sbp f4 p2 final exam 2007

17
SULIT 3472/2 [ Lihat sebelah 3472/2 SULIT 1 SEKTOR SEKOLAH BERASRAMA PENUH KEMENTERIAN PELAJARAN MALAYSIA PEPERIKSAAN AKHIR TAHUN TINGKATAN 4, 2007 ADDITIONAL MATHEMATICS Form Four Paper 2 Two hours and thirty minutes DO NOT OPEN THIS QUESTION PAPER UNTIL INSTRUCTED TO DO SO This question paper consists of 9 printed pages 3472/2 Form Four Additional Mathematics Paper 2 Oct 2007 2 ½ hours INFORMATION FOR CANDIDATES 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. 4 Show your working. It may help you to get marks. 5 The diagrams in the questions provided are not drawn to scale unless stated. 6 The marks allocated for each question and sub-part of a question are shown in brackets. 7 A list of formulae is provided on pages 2 and 3. 8 You may use a four-figure mathematical table. 9 You may use a non-programmable scientific calculator.

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Page 1: SBP F4 P2 Final Exam 2007

SULIT 3472/2

3

1

SEKTOR SEKOLAH BERASRAMA PENUHKEMENTERIAN PELAJARAN MALAYSIA

PEPERIKSAAN AKHIR TAHUN TINGKATAN 4, 2007

ADDITIONAL MATHEMATICS

Form FourPaper 2

Two hours and thirty minutes

DO NOT OPEN THIS QUESTION PAPER UNTIL INSTRUCTED TO DO SO

3472/2Form FourAdditional MathematicsPaper 2Oct 20072 ½ hours

INFORMATION FOR CANDIDATES

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.

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

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

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

7 A list of formulae is provided on pages 2 and 3.

8 You may use a four-figure mathematical table.

9 You may use a non-programmable scientific calculator.

[ Lihat sebelah472/2 SULIT

This question paper consists of 9 printed pages

Page 2: SBP F4 P2 Final Exam 2007

SULIT 3472/2

3472/2 SULIT

2

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

ALGEBRA

12 4

2

b b acx

a

2 am an = a m + n

3 am an = a m - n

4 (am) n = a nm

5 loga mn = log am + loga n

6 logan

m= log am - loga n

7 log a mn = n log a m

8 logab =a

b

c

c

log

log

9 Tn = a + (n-1)d

10 Sn = ])1(2[2

dnan

11 Tn = ar n-1

12 Sn =r

ra

r

ra nn

1

)1(

1

)1(, (r 1)

13r

aS

1, r <1

CALCULUS

1 y = uv ,dx

duv

dx

dvu

dx

dy

2v

uy ,

2vdx

dvu

dx

duv

dy

dx

,

3dx

du

du

dy

dx

dy

4. Area under a curve

= b

a

y dx or

= b

a

x dy

5 Volume generated

= b

a

y 2 dx or

= b

a

x 2 dy

5 A point dividing a segment of a line

( x,y) = ,21

nm

mxnx

nm

myny 21

6 Area of triangle

= )()(2

1312312133221 1

yxyxyxyxyxyx

1 Distance = 221

221 )()( yyxx

2 Midpoint

(x , y) =

221 xx

,

221 yy

3 22 yxr

42 2

ˆxi yj

rx y

GEOMETRY

Page 3: SBP F4 P2 Final Exam 2007

SULIT 3472/2

3

STATISTICS

TRIGONOMETRY

9 sin (A B) = sinA cosB cosA sinB

10 cos (A B) = cosA cosB sinA sinB

11 tan (A B) =BA

BA

tantan1

tantan

12C

c

B

b

A

a

sinsinsin

13 a2 = b2 + c2 - 2bc cos A

14 Area of triangle = Cabsin2

1

6 1

0

100Q

IQ

14 z =

x

3

1 Arc length, s = r

2 Area of sector , L = 21

2r

3 sin 2A + cos 2A = 1

4 sec2A = 1 + tan2A

5 cosec2 A = 1 + cot2 A

6 sin 2A = 2 sinA cosA

7 cos 2A = cos2A – sin2 A= 2 cos2A - 1= 1 - 2 sin2A

8 tan 2A =A

A2tan1

tan2

472/2

1 x =N

x

2 x =

f

fx

3 =N

xx 2)(= 2

2

xN

x

4 =

f

xxf 2)(= 2

2

xf

fx

5 m = Cf

FNL

m

2

1

71

11

w

IwI

8)!(

!

rn

nPr

n

9!)!(

!

rrn

nCr

n

10 P(AB)=P(A)+P(B)-P(AB)

11 p (X=r) = rnrr

n qpC , p + q = 1

12 Mean µ = np

13 npq

[ Lihat sebelahSULIT

Page 4: SBP F4 P2 Final Exam 2007

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Section A

[40 marks]

Answer all questions in this section.

1 Solve the simultaneous equations 3x + y = 5 and x2 – 2y + 5 = 0.Give your answer correct to three decimal places.

[5 marks]

2 The function f(x) = 2x2 + 6x – 5 has a minimum point at ( h , k ).

(a) By using the method of completing the square, express f(x) in the formf(x) = a( x + p )2 + q , where a, p and q are constants.

[3 marks](b) State the value of h and of k.

[2 marks]

(c) Find the range of values of t if 2x2 + 6x – 5 = t , where t is a constant,has no roots.

[2 marks]3 Solution to this question by scale drawing will not be accepted.

Diagram 1 shows the straight lines AN and AL.

Given that the equation of the straight line AN is 042 xy and LAN=900.

(a) Find the equation of the straight line AL.[4 marks]

(b) The straight line LA is extended until it intersects the y-axis at point B such thatLA : AB = 1 : 2 . Find the coordinates of point L.

[3 marks]

042 xy

DIAGRAM 1

y

x

O

A

L

N

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4 Table 1 shows the scores obtained by a group of 50 students in a competition.

Score Number of students

10 8

20 x

30 9

40 6

50 y

60 6

TABLE 1

(a) Calculate the value of x and of y, given the mean of the scores is 31.

[4 marks]

(b) Hence, determine the variance of the students’ score.[3 marks]

5 Diagram 2 shows the triangle AOB and the sector of a circle , AOC with thecenter O.

DIAGRAM 2

It is given that the radius of the sector is 12 cm and AOC 0.52 rad.[ Use = 3.142 ].

Find(a) the length of the arc AC ,

[2 marks](b) the area of the shaded region.

[5 marks]

B

C

OA12

0.52 rad

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6 Given that a curve y = f(x) with the gradient function,13

x

x

dx

dy.

(a) Find2

2

dx

yd.

[3 marks]

(b) The straight line 2y = kx – 7 is normal to the curve y = f(x) at x = 1, findthe value of k .

[4 marks]

Section B

[40 marks]

Answer four questions from this section.

7 Given that the function nxxg 3: and3

22:1 mxxg , where n and m are

constants.

Find

(a) the value of m and of n, [3 marks]

(b) g -1 (8) , [2 marks]

(c) g 2 (-2) , [2 marks]

(d) the value of t if g(2t) = gg -1( 3 – 2t ). [3 marks]

8 (a) Solve the equation 3 x + 1 = 7[ 2 marks]

(b) Given that x = a m and y = a n , show that loga

2

3

ay

x= nm 21

3

1 .

[3 marks]

(c) The relation between the variable x and y is given by the equation

(3 y ) ( 9 2x ) = 1 and )2(log1log 33 xy .

Find the value of x and of y which satisfy both the equations.[5 marks]

Page 7: SBP F4 P2 Final Exam 2007

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9 Diagram 3 shows a trapezium OABC.

DIAGRAM 3

The equation of the straight line AB is 123 xy and the equation of the straight line

BC is 1253 xy .

(a) Find(i) the coordinates of B ,

(ii) the area of the trapezium OABC,

(iii) the equation of the straight line that passes through point C and perpendicularto the straight line AB.

[7 marks]

(b) The straight line AC is extended to point D ( 5 , k ) such that AC = m CD.Find the value of k and of m.

[3 marks]

10 It is given that 523

2 23 xxy is an equation of a curve.

Find

(a)dx

dy, [2 marks]

(b) the turning point of the curve,[4 marks]

(c) the rate of change of y at ( 1, 4) when the rate of change of x is 3 units per second,

[2 marks](d) the approximate change in y, if x decreases from 1.0 to 0.95 .

[2 marks]

x

y

B

C

O

(3,1)

A(0,4)

Page 8: SBP F4 P2 Final Exam 2007

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11 Table 2 shows the frequency distribution of the marks obtained by 40 pupils in atest.

Marks Frequency1 – 10 611 – 20 921 – 30 1431 – 40 741 – 50 4

TABLE 2

(a) Use the graph paper to answer this question.

Using a scale of 2 cm to 10 marks on the horizontal axis and 2 cm to 2 pupils onthe vertical axis, draw a histogram to represent the frequency distribution of themarks. Find the mode mark.

[4 marks](b) Without using an ogive, calculate the median mark.

[3 marks](c) Determine the standard deviation of the marks of the pupils .

[3 marks]

Section C

[20 marks]Answer two questions from this section.

12. Table 3 shows the prices and weightages of four types of items A, B , C and D.

Price(RM) Price(RM)Item

Year 2003 Year 2005Price Index Weightage

A 7.00 8.40 w 100B 13.50 x 130 80C y 13.00 115 70D 11.00 12.10 110 z

TABLE 3(a) Calculate the value of w, x and y.

[4 marks]

(b) The composite index of these items for the year 2005 based on the year 2003is 120.Calculate the value of z.

[3 marks]

(c) The total cost of all these items is expected to increase by 20% from the year2005 to the year 2007. Find the expected composite index for the year 2007based on the year 2003.

[3 marks]

Page 9: SBP F4 P2 Final Exam 2007

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13. Diagram 4 shows two triangles ABC and CDE. The two triangles are joined at C suchthat AE and BD are straight lines. The CED is an obtuse angle.

DIAGRAM 4

(a) Calculate

(i) ACB ,

(ii) DEC .[5 marks]

(b) The straight line CE is extended to F such that DE = DF.Find the area of triangle CDF.

[5 marks]

END OF THE QUESTION PAPER

A

4 cm 7 cm

B 5 cm C D

E

9 cm

6.5 cm

Page 10: SBP F4 P2 Final Exam 2007

PEPERIKSAAN DIAGNOSTIK ADDITIONAL MATHEMATICS - 1 -SBP 2007 Paper 2

SBP 2007

SULIT3472/2AdditionalMathematicsPaper 2Okt2007

SEKOLAH BERASRAMA PENUHKEMENTERIAN PELAJARAN MALAYSIA

PEPERIKSAAN AKHIR TAHUNTINGKATAN 4

2007

ADDITIONAL MATHEMATICS

PAPER 2

MARKING SCHEME

This marking scheme consists of 8 printed pages

Page 11: SBP F4 P2 Final Exam 2007

PEPERIKSAAN DIAGNOSTIK ADDITIONAL MATHEMATICS - 2 -SBP 2007 Paper 2

SBP 2007

NO SOLUTION AND MARK SCHEME MARKSSUB

MARKS TOTAL

y = 5 – 3x or2

5

3

5 2

xyor

yx

P1

05)35(22 xx or

52

53

0523

5

2

2

xxor

yy

K1

)1(2

)5)(1(466 2 x or

)1(2

)70)(1(42828 2 y

K1

x = 0.742 , - 6.742 or y = 2.774 , 25.226 N1

1

y = 2.774 , 25.226 or x = 0.742 , - 6.742 N1

2

532)( 2 xxxf

K1

5)2

3()

2

3(2)( 22

xxf

K1

2 (a)

23 19

22 2

x

N1

3

(b)2

19,

2

3 kh

N1 , N1 2

)5)(2(462 t < 0

t <2

19

K1

N1 2

A = (4 , 0)

12,2

12121 mmusemm

P1

K1

y-0 = -2 (x - 4)K1

3 (a)

y = -2x + 8N1

4

(c)

5

7

Page 12: SBP F4 P2 Final Exam 2007

PEPERIKSAAN DIAGNOSTIK ADDITIONAL MATHEMATICS - 3 -SBP 2007 Paper 2

SBP 2007

NO SOLUTION AND MARK SCHEME MARKSSUB

MARKS TOTAL

B ( 0,8) P1

3

)1(820

3

)1(024

yor

x K1

3(b)

L (6,-4)N1 3

3150

)6(6050)6(40)9(3020)8(10

yx K1

506698 yxK1

6052

4222

yx

yx(solve the simultaneous equations)

K1

4(a)

y =6 , x = 15N1 4

6110013050)( 22 xorxxf N1

Using

f

xxf 22 )(

0r2

2

xfx

fx

=50

130500r 961

50

61100

K1

(b)

= 261N1

3

s = 0.52 x 12 K15(a)

= 6.24 N1 2

tan 0.52 radian =12

AB

AB = 6.8697

K1

12869762

1OABofAreaL1 . K1

).()(torsec 520122

1OACofAreaL 2

2

K1

Area of the shaded region = L1 - L2

= 41.2182 - 37.44K1

(b)

= 3.7782 N1 5

7

7

7

Page 13: SBP F4 P2 Final Exam 2007

PEPERIKSAAN DIAGNOSTIK ADDITIONAL MATHEMATICS - 4 -SBP 2007 Paper 2

SBP 2007

NO SOLUTION AND MARK SCHEME MARKSSUB

MARKS TOTAL

31 dx

dvand

dx

du P1

2)13(

)3)(()1)(13(

x

xx K1

6 (a)

2)13(

1

x N13

2

1,

221 m

km

P1,P1

12

1

2

k

K1

(b)

k = -4N1

4

3

3

nyx

ynx

K1

3

2

32

3

1

norh

K1

7(a)

6

1h and n = 2

N1

3

3

2)8(

3

1)8(1 g

K1(b)

= 2 N1

2)2(3)2( g or

2)23(3)2(2 xg

K1(c)

g ( -4) = -10 or 10)2(2 g N14

1(2 ) 3(2 ) 2 (3 2 ) 3 2g t t or gg t t K1

3 (2t ) + 2 = 3 – 2t K1

(d)

t =8

1 N1

3

7log3log)1( x K18(a)

x = 0.7712 N1 2

7

10

Page 14: SBP F4 P2 Final Exam 2007

PEPERIKSAAN DIAGNOSTIK ADDITIONAL MATHEMATICS - 5 -SBP 2007 Paper 2

SBP 2007

NO SOLUTION AND MARK SCHEME MARKSSUB

MARKS TOTAL

23

1

logloglog yax aaa

(Use any law of logarithm)K1

=

23

1

logloglog naa

ma aaa

K1

(b)

= nm

213

N1

3)2(22 39 xx OR 031 OR 3log1 3 P1

0)2(2 xyK1

)2(3 xy OR 32

x

y K1

042)63( xx (solve simultaneously)K1

(c)

x = 2, y = 0 N1 5

Solve simultaneously:1253

123

xy

xyK19(a)(i)

B (6,6) N1

624182

1

K1a(ii)

= 18N1

m= -3 K1

)3(31 xyK1

(iii)

103 xyN1 7

3

1

510

m

mOR

1

1

14

m

mkK1

(b)

1,2

3 km ,

N1N1 3

10 (a) xxdx

dy42 2 K1N1

2

042 2 xx K1(b)

0)2( xxK1

10

10

Page 15: SBP F4 P2 Final Exam 2007

PEPERIKSAAN DIAGNOSTIK ADDITIONAL MATHEMATICS - 6 -SBP 2007 Paper 2

SBP 2007

NO SOLUTION AND MARK SCHEME MARKSSUB

MARKS TOTAL

(0,-5) and

3

23,0 N1 ,

N14

Using

dx

dy

dt

dx

dt

dy= )42(3 2 xx

K1(c)

= - 6 N12

Using xdx

dyy (-2)(-0.05)

K1(d)

=0.1 N1 2

Histogram (Refer to Graph)All frequencies and label for x-axis correct for atleast 3 bars

N1

Correct HistogramN1

Correct methodK1

11(a)

Mode= 24.5N1

4

L = 20.5 or fm = 14 or F = 15 N1

1014

152

20

5.20

mK1

(b)

= 16.93 N13

40

960x =24

K1

40

5510 or 2)24(

40

28550

K1

( c )

= 11.74N1

3

(i) Using 1002003

2005 Q

QI

K1

w = 120N1

x = 17.55 N1

12(a)

y = 11.30 N1

4

100(*120) + 130(180) + 115(70) + 110(z) K1

z

z

250

)(110)70(115)80(130)120(*100120

K1

(b)

z = 45 N1 3

( c) 120 P1

10

10

Page 16: SBP F4 P2 Final Exam 2007

PEPERIKSAAN DIAGNOSTIK ADDITIONAL MATHEMATICS - 7 -SBP 2007 Paper 2

SBP 2007

NO SOLUTION AND MARK SCHEME MARKSSUB

MARKS TOTAL

120100

120I

K1

= 144 N13

)5)(7(2

457cos

222 ACB

K113(a)(i)

ACB = 34.050 N1

5.6

05.34sin

9

sin

CED K1

50.830 N1a(ii)

CED = 129.170 N1

5

)(2180

180

180

DEFEDFOR

CEDDEFOR

CEDACBCDE

K1

CDE = 16.780 ,DEF = 50.830 andDEF = 78.340 N1

L1 = Area of ∆ CDE = CDEsin5.692

1

OR

L2 = Area of ∆ EDF = EDFsin5.65.62

1

K1

Area of ∆ CDF = L1 + L2 K1

= 29.13 N1

5

ALTERNATIVE

)(2180

180

180

DEFEDFOR

CEDDEFOR

CEDACBCDE

K1

CDE + EDF K1

CDF = 95.120

N1

Area of ∆ EDF = CDFsin5.692

1 K1

(b)

= 29.13 N1 5

10

10

Page 17: SBP F4 P2 Final Exam 2007

PEPERIKSAAN DIAGNOSTIK ADDITIONAL MATHEMATICS - 8 -SBP 2007 Paper 2

SBP 2007

marks

2

4

6

8

12

14

Fre

qu

ency

0.5 10 .5 20.5 30.5 40.5 50.5

10