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MODUL	
  SUPER	
  SCORE	
  SBP	
   2014	
  
©Panel	
  Perunding	
  Mata	
  Pelajaran	
  Matematik	
  Tambahan,	
  	
   Page	
  1	
  
KERTAS 1
SET 1
NAMA : MARKAH
TARIKH :
Answer all questions.
Jawab semua soalan.
1. The diagram shows the relation between set X and set Y.
Rajah menunjukkan hubungan di antara set X dan set Y.
State /Nyatakan
(a) The range of the relation
Julat hubungan itu
(b) The value of x
Nilai x
[2 marks]
[2 markah]
Answer / Jawapan :
2. Given the function g : x → 5−x . Find the values of x if g(x) = 4. [2 marks]
Diberi fungsi g : x → 5−x . Cari nilai-nilai x jika g(x) = 4. [2 markah]
Answer / Jawapan :
For
examiner’s
use only
2
2
2
1
x g(x)
– 4
x
1
4 6
3
2
– 2
x
Set X Set Y
MODUL	
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  2	
  
3. Given the functions f(x) = 4x – m and
16
9
)(1
+=−
kxxf , where k and m are constants. Find the
values of k and m. [3 marks]
Diberi fungsi f(x) = 4x – m dan
16
9
)(1
+=−
kxxf , dimana k dan m adalah pemalar. Cari nilai-
nilai bagi k dan m. [3 markah]
Answer / Jawapan :
4. Diagram shows a graph of a quadratic function f(x) = ‒2(x + h)2
‒ 2 where k is a constant.
Rajah menunjukkan graf fungsi kuadratik f(x) = ‒2(x + h)2
‒ 2 dimana k ialah pemalar.
Find
Cari
(a) the value of k
nilai k
(b) the value of h
nilai h
(c) the equation of axis of symmetry.
persamaan bagi paksi simetri.
[3 marks]
[3 markah]
Answer / Jawapan :
For
examiner’s
use only
3
3
3
4
x
0
(-3, k) •
f(x) = −2(x + h)2
− 2
y
MODUL	
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5. Find the values of p if the quadratic function f(x) = 2x2
+ 2px – (p + 1) has a minimum value of
– 5 [3 marks]
Cari nilai-nilai bagi p jika fungsi kuadratik f(x) = 2x2
+ 2px – (p + 1) mempunyai nilai minimum
– 5
[3 markah]
Answer / Jawapan :
6. Find the range of values of x for xx 624)4( 2
−<− [2 marks]
Cari julat nilai x bagi xx 624)4( 2
−<− [2 markah]
Answer / Jawapan :
For
examiner’s
use only
2
6
3
5
MODUL	
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7. One of the roots of the quadratic equation 032 2
=−− kxx is – 4. Find the value of k.
[2 marks]
Satu dari punca persamaan kuadratik 032 2
=−− kxx ialah – 4. Cari nilai k. [2 markah]
Answer / Jawapan :
8. One of the roots of the equation 3x2
– 6x + p = 0 is three times the other root , find the possible
values of p. [3 marks]
Salah satu punca bagi persamaan 3x2
– 6x + p = 0 adalah tiga kali punca yang satu lagi, cari
nilai yang mungkin bagi p. [3 markah]
Answer / Jawapan :
3
8
2
7
For
examiner’s
use only
MODUL	
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9. Solve the equation 06216 42
=− +− xx
. [3 marks]
Selesaikan persamaan 06216 42
=− +− xx
[3 markah]
Answer / Jawapan :
10. Solve the equation 2x
• 5x +2
= 25000. [3 marks]
Selesaikan persamaan 2x
• 5x +2
= 25000. [3 markah]
Answer / Jawapan :
3
9
3
10
For
examiner’s
use only
MODUL	
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   2014	
  
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11. Solve the equation log2
(x – 3) = log2
4x + 1 [3 marks]
Selesaikan persamaan log2
(x – 3) = log2
4x + 1 [3 markah]
Answer / Jawapan :
12. Given that log2
x = m and log2
y = n. Express log4
(xy2
) in terms of m and n. [3 marks]
Diberi log2
x = m dan log2
y = n. Nyatakan log4
(xy2
) dalam sebutan m dan n. [3 markah]
Answer / Jawapan :
lum
3
11
4
12
For
examiner’s
use only
MODUL	
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13. Find the sum to infinity of the geometric progression 20, 10, 5, ... [2 marks]
Cari hasil tambah ketakterhinggaan janjang geometri 20, 10, 5, ... [2 markah]
Answer / Jawapan :
14. Given a geometric progression has the first term and the sum to infinity are 25 and 62.5
respectively. Find the common ratio of the progression. [2 marks]
Diberi satu janjang geometri mempunyai sebutan pertama dan hasil tambah hingga
ketakterhinggaan adalah 25 dan 62.5 masing-masing. Cari nisbah sepunya bagi janjang
tersebut. [2 markah]
Answer / Jawapan :
2
14
2
13
For
examiner’s
use only
MODUL	
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   2014	
  
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15. Write 0.01010101... as a single fraction in the lowest terms.
[3 marks]
Tulis 0.0101010... sebagai satu pecahan tunggal dalam sebutan terendah.
[3 markah]
Answer / Jawapan :
16. The diagram below shows two vectors OP and OQ.
Rajah di bawah menunjukkan dua buah vektor OP dan OQ.
Express
Ungkapkan
(a) OP in the form ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
y
x
.
OP dalam bentuk ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
y
x
.
(b) PQ in the form jyix
~~
+
PQ dalam bentuk jyix
~~
+
[4 marks]
[4 markah]
Answer / Jawapan :
3
15
4
16
For
examiner’s
use only
P(– 2 , 5)
Q(4 , – 3 )
x
y
MODUL	
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17. Given ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
=
3
4
h , ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−
=
0
2
k and ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=+
m
kha
6
, find the values of a and m. [3 marks]
Diberi ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
=
3
4
h , ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−
=
0
2
k dan ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=+
m
kha
6
, cari nilai bagi a dan m. [3 markah]
Answer / Jawapan :
18. Points A, B and C are collinear. It is given that 6 4AB a b= −
uuur
% %
and 4 (2 )BC a k b= + +
uuur
% %
, where k is
a constant. Find
Titik A, B dan C adalah segaris. Diberi bahawa 6 4AB a b= −
uuur
% %
dan 4 (2 )BC a k b= + +
uuur
% %
, dengan
keadaan k adalah pemalar. Cari
(a) the value of k
nilai k
(b) the ratio AB : BC
nisbah AB : BC
[4 marks]
[4 markah]
Answer / Jawapan :
3
17
4
18
For
examiner’s
use only
MODUL	
  SUPER	
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   2014	
  
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  Perunding	
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   Page	
  10	
  
Jawapan/Answer :
No Answer
1
(a) {– 2, 2, 3, 6}
(b) x = 0
2 x = 1, x = 9
3 k =
4
1
, m =
4
9
4
(a) k = – 2
(b) h = 3
(c) x = – 3
5 – 4, 2
6 42 <<− x
7 k = 44
8
2
1
=α ,
4
9
=p
9 x = 5
10 x = 3
11 x =
7
3
−
12
2
2 mn +
13 40
14 0.6
15
99
1
16
(a) ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−
5
2
(b)
~~
86 ji−
17 a = 2 , m = – 6
18
(a) k =
3
14
−
(b) AB : BC = 3 : 2

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1.0 modul super score kertas 1 set 1

  • 1. MODUL  SUPER  SCORE  SBP   2014   ©Panel  Perunding  Mata  Pelajaran  Matematik  Tambahan,     Page  1   KERTAS 1 SET 1 NAMA : MARKAH TARIKH : Answer all questions. Jawab semua soalan. 1. The diagram shows the relation between set X and set Y. Rajah menunjukkan hubungan di antara set X dan set Y. State /Nyatakan (a) The range of the relation Julat hubungan itu (b) The value of x Nilai x [2 marks] [2 markah] Answer / Jawapan : 2. Given the function g : x → 5−x . Find the values of x if g(x) = 4. [2 marks] Diberi fungsi g : x → 5−x . Cari nilai-nilai x jika g(x) = 4. [2 markah] Answer / Jawapan : For examiner’s use only 2 2 2 1 x g(x) – 4 x 1 4 6 3 2 – 2 x Set X Set Y
  • 2. MODUL  SUPER  SCORE  SBP   2014   ©Panel  Perunding  Mata  Pelajaran  Matematik  Tambahan,     Page  2   3. Given the functions f(x) = 4x – m and 16 9 )(1 +=− kxxf , where k and m are constants. Find the values of k and m. [3 marks] Diberi fungsi f(x) = 4x – m dan 16 9 )(1 +=− kxxf , dimana k dan m adalah pemalar. Cari nilai- nilai bagi k dan m. [3 markah] Answer / Jawapan : 4. Diagram shows a graph of a quadratic function f(x) = ‒2(x + h)2 ‒ 2 where k is a constant. Rajah menunjukkan graf fungsi kuadratik f(x) = ‒2(x + h)2 ‒ 2 dimana k ialah pemalar. Find Cari (a) the value of k nilai k (b) the value of h nilai h (c) the equation of axis of symmetry. persamaan bagi paksi simetri. [3 marks] [3 markah] Answer / Jawapan : For examiner’s use only 3 3 3 4 x 0 (-3, k) • f(x) = −2(x + h)2 − 2 y
  • 3. MODUL  SUPER  SCORE  SBP   2014   ©Panel  Perunding  Mata  Pelajaran  Matematik  Tambahan,     Page  3   5. Find the values of p if the quadratic function f(x) = 2x2 + 2px – (p + 1) has a minimum value of – 5 [3 marks] Cari nilai-nilai bagi p jika fungsi kuadratik f(x) = 2x2 + 2px – (p + 1) mempunyai nilai minimum – 5 [3 markah] Answer / Jawapan : 6. Find the range of values of x for xx 624)4( 2 −<− [2 marks] Cari julat nilai x bagi xx 624)4( 2 −<− [2 markah] Answer / Jawapan : For examiner’s use only 2 6 3 5
  • 4. MODUL  SUPER  SCORE  SBP   2014   ©Panel  Perunding  Mata  Pelajaran  Matematik  Tambahan,     Page  4   7. One of the roots of the quadratic equation 032 2 =−− kxx is – 4. Find the value of k. [2 marks] Satu dari punca persamaan kuadratik 032 2 =−− kxx ialah – 4. Cari nilai k. [2 markah] Answer / Jawapan : 8. One of the roots of the equation 3x2 – 6x + p = 0 is three times the other root , find the possible values of p. [3 marks] Salah satu punca bagi persamaan 3x2 – 6x + p = 0 adalah tiga kali punca yang satu lagi, cari nilai yang mungkin bagi p. [3 markah] Answer / Jawapan : 3 8 2 7 For examiner’s use only
  • 5. MODUL  SUPER  SCORE  SBP   2014   ©Panel  Perunding  Mata  Pelajaran  Matematik  Tambahan,     Page  5   9. Solve the equation 06216 42 =− +− xx . [3 marks] Selesaikan persamaan 06216 42 =− +− xx [3 markah] Answer / Jawapan : 10. Solve the equation 2x • 5x +2 = 25000. [3 marks] Selesaikan persamaan 2x • 5x +2 = 25000. [3 markah] Answer / Jawapan : 3 9 3 10 For examiner’s use only
  • 6. MODUL  SUPER  SCORE  SBP   2014   ©Panel  Perunding  Mata  Pelajaran  Matematik  Tambahan,     Page  6   11. Solve the equation log2 (x – 3) = log2 4x + 1 [3 marks] Selesaikan persamaan log2 (x – 3) = log2 4x + 1 [3 markah] Answer / Jawapan : 12. Given that log2 x = m and log2 y = n. Express log4 (xy2 ) in terms of m and n. [3 marks] Diberi log2 x = m dan log2 y = n. Nyatakan log4 (xy2 ) dalam sebutan m dan n. [3 markah] Answer / Jawapan : lum 3 11 4 12 For examiner’s use only
  • 7. MODUL  SUPER  SCORE  SBP   2014   ©Panel  Perunding  Mata  Pelajaran  Matematik  Tambahan,     Page  7   13. Find the sum to infinity of the geometric progression 20, 10, 5, ... [2 marks] Cari hasil tambah ketakterhinggaan janjang geometri 20, 10, 5, ... [2 markah] Answer / Jawapan : 14. Given a geometric progression has the first term and the sum to infinity are 25 and 62.5 respectively. Find the common ratio of the progression. [2 marks] Diberi satu janjang geometri mempunyai sebutan pertama dan hasil tambah hingga ketakterhinggaan adalah 25 dan 62.5 masing-masing. Cari nisbah sepunya bagi janjang tersebut. [2 markah] Answer / Jawapan : 2 14 2 13 For examiner’s use only
  • 8. MODUL  SUPER  SCORE  SBP   2014   ©Panel  Perunding  Mata  Pelajaran  Matematik  Tambahan,     Page  8   15. Write 0.01010101... as a single fraction in the lowest terms. [3 marks] Tulis 0.0101010... sebagai satu pecahan tunggal dalam sebutan terendah. [3 markah] Answer / Jawapan : 16. The diagram below shows two vectors OP and OQ. Rajah di bawah menunjukkan dua buah vektor OP dan OQ. Express Ungkapkan (a) OP in the form ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ y x . OP dalam bentuk ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ y x . (b) PQ in the form jyix ~~ + PQ dalam bentuk jyix ~~ + [4 marks] [4 markah] Answer / Jawapan : 3 15 4 16 For examiner’s use only P(– 2 , 5) Q(4 , – 3 ) x y
  • 9. MODUL  SUPER  SCORE  SBP   2014   ©Panel  Perunding  Mata  Pelajaran  Matematik  Tambahan,     Page  9   17. Given ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = 3 4 h , ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛− = 0 2 k and ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =+ m kha 6 , find the values of a and m. [3 marks] Diberi ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = 3 4 h , ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛− = 0 2 k dan ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =+ m kha 6 , cari nilai bagi a dan m. [3 markah] Answer / Jawapan : 18. Points A, B and C are collinear. It is given that 6 4AB a b= − uuur % % and 4 (2 )BC a k b= + + uuur % % , where k is a constant. Find Titik A, B dan C adalah segaris. Diberi bahawa 6 4AB a b= − uuur % % dan 4 (2 )BC a k b= + + uuur % % , dengan keadaan k adalah pemalar. Cari (a) the value of k nilai k (b) the ratio AB : BC nisbah AB : BC [4 marks] [4 markah] Answer / Jawapan : 3 17 4 18 For examiner’s use only
  • 10. MODUL  SUPER  SCORE  SBP   2014   ©Panel  Perunding  Mata  Pelajaran  Matematik  Tambahan,     Page  10   Jawapan/Answer : No Answer 1 (a) {– 2, 2, 3, 6} (b) x = 0 2 x = 1, x = 9 3 k = 4 1 , m = 4 9 4 (a) k = – 2 (b) h = 3 (c) x = – 3 5 – 4, 2 6 42 <<− x 7 k = 44 8 2 1 =α , 4 9 =p 9 x = 5 10 x = 3 11 x = 7 3 − 12 2 2 mn + 13 40 14 0.6 15 99 1 16 (a) ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛− 5 2 (b) ~~ 86 ji− 17 a = 2 , m = – 6 18 (a) k = 3 14 − (b) AB : BC = 3 : 2