math paper 2 f4 final 07

32
SEKTOR SEKOLAH BERASRAMA PENUH BAHAGIAN SEKOLAH KEMENTERIAN PELAJARAN MALAYSIA PEPERIKSAAN DIAGNOSTIK TINGKATAN EMPAT 2007 Pemeriksa Bahagian Soalan Markah Penuh Markah Diperoleh 1 4 MATEMATIK Kertas 2 Dua jam tiga puluh minit 2 4 3 5 4 4 5 3 6 4 7 6 8 6 9 4 10 7 A 11 5 12 12 13 12 14 12 15 12 B 16 12 JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU 1. Kertas soalan ini mengandungi dua bahagian : Bahagian A dan Bahagian B. Jawab semua soalan daripada Bahagian A dan empat soalan dalam Bahagian B. 2. Jawapan hendaklah ditulis dengan jelas dalam ruang yang disediakan dalam kertas soalan. Tunjukkan langkah-langkah penting. Ini boleh membantu anda untuk mendapatkan markah. 3. Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan. 4. Satu senarai rumus disediakan di halaman 2&3. 5. Anda dibenarkan menggunakan kalkulator saintifik yang tidak boleh diprogram. Jumlah Kertas soalan ini mengandungi 24 halaman bercetak. SULIT 1449/2 Matematik Kertas 2 Oktober 2007 2 1 2 jam 1449/2 1449/2 2007 Hak Cipta Sektor SBP [Lihat sebelah SULIT NAMA : TINGKATAN :

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Page 1: Math Paper 2 f4 Final 07

SEKTOR SEKBA

KEMENTER

PEPERTING

MATEMATIK

Kertas 2

Dua jam tiga puluh minit

JANGAN BUKA KERTAS SOALASEHINGGA DIBERITAHU

1. Kertas soalan ini mengandungi dua baBahagian A dan Bahagian B. Jawabsoalan daripada Bahagian A dan empadalam Bahagian B.

2. Jawapan hendaklah ditulis dengan jelaruang yang disediakan dalam kertasTunjukkan langkah-langkah penting. Imembantu anda untuk mendapatkan mar

3. Rajah yang mengiringi soalan tidakmengikut skala kecuali dinyatakan.

4. Satu senarai rumus disediakan di halam

2 & 3.

5. Anda dibenarkan menggunakan kasaintifik yang tidak boleh diprogram.

Kertas soalan in

SULIT1449/2MatematikKertas 2Oktober2007

2

12 jam

1449/2

1449/2 2007 Hak Cipta Sektor SBP

NAMA :

TINGKATAN :

OLAH BERASRAMHAGIAN SEKOLAHIAN PELAJARAN M

IKSAAN DIAGNOSTKATAN EMPAT 200

Peme

Bahagian

A

B

N INI

hagian :semua

t soalan

s dalamsoalan.

ni bolehkah.

dilukis

an

lkulator

i mengandungi 24 halama

A PENUH

ALAYSIA

IK7

riksa

SoalanMarkahPenuh

MarkahDiperoleh

1 4

2 4

3 5

4 4

5 3

6 4

7 6

8 6

9 4

10 7

11 5

12 12

13 12

14 12

15 12

16 12

Jumlah

n bercetak.

[Lihat sebelahSULIT

Page 2: Math Paper 2 f4 Final 07

SULIT 1449/2

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MATHEMATICAL FORMULAE

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

RELATIONS

1 am x an = a m+ n

2 am an = a m – n

3 ( am )n = a mn

4 A-1 =bcad

1

ac

bd

5 P ( A ) =)(

)(

Sn

An

6 P ( A ) = 1 P(A)

7 Distance = 2 21 2 1 2( ) ( )x x y y

8 Midpoint, ( x, y ) =

2,

22121 yyxx

9 Average speed =

10 Mean =

11 Mean =

12 Pythagoras Theoremc2 = a2 + b2

13 m =12

12

xx

yy

14-intercept

-intercept

ym

x

distance travelledtime taken

sum of datanumber of data

sum of (class mark × frequency)sum of frequencies

Page 3: Math Paper 2 f4 Final 07

SULIT 1449/2

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UntukKegunaanPemeriksaSHAPES AND SPACE

1 Area of trapezium =2

1× sum of parallel sides × height

2 Circumference of circle = d = 2r

3 Area of circle = r2

4 Curved surface area of cylinder = 2rh

5 Surface area of sphere = 4r2

6 Volume of right prism = cross sectional area × length

7 Volume of cylinder = r2h

8 Volume of cone =3

1r2h

9 Volume of sphere =3

4r3

10 Volume of right pyramid =3

1× base area× height

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

12arc length angle subtended at centre

circumference of circle 360

13area of sector angle subtended at centre

area of circle 360

14 Scale factor , k =PA

PA'

15 Area of image = k 2 × area of object

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Section A[52 marks]

Answer all questions in this section.

1 Solve the quadratic equation3

3

4

3

x

xx

[4 marks]

Answer :

2 Calculate the value of p and of q that satisfy the following simultaneous linear equations:

394

632

qp

qp

[4 marks]

Answer :

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3 (a) State whether the following statement is true or false.

63and326 2

(b) Write down two implications based on the sentence below.

4m 20 if and only if m 5

(c) Given the number sequence 5, 18, 39, 68, … which follows the pattern

5 = 1 + 4(12)

18 = 2 + 4(22)

39 = 3 + 4(32)

68 = 4 + 4(42)

……………..……………..

Form a general conclusion by using the induction method for the numerical sequenceabove.

[5 marks]

Answer :

(a)

(b) Implication 1 : _________________________________________________

Implication 2 : _________________________________________________

(c)

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4 Diagram 1 shows a combined solid consists of a cuboid JKLMPQRS and half cylinder

JMNTSP.

Using =7

22, calculate the volume, in cm3, of the solid

[4 marks]

Answers:

J K

LM

P Q

RS

T

N

8 cm7 cm

16 cm

DIAGRAM 1

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5 The Venn diagram in the answer space shows sets P, Q and R such that the universal set,

= P Q R.

On the diagrams in the answer space, shade

(a) P R

(b) Q R P

[3 marks]

Answer :

(a)

(b)

P

QR

P

QR

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6 Diagram 2 shows a cuboid with horizontal rectangular base, ABCD.

Calculate the angle between the plane ABE and the base ABCD.

[4 marks]Answer :

A B

CD

E

F G

H

DIAGRAM 2

10 cm

8 cm

6 cm

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7 In Diagram 3, O is the origin. KL, PQ and RS are straight lines. PQ is parallel to RS.The equation of the straight line KL is 62 xy .

Find the

(a) coordinate of P,

(b) gradient of the straight line PQ,

(c) equation of the straight line RS.

Answer:

(a)

(b)

(c)

x

K R

S

DIA

y

P

Q(3, 3)

O

9

GRAM 3

[Lihat sebelahSULIT

[6 marks]

L

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8 (a) Identify the antecedent and consequent in the following implication.

‘ If a triangle has two equal sides, then it is an isosceles triangle.’

(b) State the converse of the following implication.

If 5x , then x2 > 25

(c) Complete the premise in the following argument.

Premise 1 : If x is an angle in a semicircle, then x = 90°

Premise 2 : _________________________________________________

Conclusion : x = 90°

(d) Complete the following argument

Premise 1 : _________________________________________________

Premise 2 : M N ≠ M

Conclusion : M N

[6 marks]

Answer:

(a) Antecedent : …………………………………………………………………

Consequent : ………………………………………………………………….

(b) …………………………………………………………………………………

(c) Premise 2 : ……………………………………………………………………

(d) Premise 1 : …………………………………………………………………….

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9 Ten cards are placed in an empty box.

(a) If a card is selected at random from the box, state the probability that the card marked

C is selected.

(b) A number of cards marked C are added into the box. If a card is selected at random

from the box, the probability of selecting a card marked C is7

3.

Find the number of cards that has been added into the box.

[4 marks]Answer :

(a)

(b)

E X C E L L E N C E

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10 In Diagram 4, ABOD is a square and PQR is a semicircle with centre O.

OR = 7 cm. Using =7

22, calculate

(a) the perimeter, in cm, of the whole diagram,

(b) the area, in cm2, of the shaded region.

[7 marks]

Answer :

(a)

(b)

A B

D O

P

Q

R

DIAGRAM 4

14 cm

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11 Diagram 5 shows a right pyramid with rectangular base PQRS .

The apex V is 5 cm vertically above the point T.

Calculate the angle between the line VP and the base PQRS.[5 marks]

Answer:

R

8 cm

V

P Q

S

6 cmT

DIAGRAM 5

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Section B

[48 marks]

Answer any four questions in this section.

12 Table 1 in the answer space shows the distribution of the ages of 200 participants in a bigwalk event.

(a) Using the data in Table 1, complete the table provided in the answer space.[4 marks]

(b) By using a scale of 2 cm to 5 years on the x-axis and 2 cm to 20 participants on

the y-axis , draw an ogive for the data.[5 marks]

(c) Based on your ogive in (b),

(i) find the inter quartile,

(ii) explain briefly the meaning of the third quartile.

[3 marks]

Answer:

(a)Age (years) Frequency Cumulative

FrequencyUpper

Boundary

15 19 10

20 24 20

25 29 50

30 34 60

35 39 36

40 44 18

45 49 6

(b) Refer graph on page 15.

(c) (i)

(ii)

TABLE 1

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Graph for Question 12

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13 (a) Diagram 6 shows sectors OTQR, OABC and OCP with the same centre O.TAOP and OCR are straight lines.

OR = 2OC , OC =2

21cm and COP = 60.

Using =7

22, calculate

(i) the perimeter, in cm, of the whole diagram,[3 marks]

(ii) the area, in cm2, of the shaded region.[4 marks]

(b) Diagram 7 shows some number cards.

A card is picked at random. State the probability of choosing a card containing

(i) digit 5,

(ii) a multiple of 5,

(iii) a number such that when divided by 5, the remainder is 1.

[5 marks]

T A

BC

QR

PO

DIAGRAM 6

45 46 47 48 49 50 51 52

53 54 55 56 57 58 59 60

DIAGRAM 7

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Answer:

13 (a) (i)

(ii)

(b) (i)

(ii)

(iii)

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14 The data in Diagram 8 show the marks of 40 students for the Mathematics monthly test.

46 53 44 60 42 38 31 55

35 37 54 32 46 56 40 60

52 40 34 45 52 35 50 36

47 38 40 48 45 42 53 44

50 44 58 51 36 48 56 32

(a) Using the data in Diagram 8 and a class interval of 5 marks, complete Table 2

in the answer space.

[4 marks]

(b) Based on Table 2 in (a),

(i) state the modal class,

(ii) calculate the mean mark of the Mathematics monthly test and give your answer

correct to 2 decimal places.

[4 marks]

(c) For this part of the question, use the graph paper provided on page 19.

By using a scale of 2 cm to 5 marks on the horizontal axis and 2 cm to 1 student on the

vertical axis, draw a histogram for the data.

[4 marks]

DIAGRAM 8

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Answer:

14 (a)

Class interval Midpoint Frequency

31 – 35

36 - 40

(b) (i)

(ii)

(c) Refer to the graph on page 20.

TABLE 2

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Graph for Question 14

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15 (a) Diagram 9 is a Venn diagram showing the number of students in a class who play at least

one of the three games. Given the universal set, = F R H.

Given that set F = { students who play football }

set R = { students who play rugby } and

set H = { students who play hockey}.

(i) Find the number of students who play hockey.

(ii) If the number of students who play football is 30, find

(a) the number of students who play rugby,

(b) the number of students who play one game only.

[6 marks]

Answer:

(a) (i)

(ii) (a)

(b)

R

F

H

x

72x

5

18 – 2x10 + x 10

DIAGRAM 9

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15 (b) In Diagram 10 , TQRS, NR and OQ are straight lines drawn on a Cartesan plane. O is the

origin and OQ is parallel to NR. The gradient of NR = 2.

Find

(i) the value of p,

(ii) the coordinate of R,

(iii) the equation of the straight line ST.

[6 marks]

Answer:

(b) (i)

(ii)

(iii)

y

x

R

N(-6, 0)

Q(p, 3)

T

0

DIAGRAM 10

S

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16 (a) Solve the equation 15

32 2

w

w.

[4 marks](b) Calculate the values of x and y that satisfy the following simultaneous linear

equations:2x + y = 34x - 3y = 11

[4 marks]

(c) (i) Form a true compound statement by combining the two statementsgiven below.

(a) 52 = 10

(b)4

1= 0.25

(ii) Form a general conclusion by induction based on the numerical sequencebelow.

2, 9, 16, 23 …

2 = 2 + 7 (0)9 = 2 + 7 (1)16 = 2 + 7 (2)23 = 2 + 7 (3)……………….

(iii) Complete the following sentence using a suitable quantifier to make it a truestatement.

‘………………. prime numbers are odd numbers.’

[4 marks]

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Answer:

16 (a)

(b)

(c) (i)

(ii)

(iii)

Page 25: Math Paper 2 f4 Final 07

SULIT 1449/21449/2MatematikKertas2PeraturanPemarkahanOktober2007

SEKT

KEME

Peraturan

1449/2

OR SEKOLAH BERASRAMA PENUHBAHAGIAN SEKOLAH

NTERIAN PELAJARAN MALAYSIA

PEPERIKSAAN DIAGNOSTIKTINGKATAN 4 2007

MATEMATIK

Kertas 2

PERATURAN PEMARKAHAN

pemarkahan ini mengandungi 10 halaman.

[Lihat sebelahSULIT

Page 26: Math Paper 2 f4 Final 07

Section A[ 52 marks ]

No Marking Scheme Marks

1

3,3

4

0)3)(43(

012`53 2

xx

xx

xx K1

K1

N1N1 4

2

2

3

1

15101515

18961264

p

q

porq

qporqp K1

K1

N1

N1 4

3(a) False

(b) Implication 1 : If 4m 20, then m 5Implication 2 : If m 5, then 4m 20

(c) n + 4n2 , n = 1,2, 3, ….

P1

P1P1

P1, P1 5

4 8 x 7 x 16

162

7

2

7

7

22

2

1

8 x 7 x 16 + 162

7

2

7

7

22

2

1

1204

K1

K1

K1

N1 4

5 (a) PR (b) QRP

6

2

Identify EAD

Tan EAD =8

6

EAD = 36.87 or 3652

P1 P2

3

P1

K2

N1 4

Page 27: Math Paper 2 f4 Final 07

3

No Marking Scheme Marks

7 (a) P(3,0)

(b) mPQ =30

33

= 2

(c) mRS = 2

20

6

x

yor equivalent

y = 2x + 6

P1

K2

N1

K1

K1

N1 6

8 (a) Antecedent : A triangle has two equal sidesConsequent: It is an isoceles triangle

(b) If2x > 25, then x > 5

(c) x is an angle in a semicircle

(d) If M N, then MN = M

P1P1

P1

P1

P1 7

9 (a)5

1

(b)7

3=

x

x

10

2

x = 4

P1

K2

N1 4

10(a) 7

7

222

2

1

14(4) + 77

222

2

1

78

(b) 777

22

2

1

14147

22

4

1

777

22

2

1 + ( 14 X 14 1414

7

22

4

1 )

119

K1

K1

N1

K1

K1

K1

N1

7

Page 28: Math Paper 2 f4 Final 07

4

No Marking Scheme Marks

11 VPR or VPT

22 86 ( seen )

Tan VPR =5

5

VPR = 45°

P1

K1

K2

N15

Section B[48 marks]

No Marking Scheme Marks

12 (a)

Age (years) Frequency Cumulative frequency Upper boundary

10 – 14 0( I )

0( II )14.5

15 - 19 10 10 19.520 – 24 20 30 24.525 – 29 50 80 29.530 – 34 60 140 34.535 - 39 36 176 39.540 - 44 18 194 44.545 - 49 6 200 49.5

All values in Column ( I ) correct excluding Row I correct.All values in Column ( II ) correct excluding Row I correct.

(b) x-axis is drawn with the right direction and in uniform scale from14.5 ≤ x ≤ 49.5

y-axis is drawn with the right direction and in uniform scale from0 ≤ y ≤ 200

All eight points* plotted correctly.Note : Seven or six points* plotted correctly. 1 P1

(14.5, 0) plotted or passed throughAll the right eight points plotted correctly and ogive is drawn smoothly passingthrough all the points.

(c) (i) 35.5 – 27.58 – 9

(ii) 150 participants aged 35.5 0.5

P1P2

K1

K1

K2

K1N1

K1N1

N1 12

13(a)

360

120x 2 x

7

22x 21

360

60x 2 x

7

22x

2

21

360

120x 2 x

7

22x 21 + 21 + 21 +

360

60x 2 x

7

22x

2

21

97

K1

K1

K1

N1

Page 29: Math Paper 2 f4 Final 07

5

No Marking Scheme Marks

(b)360

120x

7

22x 21 x 21

360

120x

7

22x

2

21x

2

21

360

120x

7

22x 21 x 21

360

120x

7

22x

2

21x

2

21+

360

60x

7

22x

2

21x

2

21

4044

1

(b) (i)16

11

(ii)4

1

(iii)16

3

K1

K1

K1

N1

P1

P1

P212

14 (a)

Class interval Midpoint Frequency

31 – 35 33 6

36 - 40 38 5

41 – 45 43 10

46 – 50 48 6

51 – 55 53 8

56 - 60 58 5

Column 1Column 2Column 3

(b) (i) 41 – 45

(ii)5861056

)5(58)8(53)6(48)10(43)5(38)6(33

45.5

(c) Refer to the graphAxes drawn in the correct direction , uniform scale for 5.605.30 x and

100 y .

Horizontal axis labeled using midpoint / boundary / class interval.6 bars drawn correctly

P1P1P2

P1

K2

N1

K1

K1N2 12

15 (a) (i) 2 x + 10 + 5 + 18 - 2 x33

(ii) 7 + 10 + 12 + 1645

(iii) 6 + 6 +1628

K1N1

K1N1

K1N1

Page 30: Math Paper 2 f4 Final 07

6

No Marking Scheme Marks

(b) (i)0

03

p= 2

p =2

3

(ii) 2)6(0

0

y

R (0,12)

(iii) m ST = 6

126 xy

K1

N1

K1

N1

K1N1

12

16a 2

2w - 5 w - 3 = 0

(2 w +1)( w -3) =0

w = -2

1, 3

b 4 x + 2 y = 6 or equivalent

5 y = - 5

y = -1

x = 2

c i orii 2+7n, n = 0,1,2,3,…

iii some

K1K1

N1N1

K1K1N1N1

P1P2P1 12

Page 31: Math Paper 2 f4 Final 07

7

Graph for number 12

20

Cu

mu

lati

ve

Fre

qu

ency

40

60

80

100

120

140

160

180

200

14.5 19.5 24.5 29.5 34.5 39.5 44.5

Ages49.5

Page 32: Math Paper 2 f4 Final 07

8

Graph for number 14

0 33 38 43 48 53 Marks

1

2

3

4

5

6

7

8

9Nu

mb

erof

stu

den

ts

10

58