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1.0 modul super score kertas 1 set 1

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1.0 modul super score kertas 1 set 1

  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. 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. 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. 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. 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. 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. 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. 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. 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. 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

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