c)1 1

(3 10)4 (16 4)10 (10 22)( 16) 3062 2

26

t

t s

P1

K1

N1

K2

N1 6

11

2

3

2 3

223

7

1 4 223

2 3 7

22 1 4 22 63 3 282

7 2 3 7 7

12

t

t

cm

Note:1. Accept for K mark.2. Correct answer from incomplete working, award KK2.

K1

K1

K1

N14

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6

Question Solution and Mark Scheme Marks12(a) (i) (7,5)

Note :Award P1 if coordinate (7,5) marked on diagram orcoordinate (4,7) indicated in answer space.

[P2]

(ii) (1,5)

Note :Award P1 if coordinate (1,5) marked on diagram orcoordinate (5,5) indicated in answer space.

[P2] 4

(b) (i)

(ii)

U : Reflection in the line GH or y = 9-x

Note :1. Award P1, if the word “reflection” seen.

V : Enlargement at centre (9,4) with scale factor of 2.

Note :

1. Award P2, if “Enlargement at centre (9,4)” or“Enlargement scale factor of 2” seen.

2. Award P1, if the word “Enlargement” seen.

[P2]

[P3] 5

(c) Area of image = scale factor2 x Area of object336 = 22 x Area of object

Area of object = 336 ÷ 4= 84 unit2.

[K2]

[N1] 3

Jumlah 12

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7

Question Solution and Mark Scheme Marks13(a) 2

5.5

[K1]

[K1] 2

(b) Graph

1. Axes drawn in correct direction with uniformscales in-2.5≤ x ≤ 3 and -8≤ y ≤ 30.5

2. All points correctly plotted or curve passesthrough these points for -2.5≤ x ≤ 3

3. Smooth and continuous curve without anystraight line and passes through all correctpoints

Note :1. 6 or 7 points correctly plotted, award K1

[P1]

[K2]

[N1] 4

(c) (i) 9 ≤ y ≤ 10 [P1]

(ii) -2.2≤ x ≤ -2.1 [P1] 2

(d)y = 4x2 – 5x – 7

- 0 = 4x2 + 2x – 2y = – 7x – 5

Straight line y = – 7x – 5 correctly drawn

0.4≤ x ≤ 0.6-1.1≤ x ≤ -0.9

Note :1. Allow P mark or N mark if values of y and x

shown on graph.2. Values of y and x obtained by computations,

award P0 or N0.

[K1]

[K1]

[N1][N1] 4

Jumlah 12

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×2Gr

×

--2-3

×

×

SMS MUZ

http://tutorm
8
aph for Question 13


11 0

5

10

15

20

25

30

-5

-10

×

××

×

×

X

Y

y = 4x2-5x-7

y =

MOZ@C

AFFAR SYAH , MELAKA

ansor.wordpress.com

×
2
1

-7x-5

/

31

9

Question Solution and Mark Scheme Marks14(a) Mass

(kg)Frequency Mid Point

1-5 5 3 I

6-10 4 8 II

11-15 5 13 III

16-20 6 18 IV

21-25 9 23 V

26-30 7 28 VI

31-35 3 33 VII

36-40 1 38 VIII

Class interval : II - VIIIFrequency : II – VIIIMidpoint : II - VIII

Note :Allow 1 mistake in Frequency for P1

[P1][P2][P1]

4

(b) (3x5)+(8x4)+(13x5)+(18x6)+(23x9)+(28x7)+(33x3)+(38x1)40

76040

= 19Note :

1. Allow 2 mistakes for K1Award KK2, if incomplete working shown.

[K2]

[N1]

3(c) Graph

1. Axes drawn in correct direction with uniform scalesfor0.5≤ x ≤ 40.5 and 0≤ y ≤ 9Accept midpoint/class interval for the horizontalaxis.

2. All bar correctly drawnNote :6 or 7 bar correctly drawn, award K1

[P2]

[K2]

4

(d) The modal class is 21-25 [P1] 1

Jumlah 12

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4

5

6

7

Frequency

8

9

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10

5.5 105 15.5 20.5 25.5 30.5 35.5 40.5

1

2

3

Mass (kg)

0.5



11

Question Solution and Mark Scheme Marks15(a)

J /F I H/Q/G P/ N

6cm

E/A D C/M/ B L/ K2cm 2cm 3 cm

Correct shape with rectangles EDIJ, DCHI and CLPH.All solid lines.

ED=DC, EL>JE>CL>ED

Measurement correct to ± 0.2 cm (one way) and all anglesat the vertices of rectangles = 90o ± 1o.

[K1]

[K1] depK1

[N1] depK1K1

3

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12

Question Solution and Mark Scheme Marks15(b)(i)

E C

D

6cm

5cm

R7cm

M

2cm

L

4 cm

A 4cm B 3 cm K

Correct shape with pentagon ABCDE and BKLRM.All solid lines.

AE>CM>AB>CD>BK >RLBK=LK

Measurement correct to ± 0.2 cm (one way) and angle A,angle B and angle K= 90o ± 1o.

[K1]

[K1] depK1

[N2] depK1K1 4

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13

Question Solution and Mark Scheme Marks15(b)(ii)

C H

1cmD I

2cm

R S2 cm

L P2cm

K N6cm

Correct shape with rectangle KLPN, RLPS and CRSHAll solid lines.

Note :Ignore dotted line DI

D and I joined with dotted line to form rectangle CDIH

CK>KN, IS>PN>PS >IH

Measurement correct to ± 0.2 cm (one way) and all angle atthe vertices of rectangles = 90o ± 1o.

[K1]

[K1]dep K1

[K1]dep

K1K1

[N2]dep

K1K1K1

5

Jumlah 12

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14

Question Solution and Mark Scheme Marks16(a) 175oE

Note :a. 175o or E, award P1

[P2]

2

(b) (180o-54o-54o)x60=72 x 60=4320 n.m.

[K2]

[N1]

3

(c) (5+25) x 60 x Cos 54o

= 1058 n.m

Note:Award K1, if

a. (5+25) orb. Cos 54o used correctly.

[K2]

[N1]

3

(d) 550 = Distance6

Distance = 550 x 6= 3300 n.m

330060

=55o

55o-54o

=1oN

[K2]

[K1]

[K1]

4

Jumla

h

12

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