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Spm 2014 add math modul sbp super score [lemah] k1 set 2 dan skema

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Bahan Pecutan Akhir Add Math SPM

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Spm 2014 add math modul sbp super score [lemah] k1 set 2 dan skema

  1. 1. MODUL SUPER SCORE SBP 2014 Page 1 KERTAS 1 SET 2 NAMA : MARKAH TARIKH : Answer all questions. Jawab semua soalan. 1. Diagram below show a relation between set P and set Q in the graph form. Rajah di bawah menunjukkan hubungan antara set P dan set Q dalam bentuk graf. State Nyatakan (a) the object of 4, objek bagi 4, (b) the range of the relation, julat hubungan tersebut, (c) the type of the relation. jenis hubungan tersebut [3 marks] [3 markah] Answer / Jawapan : For examiner’s use only 3 1 J K L M N 3 2 1 Set P Set Q 4
  2. 2. MODUL SUPER SCORE SBP 2014 Page 2 2. Given the function 6) ( x x f . Find the values of x such that f (x) = 10 Diberi fungsi 6) ( x x f . Cari nilai-nilai x dengan keadaan f (x) = 10. [2 marks] [2 markah] Answer / Jawapan : 3. Given the function f(x) = x – 8 , find Diberi fungsi f(x) = x – 8, cari a) f (4) b) the value of h if 2 ( ) (4) 1 f x  f  nilai h jika 2 ( ) (4) 1 f x  f  [3 marks] [3 markah] Answer / Jawapan : For examiner’s use only 2 2 3 3
  3. 3. MODUL SUPER SCORE SBP 2014 4. Given that the roots of quadratic equation ( 2) 3 1 2 x  a  x  b  are a and b. Find the values of a and b. [3 marks] Diberi bahawa punca-punca persamaan kuadratik ( 2) 3 1 2 x  a  x  b  ialah a dan b. Cari nilai-nilai bagi a dan b. [3 markah] Page 3 Answer / Jawapan : 5. Given the equation 3 ( 2) 25 2 x  b  x  b  has one of the root is negative to the other. Find, Diberi persamaan 3 ( 2) 25 2 x  b  x  b  mempunyai satu punca yang negatif kepada punca yang satu lagi. Cari a) the value of b nilai b b) the roots of the equation punca-punca persamaan [3 marks] [3 markah] Answer / Jawapan : For examiner’s use only 3 5 3 4
  4. 4. MODUL SUPER SCORE SBP 2014 Page 4 6. Diagram shows a graph of a quadratic function y = b ax x    2 2 . Rajah menunjukkan graf fungsi kuadratik y =  x  ax  b 2 2 . Find the values of a and b. [3 marks] [3 markah] Answer / Jawapan : 7. Given that a quadratic function ( ) 2 14 2 f x   x  x  . Find the range of values of x for which f (x)  8 . [2 marks] Diberi satu fungsi kuadratik ( ) 2 14 2 f x   x  x  . Cari julat nilai x bagi f (x)  8 [2 markah] Answer / Jawapan : examiner’s use only 2 For 7 3 6  ( 2, 10) x y 0
  5. 5. MODUL SUPER SCORE SBP 2014 8. Given the quadratic equation ( 1) 3 0 2 x  p x    has no real roots. Find the range of the values of p. [2 marks] Diberi persamaan kuadratik ( 1) 3 0 2 x  p x    tidak mempunyai punca nyata. Cari julat nilai p [2 markah] Page 5 Answer / Jawapan : 9. Solve the equation 1875 5 3 3   x . [3 marks] Selesaikan persamaan 1875 5 3 3   x [3 markah] Answer / Jawapan : 10. Solve the equation 2 3 4 2 3     x x x [3 marks] Selesaikan persamaan 2 3 4 2 3     x x x . [3 markah] Answer / Jawapan : examiner’s use only 2 For 8 3 9 3 10
  6. 6. MODUL SUPER SCORE SBP 2014 11. Given that log2 5 = p and log2 9 = q, express log2 0.12 in terms of p and q . [4 marks] Diberi log2 5 = p dan log2 9 = q, nyatakan log 2 0.12 dalam sebutan p dan q. [4 markah] Page 6 Answer / Jawapan : 12. Solve log 5 log ( 3) 1 0 2 2 x  x    . [3 marks] Selesaikan log 5 log ( 3) 1 0 2 2 x  x    . [3 markah] Answer / Jawapan : examiner’s use only 4 For 11 3 12
  7. 7. MODUL SUPER SCORE SBP 2014 Page 7 13. The sum to infinity of a geometric progression 1, m2, m4, m6, ... is 2. Hasil tambah hingga ketakterhinggaan bagi janjang geometri 1, m2, m4, m6, ... ialah 2. Find Cari (a) the commom ratio in terms of m. nisbah sepunya dalam sebutan m. (b) the positive value of m. nilai positif bagi m. [3 marks] [3 markah] Answer / Jawapan : 14. Given h  3,9,h  5are three consecutive terms of an arithmetic progression. Find the common difference of the progression. [3 marks] Diberi h  3,9,h  5 ialah tiga sebutan berturutan bagi sebuah janjang aritmetik. Cari beza sepunya bagi janjang itu. [3 markah] Answer / Jawapan : For examiner’s use only 13 14 3 3
  8. 8. MODUL SUPER SCORE SBP 2014 15. The fifth term of an arithmetic progression is 22. The difference of the sixth term and the third term is 18. Sebutan kelima bagi suatu janjang aritmetik ialah 22. Beza sebutan keenam dan sebutan ketiga ialah 18. Page 8 Find Cari (a) the first term and the common difference, sebutan pertama dan beza sepunya. (b) the sum of the fourth term to the seventh term. hasil tambah bagi sebutan keempat hingga sebutan ketujuh. [ 4 marks] [4 markah] Answer / Jawapan : ~  4  a  and   16. Given       2   6 ~      p b , find the value of p ~ 4  a  dan   Diberi       2   6 ~      p b , cari nilai bagi p ~ are parallel to each other. a) if a~ and b ~ selari antara satu sama lain. jika a~ dan b ~ are perpendicular to each other. b) if a~ and b ~ berserenjang antara satu sama lain jika a~ dan b [4 marks] [4 markah] Answer / Jawapan : examiner’s use only 4 For 15 4 16
  9. 9. MODUL SUPER SCORE SBP 2014 Page 9 17. Diagram shows two vectors , OA and OB , in a Cartesian plane. Diagram menunjukkan dua vektor , OA danOB , dalam satah Cartesian. Express Ungkapkan   y (a) AB in the form .    x   y AB dalam bentuk .    x ~ ~  . (b) the unit vector in the direction of AB in the form j y i x ~ ~  vektor unit dalam arah AB dalam bentuk j y i x [4 marks] [4 markah] Answer / Jawapan : ~ 5 ~ ~   where k is a constant. ~  7  and b i kj 18. Given vector a i j ~ ~ 2 ~ 5 ~ ~   dengan k ialah pemalar. ~  7  danb i kj Diberi vektor a i j ~ ~ 2 ~ ~  in term of k , i a) Express the vector a b ~ ~ and j ~ ~  dalam sebutan k, i Ungkapkan vektor a b ~ dan j ~ ~ ~ a  b  units, find the values of k. b) If 13 ~ ~ a  b  unit, cari nilai-nilai bagi k. Jika 13 [4 marks] [4 markah] Answer / Jawapan : examiner’s use only 4 For 17 4 18 A(4, 7) B(2,– 3 ) y x 0
  10. 10. MODUL SUPER SCORE SBP 2014 Page 10 Jawapan/Answer : No Answer 1 (a) K and L (b) {1 , 2, 3, 4} (c) many to one 2 4; 16x x 3 a) – 4 b) x = – 10 4 b = 2 , 5  a 2 5 a) b = – 2 b) 3 , – 3 6 a = 8 , b = 2 3 7   x  2 2 8 2 6    p 9 x = 1 10 1.29 11 4 p q  2 q Or 2 p 2  12 x = 2 13 (a) m 2 (b) m = 0.7071 or 1 2  m 14 4 15 (a) a = – 2 , d = 6 (b) 100 16 a) 3 b) – 12 17    a)        2 10 b) 104 ~ 10 ~  2i  j or  i5 j ~ ~ 26 18 ~ ~ 5   a) i (5 k ) j b) 17 or – 7

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