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SULIT 3472/2 ADDITIONAL MATHEMATICS PAPER 2 AUGUST 2008 2 ½HOURS

JABATAN PELAJARAN NEGERI SABAH

SIJIL PELAJARAN MALAYSIA TAHUN 2008 EXCEL 2

___________________________________________________________________________

ADDITIONAL MATHEMATICS PAPER 2 (KERTAS 2)

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

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

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 for 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 to 4.

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

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

___________________________________________________________________________ This question paper consists of 13 printed pages.

(Kertas soalan ini terdiri daripada 13 halaman bercetak.) [Turn over (Lihat sebelah) The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used.

NAMA : _____________________ KELAS : _____________________ NO K.P : _____________________ A. GILIRAN : _________________-

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ALGEBRA

1. 2 4

2b b acx

a− ± −

=

2. m n m na a a +× =

3. m n m na a a −÷ =

4. ( )m n mna a=

5. log log loga a amn m n= +

6. log log loga a am m nn= −

7. log logna am n m=

8. logloglog

ca

c

bba

=

9. ( 1)nT a n d= + −

10. [2 ( 1) ]2nnS a n d= + −

11. 1nnT ar −=

12. ( 1) (1 ) , 11 1

n n

na r a rS r

r r− −

= = ≠− −

13. , 11

aS rr∞ = <

−

CALCULUS

1. , dy dv duy uv u vdx dx dx

= = +

2. 2,

du dvv uu dy dx dxyv dx v

−= =

3. dy dy dudx du dx

= ×

4. Area under a curve

=b

a

y dx∫ or

= b

a

x dy∫

5. Volume generated

= 2b

a

y dxπ∫ or

= 2b

a

x dyπ∫

STATISTICS

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1. x

xN

= ∑

2. fx

xf

= ∑∑

3. 2 2

2( )x x xx

N Nσ

−= = −∑ ∑

4. 2 2

2( )f x x fxx

f fσ

−= = −∑ ∑

∑ ∑

5.

12

m

N Fm L c

f

⎛ ⎞−⎜ ⎟= + ⎜ ⎟

⎜ ⎟⎝ ⎠

6. 1 100o

QIQ

= ×

7. i i

i

W II

W=∑

∑

8. ( )

!!r

nnn rP =−

9. ( )

!! !r

nnn r rC =−

10. ( ) ( ) ( ) ( )P A B P A P B P A B∪ = + − ∩

11. ( ) , 1n r n rrP X r C p q p q−= = + =

12. Mean, µ = np

13. npqσ =

14. xZ µσ−

=

GEOMETRY

1. Distance

= ( ) ( )2 21 2 1 2x x y y− + −

2. Midpoint

( ) 1 2 1 2, ,2 2

x x y yx y + +⎛ ⎞= ⎜ ⎟⎝ ⎠

3. A point dividing a segment of a line

( ) 1 2 1 2, ,nx mx ny myx ym n m n+ +⎛ ⎞= ⎜ ⎟+ +⎝ ⎠

4. Area of triangle =

1 2 2 3 3 1 2 1 3 2 1 31 ( ) ( )2

x y x y x y x y x y x y+ + − + +

5. 2 2r x y= +

6. 2 2

ˆxi yj

rx y

+=

+

TRIGONOMETRY

1. Arc length, s rθ= 8. sin ( ) sin cos cos sinA B A B A B± = ±

9. cos ( ) os os sin sinA B c Ac B A B± = ∓

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2. Area of sector, 212

A r θ=

3. 2 2sin cos 1A A+ =

4. 2 2sec 1 tanA A= +

5. 2 2cosec 1 cotA A= +

6. sin 2 2sin cosA A A=

7. 2 2cos 2 cos sinA A A= −

2

2

2 os 11 2sin

c AA

= −

= −

10. tan tantan ( )1 tan tan

A BA BA B±

± =∓

11. 2

2 tantan 21 tan

AAA

=−

12. sin sin sin

a b cA B C= =

13. 2 2 2 2 cosa b c bc A= + −

14. Area of triangle 1 sin2

ab C=

Section A

[40 marks]

Answer all questions.

1 Solve the simultaneous equations 24 3x y x x y+ = + − = − . [5 marks]

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2 Diagram 1 shows a straight line CD which meets a straight line AB at point D. The

point C lies on the y-axis.

Diagram 1

(a) State the equation of AB in the intercept form. [1 mark]

(b) Given that 2AD = DB, find the coordinates of D. [3 marks]

(c) Given that CD is perpendicular to AB, find the y-intercept of CD. [3 marks]

3 (a) Sketch the graph of 3sin 2 for 0 2y x x π= − ≤ ≤ . [4 marks]

(b) Hence, using the same axes, sketch a suitable straight line to find the number

of solutions for the equation 3sin 2 =1 for 0 2xx x ππ

+ ≤ ≤ . State the number

of solutions. [3 marks]

4 Given that the gradient of the tangent to the curve 3 22 6 9 1y x x x= − + − at point P is

3, find

(a) the coordinates of P, [2 marks]

(b) the equation of the tangent and normal to the curve at P. [4 marks]

5 Table 1 shows the distribution of the ages of 100 teachers in a secondary school.

Age

(years) <30 <35 <40 <45 <50 <55 <60

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Number of

teachers 8 22 42 68 88 98 100

Table 1

(a) Based on Table 1, copy and complete Table 2.

Age

(years) 25 - 29

Frequency

Table 2

[2 marks]

(b) Without drawing an ogive, calculate the interquartile range of the distribution.

[5 marks]

6 The first three terms of a geometric progression are also the first, ninth and eleventh

terms, respectively of an arithmetic progression.

(a) Given that all the term of the geometric progressions are different, find the

common ratio. [4 marks]

(b) If the sum to infinity of the geometric progression is 8, find

(i) the first term,

(ii) the common difference of the arithmetic progression. [4 marks]

Section B

[40 marks]

Answer four questions.

7 Use graph paper to answer this question.

Table 3 shows the values of two variables, x and y, obtained from an experiment.

Variables x and y are related by the equation xy ab−= , where a and b are constants.

x 1 2 3 4 5 6

y 41.7 34.7 28.9 27.5 20.1 16.7

Table 3

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(a) Plot 10log y against x by using a scale of 2 cm to 1 unit on the x-axis and 2 cm

to 0.2 unit on the 10log y -axis.

Hence, draw the line of best fit. [4 marks]

(b) Use your graph from (a) to find

(i) the value of y which was wrongly recorded, and estimate a more

accurate value of it,

(ii) the value of a and of b,

(iii) the value of y when x = 3.5. [6 marks]

8 Diagram 2 shows a trapezium PQRS. U is the midpoint of PQ and 2PU SV= . PV and

TU are two straight lines intersecting at W where TW : WU = 1 : 3 and PW = WV.

Diagram 2

It is given that 12 , 18 and QR 18 5PQ a PS b b a= = = − .

(a) Express in terms of and/or a b ,

(i) SR ,

(ii) PV ,

(iii) PW . [5 marks]

(b) Using PT : TS = h : 1, where h is a constant, express PW in terms of h,

and/or a b and find the value of h. [5 marks]

9 Diagram 3 shows a circle with centre C and of radius r cm inscribed in a sector OAB

of a circle with centre O and of radius 42 cm. [Use π = 3.142]

S R

P Q

V

T W

U

CONFID

G

(a

(b

(c

10 D

T

gr

(a

(b

DENTIAL

Given that ∠

a) the va

b) the pe

c) the are

Diagram 4 sh

The curve int

radient of th

a) Find t

b) Hence

(i)

ra3

AOB π=

alue of r,

erimeter, in c

ea, in cm2, o

hows part of

tersects the s

he curve at th

the value of

e, calculate

area of the

Dia

ad , find

cm, of the sh

of the shade

the curve y

Dia

straight line y

he point A is

k.

e shaded regi

y

O

R

8

agram 3

haded region

d region.

1x= − .

agram 4

y = k at poin

14

.

ion R : area

A

S

n,

nt A, where k

of the shade

x

y x=y = k

k is a constan

ed region S.

1x −

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[2 marks]

[4 marks]

[4 marks]

nt. The

[3 marks]

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(ii) the volume generated, in terms of π, when the region R which is

bounded by the curve, the x-axis and the y-axis, is revolved through

360o about the y-axis. [7 marks]

11 (a) A committee of three people is to be chosen from four married couples. Find

how many ways this committee can be chosen

(i) if the committee must consist of one woman and two men,

(ii) if all are equally eligible except that a husband and wife cannot both

serve on the committee. [5 marks]

(b) The mass of mango fruits from a farm is normally distributed with a mean of

820 g and standard deviation of 100 g.

(i) Find the probability that a mango fruit chosen randomly has a

minimum mass of 700 g.

(ii) Find the expected number of mango fruits from a basket containing

200 fruits that have a mass of less than 700 g. [5 marks]

Section C

[20 marks]

Answer two questions.

12 A particle moves along a straight line and passes through a fixed point O. Its velocity,

v m s–1, is given by 2 16v pt qt= + − , where t is the time, in seconds, after passing

through O, p and q are constants. The particle stops momentarily at a point 64 m to

the left of O when t = 4.

[Assume motion to the right is positive.]

Find

(a) the initial velocity of the particle, [1 mark]

(b) the value of p and of q, [4 marks]

(c) the acceleration of the particle when it stops momentarily, [2 marks]

(d) the total distance traveled in the third second. [3 marks]

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13 Table 4 shows the prices of four types of book in a bookstore for three successive

years.

Book Price in year (RM) Price index in

2001

based on 2000

Price index in 2002

based on 2000 Weightage

2000 2001 2002

P w 20 30 150 225 6

Q 50 x 65 115 130 5

R 40 50 56 125 140 3

S 80 z 150 y y 2

Table 4

(a) Find the values of w, x, y and z. [4 marks]

(b) Calculate the composite index for the year 2002 based on the year 2001.

[4 marks]

(c) A school spent RM4, 865 to buy books for the library in the year 2002. Find

the expected total expenditure of the books in the year 2003 if the composite

index for the year 2003 based on the year 2002 is the same as for the year

2002 based on the year 2001.

[2 marks]

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14 Use graph paper to answer this question.

A farmer wants to plant x-acres of vegetables and y-acres of tapioca on his farm.

Table 5 shows the cost of planting one acre and the number of days needed to plant

one acre of vegetable and one acre of tapioca.

Vegetables Tapioca

Cost of planting

per acre RM100 RM 90

Number of days

needed per acre 4 2

Table 5

The planting of the vegetables and tapioca is based on the following constraints:

I The farmer has a capital of RM1800.

II The total number of days available for planting is 60.

III The area of his farm is 20 acres.

(a) Write down three inequalities, other than 0 and 0x y≥ ≥ , which satisfy all the

above constraints. [3 marks]

(b) By using a scale of 2 cm to 4 acres on both axes, construct and shade the

region R that satisfies all the above constraints. [3 marks]

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

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(i) the maximum area of tapioca planted if the area of vegetables planted

is 10 acres,

(ii) the maximum profit that the farmer can get if the profit for one acre of

vegetables and one acre of tapioca planted are RM60 and RM20

respectively. [4 marks]

15 Diagram 5 shows a quadrilateral ABCD such that ABC∠ is acute.

Diagram 5

(a) Calculate

(i) ABC∠ ,

(ii) ADC∠ ,

(iii) the area, in cm2, of quadrilateral ABCD. [8 marks]

(b) A triangle AB’C has the same measurement as triangle ABC, that is, AC = 15

cm, CB’ = 9 cm and ' 30B AC∠ = , but is different in shape to triangle ABC.

(i) Sketch the triangle AB C′ .

(ii) State the size of 'AB C∠ . [2 marks]

END OF QUESTION PAPER

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NO. KAD PENGENALAN

ANGKA GILIRAN Arahan Kepada Calon

1 Tulis nombor kad pengenalan dan angka giliran anda pada ruang yang disediakan. 2 Tandakan (√ ) untuk soalan yang dijawab.

3 Ceraikan helaian ini dan ikat sebagai muka hadapan bersama-sama dengan buku jawapan.

Kod Pemeriksa

Bahagian Soalan Soalan Dijawab

Markah Penuh

Markah Diperoleh (Untuk Kegunaan Pemeriksa)

A

1 5

2 6

3 5

4 9

5 7

6 8

B

7 10

8 10

9 10

10 10

11 10

C

12 10

13 10

14 10

15 10

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Jumlah

EXCEL 2

PAPER 2 MARKING SCHEME

No. Solution and Mark Scheme Sub Marks

Total Marks

1

4 3y x= − − or equivalent 34

yx − −=

Eliminate x or y 2 ( 4 3) 3x x x+ − − − = − or

23 3 34 4y y y− − − −⎛ ⎞ ⎛ ⎞+ − = −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Solve the quadratic equation 2 5 6 0

( 2)( 3) 0x xx x+ + =+ + =

2 14 45 0

9 5 0y y( y )( y )

− + =− − =

3, 2x = − − for both values of x. y = 5, 9

9,5y = x = −3, −2

5

5

2

(a) 16 3x y− =

(b) : 1: 2AD DB =

1(6) 2(0) 1(0) 2( 3),3 3+ + −⎛ ⎞

⎜ ⎟⎝ ⎠

( )2, 2− (c) 2CDm = − ( 2) 2( 2)y x− − = − − 2 2y x= − + intercept 2y − =

1

3

3

7

K1

N1

K1

K1

N1

P1

P1

K1

K1

N1

P1

N1

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3

(a)

Shape of sin x Maximum = 3, minimum = –3 2 periods for 0 2x π≤ ≤ Inverted sin x

(b) 1xyπ

= − or equivalent

Draw the straight line 1xyπ

= −

No. of solutions = 5

4

3

7

4

(a) 26 12 9 3dy x xdx

= − + =

2 2 1 0x x− + =

( 1)( 1) 0x x− − = 1x =

3 22(1) 6(1) 9(1) 1

4y = − + −=

(1, 4)P

(b) Equation of tangent:

2

P1

P1

P1

P1

x O 2π –3

3 1xyπ

= −

N1

K1

N1

y

N1

K1

K1

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4 3( 1)y x− = −

3 1y x= + Equation of normal:

14 ( 1)3

y x− = − −

3 13y x= − +

4

6

5 (a)

Age (years) Frequency, f 25 – 29 8 30 – 34 14 35 – 39 20 40 – 44 26 45 – 49 20 50 – 54 10 55 – 59 2

(b) 1 Q1

1 Q1

L 34.5, F 22 or L =34.5 , f 20

= =

=

3 Q3

3 Q3

L 44.5, F 68 or L 44.5, f 20

= =

= =

Use 1

Q141 1

Q1

N-FQ L C

f⎡ ⎤

= + ⎢ ⎥⎢ ⎥⎣ ⎦

or 3

Q343 3

Q3

N-FQ L C

f⎡ ⎤

= + ⎢ ⎥⎢ ⎥⎣ ⎦

Interquartile Range = 46.25 – 35.25 = 11

2

5

7

6 (a) GP : T1 = a, T2 = ar, T3 = ar2

AP : T1 = a, T9= a + 8d, T11 = a + 10d

N1

K1

N1

N1

K1

K1

N1 N1

P1

P1

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ar = a + 8d or ar2 = a + 10d

a(r2 – 1) = 10d or a(r−1) = 8d or ar(r−1)=2d

2 1 101 8

rr−

=−

14

r =

(b) (i) 1

41

8 a

−=

a = 6

(ii) 16( ) 6 84

d= + or 216( ) 6 104

d= +

916

d = −

4

4

8

7 (a)

x 1 2 3 4 5 6

10log y 1.620 1.540 1.461 1.439 1.303 1.223

Plot 10log y against x (Correct axes and correct scales) 6 points plotted correctly Draw line of best fit

(b) (i) y = 27.5 should be y = 24.0 (ii) 10 10 10log (log ) logy b x a= − +

a = 50 b = 1.2

4

K1

P1

N1

K1

N1

N1

K1

K1

N1

P1

K1

N1

N1

N1

N1

N1

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(iii) 10log 1.42y = y = 26.3

6

10

8 (a) (i) Use Triangle Law to find or SR PV

7SR a=

(ii) 3SV a=

3 18PV a b= +

(iii) ( )1 3 182

PW a b= +

(b) 34

PW PU UW PU UT= + = + or equivalent

6 (18 )1

hUT UP PT a bh

= + = − ++

3 186 ( 6 )4 ( 1)

hPW a a bh

= + − ++

3 272 2( 1)

hPW a bh

= ++

(2)

Comparing (1) & (2) 2792( 1)

hh

=+

h = 2

5

5

10

N1

N1

N1

N1

N1

K1

N1

K1

K1

P1

K1

K1

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9 (a) sin 30

42r

r=

−

14r =

(b) 143π

× or 2143π

×

2 228 14− Perimeter = 24.249 + 24.249 + 29.325 = 77.823 (accept 77.82)

(c) 21 142 3

π× ×

1 14 5882× ×

Area = 2 ( 169.741 – 102.639) = 134.204 Accept 134.2

2

4

4

10

10

(a) 12 1

dydx x

=−

1 1

42 1x=

−

x = 5, k = 2

(b) (i) Area of R or Area of S

=2

2

0

( 1)y dy+∫ 5

11 x dx= −∫

3

P1

K1

N1

K1

K1

N1

K1

K1

K1

N1

K1

N1

K1

K1

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= 23

03y y

⎡ ⎤+⎢ ⎥

⎣ ⎦ = ( )

532

1

132

x⎡ ⎤−⎢ ⎥

⎢ ⎥⎢ ⎥⎣ ⎦

= 243

or 15 3

Area of S or Area of R

= 22 5 43

× − = 12 5 53

× −

= 153

= 24 3

Area of R : Area of S = 7 : 8

(ii) 2

2 2

0

( 1)V y dyπ= +∫

253

0

25 3

111315

yV y y

V

π

π

⎡ ⎤= + +⎢ ⎥

⎣ ⎦

=

7

10

11 (a) (i) 4 4

1 2C C = 24

(ii) If 4 4 4 2 4 3 4 4

3 0 2 1 1 2 0 3C or or or CC C C C C C is shown

4 4 4 2 4 3 4 43 0 2 1 1 2 0 3C + + + CC C C C C C

= 32

or

8 6 4 3!8 6 4

3!32

or× ×× ×

=

(b) (i) 700 820100−

P( 1.2)X > −

5

N1

K1

N1

K1

K1

K1

K1

N1

K1

N1

K1

N1

K1

K1

K1

N1

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=1 – 0.1151 = 0.8849

(ii) 200 x 0.1151

= 23

5

10

12 (a) –16 m s–1 (b) Integrate 2 16pt qt+ − with respect to t

3 2 163 2p qs t t t= + −

t = 4, v = 0

16p + 4q = 16 or 64 8 03

p q+ =

p = 3 q = –8

(c) a = 6t – 8 t = 4, a = 16

(d) 3 24 16s t t t= − −

Find 3

2 dtv∫ or 3 2t tS S= =−

Substitute 2 or 3 into st t= = d = |[ 3 23 4(3 ) 16(3)s = − − ] – [ 3 22 4(2 ) 16(2)s = − − ]|

d = 17 m

1

4

2

3

10

K1

N1

N1

N1

K1

N1

N1

N1

K1

K1

N1

K1

K1

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13 (a) w = 13.33 x = 57.50 y = 187.5 z = 150 (b) I2002 / 2001 : 150 , 113.04, 112, 100

Use i i

i

W II

W= ∑∑

150 6 113.04 5 112 3 100 26 5 3 2

I × + × + × + ×=

+ + +

2001.216

=

= 125.08

(c) 125.08 4865100

×

=6085.14

4

4

2

10

14 (a) 100 90 1800x y+ ≤ or equivalent 4 2 60x y+ ≤ or equivalent

20x y+ ≤ or equivalent

(b) Draw correctly at least one straight line

Draw correctly all the three straight lines Region R shaded correctly

(c) (i) y = 8.0 – 9.0

(ii) maximum point (15, 0)

3

3

N1

N1

K1

N1

N1

N1

N1

N1

N1

N1

N1

N1

K1

K1

P1

N1

K1

N1

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RM15× 60 + RM20×0 = RM900

4

10

15

(a) (i) sin 30sin 159

o

ABC∠ = ×

'56.44 56 27o oABC or∠ = (ii) 2 2 215 10 8 2(10)(8)cos ADC= + − ∠ 112.41 or 112 25'ADC∠ =

(iii) 1area of 10 8 sin112.412

ACD∆ = × × ×

1area of 15 9 sin(180 56.44 30 )2

ABC∆ = × × × − −

area of quadrilateral ABCD = 36.98 + 67.37 = 104.35 (b) (i)

'AB C∠ must be obtuse

(ii) 123.56 or 123° 33’

8

2

10

K1

N1

N1

K1

K1

K1

K1

N1

N1

K1

N1

N1

B′

A C

GRAPH FOR QUESTION 7

10log y

× 1.6

1.8

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y

32

GRAPH FOR QUESTION 14

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CO-ORDINATOR JUDY LIAN YEE LING

SEKTOR PENGURUSAN AKADEMIK

LIST OF PANEL MEMBERS

0

4

8

12

16

20

24

28 2x+ y = 30

x 4 8 12 16 20 10x+9 y = 180

x+ y = 20

R

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1. Shirney Chua (Leader) SMK Kolombong, Inanam

2. Aileen Beh Chik Heang SMK Datuk Peter Mojuntin

3. Farah Ibrahim SMK Agama Kota Kinabalu

4. Phoon Chiew Fun SM Chung Hwa, Tenom

5. Seak Sain Yon SM Konven St. Ursula, Tawau

6. Sudirman Darise SMK Bongawan, Papar

7. Surianih Sewan SMK Sri Nangka, Tuaran

8. Ting Kai Chu SM St. Michael, Penampang