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3472/1/2 Name : ……………………………...US217 Aug 2010 Class: Form 4 ……………………..

UJIAN SETARA 2(ANSWER & MARKING SCHEME)

ADDITIONAL MATHEMATICS

Paper 1 & 2

One hour

SEKOLAH MENENGAH KEBANGSAAN TINGGI KAJANG

INFORMATION FOR CANDIDATES:

1. This paper consists of two parts, namely Paper 1 and Paper 2. Answer all questions in both parts in this booklet.3. The diagrams shown in the questions are not drawn to scale unless stated.4. You may use a non-programmable scientific calculator.5. The following formulae may be helpful in answering the questions. The symbols given

the ones commonly used.

5.1.a

acbb x

2

42 −±−=

5.2. nmnm aaa +=×

5.3. am ÷ an = am-n

5.4. ( am ) n = am n

5.5. loga mn = loga m + loga n

5.6. loga n

m

= loga m – loga n5.7. loga mn= n loga m

This question paper consist of 7 printed pages

Prepared by: ………………………….. Verified by: ………………………

Paper 1

5.8.a

bb

c

c

a

log

loglog =

5.9. A

asin

=

Bb

sin =

C c

sin

5.10. a2 = b2 + c2 – 2bc cos A

5.11. Area of triangle=

2

1ab sin C

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2

Answer all questions. Jawabsemua soalan.

1. The following information refers to set K and set L.Maklumat berikut adalah berkaitan dengan set K dan set L.

K = { -1, 0, 1, 2 } L= { 2. 3. 6 }

The relation between set K and set Lis defined by the following set of ordered pairs{ (-1, 3), (0, 2), (1, 3), (2, 6) }. Hubungan di antara set K dan set L ditakrifkan oleh set pasangan tertib yang berikut { (-1, 3), (0, 2), (1, 3), (2, 6) }.

State Nyatakan

(a) the image of 1,imej bagi1,

(b) the type of the relation. [2marks] jenist bagi hubungan tersebut .

3 √ 1Answer : (a) ………………………...

many-to-one √ 1 (b) ……...……………........

2. Given

2

1and –2 are the roots of a quadratic equation. Write the quadratic equation

general form. [2marks] Diberi ½ dan –2 adalah punca-punca bagi satu persamaan kuadratik. Tuliskan persamaan kuadratik idalam bentuk am.

(x – ½)(x + 2) = 0 OR S.O.R = ½ + (-2) = −3/2√ P1 P.O.R = ½ x (-2) = −1 √ P1

2x2 − 3x – 2 = 0 √2

Answer : …..………………………...

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3. The quadratic equation x2 – 5x+ p = 0has two distinctive different roots. Find the rangeof value p. Persamaan kuadratik x2 – 5x + p = 0 mempunyai dua punca berlainan yang nyata. Cari julat nilai p.

[2 marks]

a = 1, b = −5, c = p

b2 – 4 ac > 0

(-5)2 – 4(1)(p) > 0 √ P1

25 > 4p

p < 25/4 √ 2Answer : ….………………………...

4. Diagram 1 shows the graph of function y = (x – p)2

+ 5. Rajah 1 menunjukkan graf fungsi y = (x – p)2 + 5. [3 marks]State y Nyatakan

(a) the value of p,nilai p.

(b) the equation of axis of symmetry. persamaan bagi paksi simetri,

(c) the maximum value. 0 4nilai maksimum.

Answer : (a) p = …………………...

(b) ……...…………….......

(c) ……………………….

5. Solve the equation.Selesaikan persamaan

32x + 1 = 92x [3 marks]

32x + 1= (32 )2x √ P1 2x + 1 = 4x √ P2

Answer: x =……………………….

x

x = 2 √ 1

5 √ 1

2 √ 1

½ √ 3

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4 x y

42y x

22 y

x

12)(y

xlog

12)(ylog xlog

12)(ylog 2

xlog

12)(ylog 4log

xlog

1/2

1/2

11/2

1/2

2

21/2

2

22

22

2

−=

+=

=+

=+

=+−

=+−

=+−

2

4 x y

−=

2

4 x y

−=

x

4

6. Givenlog 4 x – log 2(y + 2) = 1.Expressy in termsof x. Diberi log 4 x – log 2(y + 2) = 1. Nyatakan y dalam sebutan x . [4 marks]

√ P1

√ P2

√ P3

√ 4 OR

√ 4Answer : ……………………………

7. Find the range of values of x for which x(2x – 7)≥ 7 – 2x. [4marks]

Cari julat nilai x bagi x(2x – 7)≥ 7 – 2x.

2x2 – 5x – 7 ≥ 0 √ P1 √ P3

Let 2x2 – 5x – 7 = 0

(2x – 7)(x + 1) = 0√ P2 −1 7/2

x = 7/2, x = − 1 √ P3

x ≤ − 1, x ≥ 7/2 √4

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)3( 2

)2 )( 3( 49 )9( 2 −±−−

5

Answer : ……………………………

Paper 2

Answer all questions. Jawabsemua soalan

1. Solve the simultaneous equations x + 2y = 3 and x2 + y2 + xy = 7 . Give the answerscorrect to three decimal places.Selesakan persamaan serentak x + 2y = 3 dan x2 + y2 + xy = 7. Beri jawapan betul kepada tiga angka perpuluhan.

[5 marks]

x = 3 – 2y √ M1

(3 – 2y)2 + y2 + (3 – 2y)y = 7 √ M13y2 – 9y + 2 = 0

y = √ M1

y1 = 2.758, y2 = 0.2417 √ A1

x1= −2.516, x2=2.517 √ A1

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23 x −

2

2 x2 x 2 −+

28 x2 −

6

2. Given that f(x) = 2x + 3and g(x) = x2 + 2x + 1. Diberi f(x) = 2x + 3 dan g(x) = x2 + 2x + 1.Find,Cari,

(a) f -1(x) [1 mark ](b) f -1 g(x) [2 marks](c) h(x)such that fh(x) = 2x – 5. [2 marks]

(a) f −1(x)= √ P1

(b) f −1 g(x) = f −1(x2+ 2x + 1) √ M1

= √ A1

(c) Let h(x) = y f(y) = 2x – 5

f(y) = 2y + 32y + 3 = 2x – 5 √ M1

y =

h(x) = x – 4 √ A1

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16 .24

98 sin

6 .7

CBD sin o

=∠

6 .7

15.18 sin

BD

15.116 sin oo

=

7

3. Diagram 2 shows a quadrilateral ABCD.A 10 cm B

(a) Calculate, 52o

(i) Length of BC. (ii) ∠ BCD [5 marks] 29 cm

(b) Point C’ lies on BC such that 98o DC = DC’. D

7.6cm(i) Sketch the triangle BC’D. C

(ii) Calculate the area, in cm2, triangle BC’D. DIAGRAM 2 [5 marks]

(a) (i) BC 2 = 102 + 292 – 2(10)(29) cos52o √ M1 BC = 24.16 √ A1

(ii) √ M1

∠ CBD = 18.15o √ P1

∠ BCD = 63.85o

√ A1

(b) (i) B

18.15o(ii) ∠ BC’D = 116.15o, ∠ BDC’= 45.70o √ P1

√ M1

√ P1

C’ 45.70o

BD = 21.9 7.6 cm D Area BC’D = ½(7.6)(21.9) sin 45.70o √ M1

= 59.56 √ A1