1. 3 6 3 3x Term in x C 3 2 2 Note: 3 (M1)(A1)(A1)(A1) Award M1 for recognizing Binomial Theorem and A1 for each correct element. 27 x 3 8 = 20 8 A1 = 540 x 3 A1 (Coefficient of x = 540) 3 2. 3 3 3 2 3 3 2 3 3 3 2 2 1 [6] 2 2 3 M1A1 = 3 3 18 12 3 8 = 15 3 26 (M1)A1 (accept a = 15, b = 26) A1A1 N3 [6] 3. (a) (b) 5 5 4 3 2 2 3 4 5 (2 + x) = 2 + 5(2) (x) + 10(2) x + 10(2) x + 5(2)x + x (M1)(A1) = 32 + 80x + 80x2 + 40x3 + 10x4 + x5 (A1) (C3) Note: Award (C2) for 5 correct terms, (C1) for 4 correct terms. Let x = 0.01 = 10–2 (2.01)5 = 32 + 0.8 + 0.008 + 0.000 04 + 0.000 0001 + 0.000 000 0001(M1)(A1) = 32.808 040 1001 (A1) (C3) [6] 4. 8 – 1 3 The coefficient of x3 is 3 2 (M2) (A2) The coefficient of x3 is –7 (A2) (C6) [6] 5. The total number of four-digit numbers = 9 × 10 × 10 × 10 = 9000. The number of four-digit numbers which do not contain a digit 3 = 8 × 9 × 9 × 9 = 5832. Thus, the number of four-digit numbers which contain at least one digit 3 is 9000 – 5832 = 3168. (A1) (A1) (A1) (C3) [3] 6. METHOD 1 Consider the group as two groups – one group of the two oldest and one group of the rest Either one of the two oldest is chosen or neither is chosen Then the number of ways to choose the committee is 6 2 6 2 3 1 4 0 = 40 + 15 = 55 ways METHOD 2 The number of ways to choose a committee of 4 minus the number of ways to have both the oldest 8 6 4 2 = 70 – 15 = 55 ways (M2) (M1) (M1)(M1) (A1) (C6) (M3) (M1)(M1) (A1) (C6) [6] 1 7. 9 The first car can be filled in C 3 ways. The second car can be filled in The third car can be filled in Number of combinations = 9 3 6 M1A1 C 3 ways. A1 C 3 ways. (A1) C 3 C 3 C 3 = 84 20 1 = 1680. 6 3 M1A1 N4 [6] 2