Your Federal Quarterly Tax Payments are due April 15th

# Add Maths - Questions, Answers & Marking Scheme - UJian Setara 2/2010 by nklye

VIEWS: 964 PAGES: 7

• pg 1
1

3472/1/2                                                      Name : ……………………………...
US2
17 Aug 2010                                                   Class: Form 4 ……………………..

UJIAN SETARA 2

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 are
the ones commonly used.

 b  b 2  4ac                                         log c b
5.1. x                                           5.8. log a b 
2a                                                log c a

5.2. a m  a n  a mn                                     a     b       c
5.9.              
sin A sin B   sin C
5.3. a m ÷ a n = a m-n
5.10. a 2 = b 2 + c 2 – 2bc cos A
5.4. ( a m ) n = a m n
5.11. Area of triangle
5.5. loga mn = loga m + loga n
1
= ab sin C
m                                              2
5.6. loga     = loga m – loga n
n
5.7. loga mn = n loga m

This question paper consist of 7 printed pages

Prepared by: …………………………..                                     Verified by: ………………………...
2

Paper 1

Jawab semua 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 L is 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 bagi 1,
(b)     the type of the relation.                                                        [2 marks]
jenist bagi hubungan tersebut.

3          √1
many-to-one √1
(b) ……...……………........

1
2.   Given     and –2 are the roots of a quadratic equation. Write the quadratic equation in
2
general form.                                                                [2 marks]
dalam 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
3

3.   The quadratic equation x2 – 5x+ p = 0 has two distinctive different roots. Find the range
of 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 √2

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

2     √1
Answer : (a) p = …………………...
x=2      √1
(b) ……...…………….......
5     √1
(c) ……………………….

5.   Solve the equation.
Selesaikan persamaan

32x + 1 = 92x                                                                             [3 marks]

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

½      √3
4

6.    Given log4 x – log2 (y + 2) = 1. Express y in terms of x.
Diberi log4 x – log2 (y + 2) = 1. Nyatakan y dalam sebutan x .                             [4 marks]

log 2 x
√ P1            log 2 (y  2)  1
log 2 4
log 2 x
 log 2 (y  2)  1
2
log 2 x 1/2  log 2 (y  2)  1
√ P2              x 1/2
log 2           1
(y  2)
√ P3    x 1/2
 21
y2
x 1/2  2y  4
x 1/2  4                                         x 4
√4      y                             OR                y
2                                             2

x 4 √4
y
2

7.    Find the range of values of x for which x(2x – 7)  7 – 2x.                                [4 marks]
Cari julat nilai x bagi x(2x – 7)  7 – 2x.

2x2 – 5x – 7  0         √ P1                                               √ P3

Let 2x2 – 5x – 7 = 0
x
(2x – 7)(x + 1) = 0 √ P2                            −1           7/2

x = 7/2, x = − 1        √ P3

x ≤ − 1, x  7/2        √4

5

Paper 2

Jawab semua soalan

1.   Solve the simultaneous equations x + 2y = 3 and x2 + y2 + xy = 7. Give the answers
correct 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      √ M1

3y2 – 9y + 2 = 0

 ( 9 )  9 2  4( 3 )( 2 )
y=                                  √ M1
2( 3 )

y1 = 2.758,    y2 = 0.2417         √ A1

x1 = −2.516, x2 =2.517             √ A1
6

2.   Given that f(x) = 2x + 3 and g(x) = x2 + 2x + 1.
Diberi f(x) = 2x + 3 dan g(x) = x2 + 2x + 1.
Find,
Cari,
(a)     f -1(x)                                    [1 mark]
-1
(b)     f g(x)                                     [2 marks]
(c)     h(x) such that fh(x) = 2x – 5.             [2 marks]

x3
(a)     f −1(x) =                        √ P1
2

(b)     f −1g(x) = f −1(x2 + 2x + 1) √ M1
x2  2x  2
=                   √ A1
2

(c)     Let h(x) = y
f(y) = 2x – 5
f(y) = 2y + 3
2y + 3 = 2x – 5                √ M1
2x  8
y=
2
h(x) = x – 4             √ A1
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)    BC2 = 102 + 292 – 2(10)(29) cos52o √ M1
BC = 24.16                         √ A1
sin CBD sin 98 o
(ii)                                            √ M1
7.6   24.16

 CBD = 18.15o                            √ P1

 BCD = 63.85o                            √ A1

(b)   (i)      B

18.15o
(ii)  BC’D = 116.15o,  BDC’= 45.70o √ P1
sin 116.15 o sin 18.15 o
                       √ M1
√ P1                       BD          7.6

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

To top