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Spm 2014 add math modul sbp super score [lemah] k1 set 4 dan skema

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Spm 2014 add math modul sbp super score [lemah] k1 set 4 dan skema

  1. 1. MODUL SUPER SCORE SBP 2014 1 KERTAS 1 SET 4 NAMA MARKAH TARIKH Answer all questions. Jawab semua soalan. 1. The relation between set P and set Q is defined by the set of ordered pairs {(1, a), (1, c), (3, a),(4, b),(4, d), (6, d)}. Hubungan antara set P dan set Q ditakrifkan oleh set pasangan bertertib {(1, a), (1, c), (3, a),(4, b), (4, d), (6, d)}. State Nyatakan (a) the image of 4, imej bagi 4, (b) the type of the relation, jenis hubungan tersebut, (c) the relation in the graph form. hubungan dalam bentuk graf. [3 marks] [3 markah] Answer/ Jawapan : (a) (b) (c) For examiner’s use only 3 1
  2. 2. MODUL SUPER SCORE SBP 2014 2 2. The following information refers to the function f and g. Maklumat berikut merujuk kepada fungsi f dan g. f : x  4x – 6 g : x  2x – 3 Find gf–1(x). Cari gf–1(x). [3 marks] [3 markah] Answer/ Jawapan : 3. Given the function h(x) = 4x + 5 and the composite function hg(x) = 8x + 7. Diberi fungsi h(x) = 4x + 5dan fungsi gubahan hg(x) = 8x + 7. Find Cari (a) hg(–1) (b) g(x) [3 marks] [3 markah] Answer/ Jawapan : For examiner’s use only 3 2 3 3
  3. 3. MODUL SUPER SCORE SBP 2014 3 4. Find the roots for the quadratic equation x(x – 3) = 9 – x. Give your answer correct to three significant figures. Cari punca-punca bagi persamaan kuadratik x(x – 3) = 9 – x. Beri jawapan anda betul kepada tiga angka bererti. [3 marks] [3 markah] Answer/ Jawapan : 5. It is given that 4 and m + 1 are the roots of the quadratic equation x2 + (2 – n)x + 8 = 0, where m and n are constants. Find the value of m and n. Diberi bahawa 4 dan m + 1 ialah punca-punca bagi persamaan kuadratik x2 + (2 – n)x + 8 = 0, dengan keadaan m dan n ialah pemalar. Cari nilai m dan n. [3 marks] [3markah] Answer/ Jawapan : For examiner’s use only 3 5 3 4
  4. 4. MODUL SUPER SCORE SBP 2014 4 6. The diagram shows the graph of a quadratic function y = –(x – 4)2 – 1. Rajah menunjukkan graf fungsi kuadratik y = –(x – 4)2 – 1. Find Cari (a) the value of k, nilai k, y (b) the equation of the axis of symmetry, persamaan paksi simetri, (c) the coordinate of the maximum point. koordinat titik maksimum. [3 marks] [3 markah] Answer/ Jawapan : 7. The quadratic function f(x) = p(x + q)2 + r, where p, q and r are constants, has maximum value of 10. The equation of axis of symmetry is x = 2. Fungsi kuadratik f(x) = p(x + q)2 + r, dengan keadaan p, q dan r ialah pemalar, mempunyai nilai maksimum 10. Persamaan paksi simetri ialah x = 2. State Nyatakan (a) the range of value of p, julat nilai p, (b) the value of q, nilai q, (c) the value of r. nilai r. [3 marks] [3 markah] Answer/ Jawapan : 3 7 3 6 –3 (k, –3) 0 x For examiner’s use only
  5. 5. MODUL SUPER SCORE SBP 2014 5 8. Find the range of values of x for x2 – 2(2x + 1)  2x2 – x. Cari julat nilai-nilai x bagi x2 – 2(2x + 1)  2x2 – x. [4 marks] [4markah] Answer/ Jawapan : 9. Solve the equation(27x)x × 1 x 243 = 9. Selesaikan persamaan (27x)x × 1 x 243 = 9. [4 marks] [4markah] Answer/ Jawapan : 10. Solve the equation3x– 1 +3x + 2= 2. Selesaikan persamaan 3x – 1 + 3x + 2 = 2. [3 marks] [3markah] Answer/ Jawapan : For examiner’s use only 4 8 4 9 3 10
  6. 6. MODUL SUPER SCORE SBP 2014 6   9a 11. Given that loga 2 = x and loga 3 = y, express loga   8 in terms of x and y.   a9 Diberi loga 2 = x dan loga 3 = y, nyatakan loga     8 dalam sebutan x dan y. [4 marks] [4 markah] Answer/ Jawapan : 12. Solve the equation 2 + log4 x = log22x. Selesaikan persamaan 2 + log4 x = log2 2x. [4 markah] [4 marks] Answer/ Jawapan : examiner’s use only 4 For 11 4 12
  7. 7. MODUL SUPER SCORE SBP 2014 7 13. In a geometric progression, the first term is 1 2 and the fourth term is 4  . 27 Dalam suatu janjang geometri, sebutan pertama ialah 1 2 dan sebutan keempat ialah 4  . 27 Calculate Hitung (a) the common ratio, nisbah sepunya, (b) the sum to infinity of the geometric progression. hasil tambah hingga sebutan ketakterhinggaan bagi janjang geometri itu. [3 marks] [3 markah] Answer/ Jawapan : 14. The 3rd term of a geometric progression is 18. The sum of the 3rd term and 4th term is 27. Sebutan ke-3 suatu janjang geometri ialah 18. Hasil tambah sebutan ke-3 dan sebutan ke-4 ialah 27 Find Cari (a) the first term and the common ratio of the progression, sebutan pertama dan nisbah sepunya janjang itu, (b) the sum to infinity of the progression. hasil tambah hingga sebutan ketakterhinggaan janjang itu. [4 marks] [4 markah] Answer/ Jawapan : For examiner’s use only 13 14 3 4
  8. 8. MODUL SUPER SCORE SBP 2014 8 15. The nth term of a geometric progression, Tn, is given by Tn = 2 2 2 3      n , find Sebutan ke-n bagi suatu janjang geometri, Tn, diberi oleh Tn = 2 2 2 3      n , cari (a) the first term and the common ratio, sebutan pertama dan nisbah sepunya. (b) the sum to infinity of the progression. hasil tambah hingga sebutan ketakterhinggaan bagi janjang itu. [ 3 marks] [3 markah] Answer/ Jawapan : 16. The diagram shows two vectors, OA and OB. Rajah menunjukkan dua vektor, OA dan OB. Express Ungkapkan   y (a) AO in the form       x ,   y OA dalam bentuk       x , (b) AB in the form xi + yj. AB dalam bentuk xi + yj. [3 marks] [3markah] Answer/ Jawapan : examiner’s use only 3 For 15 3 16 y x  A(4, 10)  B(2, –6 ) O
  9. 9. MODUL SUPER SCORE SBP 2014 ~ ~ use only 9   12 17. Given p =       5  m 1   and q =       3 , find   12 Diberi p =       5  m 1   dan q =       3 , cari (a) | p |, (b) the value of m such that p + q is parallel to y-axis. nilai m dengan keadaan p + q adalah selari dengan paksi y. [4 marks] [4 markah] Answer / Jawapan : 18. The diagram shows the vectors OA, OB and OP drawn on a grid of equal squares with sides of 1 unit. Rajah menunjukkan vektor OA, OB dan OP dilukis pada grid segiempat sama yang sama besar bersisi 1 unit. Determine Tentukan (a) | OP |, (b) OP in terms of a and b. ~ ~ OP dalam sebutan a and b. [2 marks] [2 markah] Answer/ Jawapan : For examiner’s 4 17 2 18 ~ ~ ~ A B P a ~ b ~ ~ ~ O
  10. 10. MODUL SUPER SCORE SBP 2014 10
  11. 11. MODUL SUPER SCORE SBP 2014 11 Jawapan/Answer : No Answer No Answer 1 (a) b and d (b) many to many (c) 16    (a)        4 10 (b) –2i – 16j 2 x 2 17 (a) 13 (b) –6 3 d c b (a) –1 (b) 2x + 1 2 18 (a) 20 (b) 2b – a 4 4.16 and –2.16 5 m = 1 and n = 8 6 (a) 8 (b) x = 4 (c) (4, –1) 7 (a) p < 0 (b) q = –2 (c) r = 10 8 –2  x  –1 9 x = 2, 1  3 10 x = –1.4022 11 3x – 2y – 1 12 x = 0 and x = 4 13 (a) 2  3 (b) 3 10 14 (a) a = 72, r = 1 2 (b) 144 15 (a) a = 3, r = 2 3 (b) 9 1 3 4 6 Set P Set Q a

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