Document Sample

```					SULIT                                                                                                          3472/1/2

FORM IV 2012

SEKOLAH MENENGAH KEBANGSAAN TINGGI KAJANG
JALAN SEMENYIH, 43000 KAJANG, SELANGOR

PEPERIKSAAN PERTENGAHAN TAHUN 2012

Prepared by :                                    Checked by:                                    Verified by:

........................................        ...........................................    ..…………………………
( NG KOK LYE )                               (NOORAIDA MOHD ZIN)                              ( SOH BOON CHUAN )

This question paper consists of 7 printed pages and 1 blank page

3472/1/2                                                                                                       SULIT
For          SULIT                                                                                                      3472/1/2
examiner’s                                                         SECTION 1
use only

BAHAGIAN 1
[ 44 marks/markah ]
Answer all questions /Jawab semua soalan.

1. Diagram 1 shows the relation between set A and set B.
Rajah 1 menunjukkan hubungan antara set A dan set B.
Set A                     Set B
(a) Express the relation in the form of ordered pairs.
Ungkapkan hubungan itu dalam bentuk pasangan tertib.          1                             p
(b) State the type of the relation.                                                               q
Nyatakan jenis hubungan itu          [ 3 marks/markah ]         2
3                             r
4                             s
(a) {(1,p), (2,r), (3,s), (4,p)} √ 2

Any one pair correct √ B1                                                 DIAGRAM 1
Rajah 1

1
3                  (b) many – to – one √ 1
3

2.    Diagram 2 shows the linear function f.
Rajah 2 menunjukkan fungsi linear f.                                       f
x                            f(x)
(a) State the value of w.
Nyatakan nilai bagi w.
0                            2

(b) Using the function notation,                            1                            3
express f in terms of x.
5                            w
Dengan menggunakan tatatanda
fungsi, ungkapkan f dalam sebutan x.
8                             10
[ 2 marks/markah ]
Rajah 2

(a) 7 √ 1

(b) f(x) = x + 2     or    f:x          x+2 √1
2
2
2

5
3472/1/2                                               2                                                    SULIT
SULIT                                                                                                     3472/1/2      For
examiner’s
use only
3.   Diagram 3 shows the graph of the function f(x) = |3x – 2|, for the domain 0 ≤ x ≤ 5.
Rajah 3 menunjukkan graf bagi fungsi f(x) = |3x – 2| untuk domain 0 ≤ x ≤ 5.    [ 3 marks/markah ]
State / Nyatakan
(a) the value of m.                                                       y
nilai m.

(b) the range of f(x) corresponding to the given domain.                                      f(x) = |3x – 2|
julat bagi f(x) sepadan dengan domain yang diberikan.

2
2
(a)       √1
3
0        m               5
3
(b)     0 ≤ f(x) ≤ 13 √ 2                                                      DIAGRAM 3
Rajah 3
3
13 √ B1                                                                                                       3

4.   Two functions are defined by f : x 2x + 1 and g :x x2 + 2x – 6.
Given that gf : x 4x2 + px + q, find the value of p and of q.
Dua fungsi ditakrifkan sebagai f : x    2x + 1 dan g : x         x 2 + 2x – 6.
Diberi gf : x   4x2 + px + q, cari nilai p dan nilai q.                                   [ 3 marks/markah ]

p = 8, q = −3          √ 3 (both correct)

4x2 + 8x – 3 √ B2                                                                                                   4
(2x + 1)2 + 2(2x + 1) – 6 √ B1                                                                                  3
3

x5
5.   The function of f is defined as f (x)                  , x  m . Find,
3  2x
x5
Fungsi f ditakrifkan sebagai f(x) =            , x  m. Cari                              [ 3 marks/markah ]
3  2x
(a) the value of m,
nilai m,

(b) f 1( x ) .

3
(a)        √1
2                                                                                                          5
5  3x      1                         y5
(b)      f -1(x) =          ,x    √2              x            √ B1
3
2x  1      2                        3  2y                                                        3

9
3472/1/2                                                 3                                                  SULIT
For          SULIT                                                                                          3472/1/2
examiner’s      6.                                                     2
Determine the roots of the quadratic equation 5x = 3x + 2.
use only
Tentukan punca-punca bagi persamaan kuadratik 5x2 = 3x + 2.                     [ 3 marks/markah ]

2
x=             or − 0.4 , x = 1 √ 3 (both correct)
5
 ( 3)  ( 3)2  4(5)( 2)
(5x + 2) (x – 1) = 0 √ B2        or    x                                 √ B2
2(5)
6
3                5x2 – 3x – 2 = 0 √ B1
3

7.   Find the range of values of k if the following quadratic equation (k + 1)x2 + 6x + 3 = 0 which
has two different roots.
Cari julat bagi nilai k jika persamaan kuadratik berikut (k + 1)x2 + 6x + 3 = 0 mempunyai dua punca
yang berbeza.
[ 3 marks/markah ]

k<2 √3

6 2 – 4(k + 1)(3) > 0 √ B2

62 – 4(k + 1)(3)   or b2 – 4ac > 0     or    a = (k+1), b = 6, c = 3 √ B1

7
3
3

8. Given that  and  are the roots of the quadratic equation 2x2 – 5x + 3 = 0. Form the quadratic
equation whose roots are 2 and 2 .
Diberi  dan  adalah punca-punca bagi persamaan kuadratik 2x2 – 5x + 3 = 0. Bentuk satu persamaan
kuadratik yang mempunyai punca-punca 2  dan 2  .
[ 3 marks/markah ]

x2 – 5x + 12 = 0 √ 3

4αβ = 12
2(α + β) = 5    √ B2 (both)

8                        5
α+β=
3                            2   √ B1 (both correct)
3           αβ = 3

9

3472/1/2                                           4                                            SULIT
SULIT                                                                                                           3472/1/2
For
9. Diagram 3 shows the graph of a curve y = a(x + p) ² + q that passes through the point (0, 3)                            examiner’s
and has the minimum point (2, −1). Find the values of a, p and q.                                                        use only
Rajah 3 menunjukkan geraf bagi satu lengkung y = a(x + p) 2 + q yang melalui satu titik (0.3) dan
mempunyai titik minima (2 , −1). Cari nilai-nilai a, p dan q.
y
[ 3 marks/markah ]

(0, 3)
x              DIAGRAM 3
0
(2, −1)                                   Rajah 3

p = −2 √ 1
9
q = −1    √1
3
a=1       √1                                                                                                                     3

10. Find the range of values of x for which x(x − 3) ≤ 4.
Cari julat bagi nilai x yang mana x(x − 3) ≤ 4.                                               [ 3 marks/markah ]

−1 ≤ x ≤ 4          or           x ≥ −1 , x ≤ 4         √3

(x + 1)(x – 4) ≤ 0 √ B2              or                                     or
√ B2             +    −        − √ B2
x
+    +        − x
-1                 4                              4                           10
+
-1   −        +
x2 – 3x – 4 ≤ 0 √ B1                                                                                                   3
3

81 x 1
11. Solve             243
3x
[ 4 marks markah ]
x=3 √4
4x – 4 = 5 + x      or            3x – 4 = 5 √ B3
34x–4 = 3 5+x       or            33 x – 4 = 35 √ B2                                                                      11
4       5
3 or 3 √ B1                                                                                                           4
3

10
3472/1/2                                                    5                                                    SULIT
For        SULIT                                                                                        3472/1/2
examiner’s
use only
12.   Given that lg 2  0  3 and lg 5 = 0.7. Find, without using scientific calculator or mathematical
tables, the value of log 2 20 .
Diberi lg 2 = 0.3 dan lg 5 = 0.7. Cari, tanpa menggunakan kalkulator saintifik atau jadual sifir
matematk, nilai bagi log2 20.
[ 4 marks/markah ]

13
3 or 4.333 √ 4

2(0.3) 0.7
√ B3
0.3
2lg2  lg5
12                 lg2            √ B2

4
4         lg20
√ B1
lg2

13.   Given that log5 x = k, find logx 125x2 in terms of k.
Diberi log5 x = k, find logx 125x2 dalam sebutan k.
[ 4 marks/markah ]

3  2k
√4
k
3log5 5  2log5 x
√ B3
log5 x

log5 125  log5 x 2
√ B2
13                    log5 x
4
log5 125x 2
4                      √ B1
log5 x

14.   Express 2n 3  2n  5(2n 1 ) in the simplest form.                          [ 3 marks/markah ]
Ungakapkan 2n + 3 −2n + 5(2n – 1 ) dalam sebutan paling ringkas.

2 n -1(19) √ 3
5
2 n(8 – 1 +        ) √ B2
2
 2n              
14              2 .2 –2 +5
n    3
 2
n              √ B1

3                                          
3

11
3472/1/2                                               6                                      SULIT
SULIT                                                                                                 3472/1/2
SECTION 2
BAHAGIAN 2
[ 26 marks/markah ]
Answer all questions / Jawab semua soalan.

1.   Solve the simultaneous equations. Give your answers correct to three decimal places.
Selesaikan persamaan serentak. Berikan jawapan anda betul kepada tiga angka perpuluhan.

2x  y  1
x 2  2 y  xy  5 .                                              [ 5 marks/markah ]

y1
y = 2x – 1 √ P1                                      or         x=          √ P1
2
x2 + 2(2x – 1) + x(2x – 1) = 5 √ K1                                      2       y  1
 y  1
       + 2y +  2  y = 5 √ K1
3x2 + 3x – 7 = 0                                                 2                  

3y2 + 12y – 19 = 0
 3  3 2  4(3)( 7)                √ K1
x
2(3)
 12  12 2  4(3)( 19)
y                            √ K1
x = 1.107, - 2.107 √ N1                                                       3(3)

y = 1.214, - 5.214           √ N1                              y = 1.214, - 5.214    √ N1
5                                                           5
x = 1.107, - 2.107      √
N1

2.   Given that f : x       x2 − 2 and g : x              3x + 4.
2
Diberi f : x    x − 2 dan g : x            3x + 4.

−1
(a) (i) Determine f         (x).                                                      [ 2 marks/markah ]
−1
Tentukan f     (x).
−1
(ii) State whether the f       (x) exist. Give a reason to your answer by showing the evidence of
the reason given.
Nyatakan sama ada f −1(x) itu wujud. Berikan sebab kepada jawapan anda dengan
menunjukkan bukti kepada sebab yang anda berikan .            [ 3 marks/markah ]
(b) Given gh(x) = 6x + 7, determine h(x).
Diberi gh(x) = 6x + 7, tentukan h(x).                                            [ 2 marks/markah ]

(b) Let h(x) = y
(a) x = y2 - 2 √ K1                                                          g(y) = 6x + 7
y2 = x + 2                                                             g(y) = 3y + 4
f -1(x) =  x  2 √ N1                                               3y + 4 = 6x + 7 √ K1
y = 2x + 1
Let x = any value,
f -1(x) =  the value after substitution into x  2 , √ K1             h(x) = 2x + 1 √ N1
an object has two images or
it’s a one-to-many relation or the function undefined
7
Therefore the invest function does not exist. √ N1
3472/1/2                                     7                                                         SULIT
SULIT                                                                                                    3472/1/2

3.   (a) Simplify:
Permudahkan:
4 x + 2 – 2 2x + 3                                             [ 4 marks/markah ]
(b) Hence, solve the equation:
Seterusnya, selesaikan persamaan;

4 x + 2 – 2 2x + 3 = 64                                        [ 2 marks/markah ]

(a) 2 2(x + 2) – 2 2x + 3             √ P1                   (b) 2 2x + 3 = 2 6 √ P1
2 2x . 2 4 − 2 2x . 2 3 √ K1                               2x + 3 = 6
2 2x (2 4 – 2 3) √ K1                                                  3
x=     √ N1           2
2x                                                                 2
2        (8)
2x + 3                                  4                                           6
2              √ N1

4.   The curve of a quadratic function f(x) = x2 + 2hx – 5 has minimum point of (2, k).
Lengkung bagi satu fungsi kuadratik f(x) = x2 + 2hx – 5 mempunyai titik minimum (2, k).

(a) State the equation of the axis of symmetry of the curve.
Nyatakan pesamaan paksi simmetri bagi lengkung itu.                                  [ 1 mark/markah ]
(b) By using the method of completing the square, determine the value of h and of k.
Dengan menggunakan kaedah penyempurnaan kuasadua, tentukan nilai bagi h dan bagi k.
[ 4 marks/markah ]
(c) Hence, sketch the graph of the curve.
Seterusnya, lakarkan graf bagi lengkung itu.                                         [ 3 marks/markah ]

(a) x = 2 √ N1                                                                         1
(b) f(x) = x2 + 2hx + h2 – h2 – 5 √ K1
= (x + h)2 – h2 – 5 √ K1
h = −2 √ N1
k = −h2 – 5
= −4 – 5
4
= −9 √ N1
f(x)

(c)                                                       √ P1
x
0          2
3
-5
√ P1
(2, −9) √ P1                          8

END OF THE ANSWER & MARKING SCHME
3472/1/2                                                      8                                           SULIT
SULIT                                            3472/1/2
SKEMA PEMARAKAHAN DAN JAWAPAN TAMAT

3472/1/2                    9                     SULIT

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 780 posted: 5/29/2012 language: pages: 9