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ADDITIONAL MATHEMATICS SPM 2012 PAPER 1 MY ANSWER & MARKING (This is my own answer and marking scheme and it has nothing to do with any scheme created by anyone else) PAPER 1 1. Diagram 1 shows the relation between set A and set B. State −3 (a) the object of -1. 5 −1 6 1 the relation, 3 (b) the range of the relation. 7 4 Set A Set B Answer: Diagram 1 (a) 5 √ 1 (b) { -3, -1, 1, 3} √1 2. Given that f(x) = 3x + 4 and fg(x) = 6x + 7, find (a) fg(4), (b) g(x). Answer: (a) fg(4) = 6(4) + 7 = 31 √1 (b) f[g(x)] = 3g(x) + 4 f(x) = 3x + 4 3g(x) + 4 = 6x +7 √ M1 fg(x) = 6x + 7 6x + 3 g(x) = 3 = 2x + 1 √2 3. Given that f : x x + 5, find (a) f(3) (b) the value of h such that 2 f -1(h) = f(3), Answer: (a) f (3) = 3 + 5 = 8 √1 Using the principle: (b) f -1(h) = h − 5 √ M1 ax + b c = cx – b a 2(h − 5) = 8 h=9 √2 4. It is given that 3 and m + 4 are the roots of the quadratic equation x2 + (n – 1)x + 6 = 0, where m and n are constants. Find the value of m and of n. Answer: (a) (x – 3) [x – (m + 4)] = 0 x2 – 3x – (m + 4)x + 3(m + 4) = 0 x2 – (m + 7)x + 3(m + 4) = 0 √ M1 Compare to x2 + (n – 1)x + 6 = 0. 3m + 12 = 6 ; – (m + 7) = n -1 m=–2 –5=n–1 √2 n = – 4 √ M1 if only one correct If both correct 5. A quadratic equation x(x – 4) = p – 2q, where p and q are constants, has two equal roots. Express p in terms of q. Answer: x2 – 4x – p + 2q = 0 √ M1 In general form a = 1, b = –4, c = – p + 2q (or) 2q – p b2 – 4ac = 0 (– 4)2 – 4(1)(2q – p) = 0 √ M2 roots’ condition 16 − 8q + 4p = 0 p = 2q – 4 √3 6. Given that f(x) = –3x2 + 2x + 13, find the range of values of x for f(x) ≤ 5 Answer: a = 3 > 0 minimum graph –3x2 + 2x + 13 ≤ 5 –3x2 + 2x + 8 ≤ 0 √ M1 3x2 – 2x – 8 ≥ 0 x Let 3x2 – 2x – 8 = 0 - 4/3 2 (3x + 4)(x – 2) = 0 √ M2 √ M2 x = – 4/ 3 ; x = 2 ? x≤−43 ; x≥2 √3 7. Solve the equation: 27(32x + 4) = 1 Answer: 33 (3 2x + 4) = 1 √ M1 same base 32x + 7 = 1 30 = 1 use law of indices No. 1 Use the concept of the rules of 2x + 7 = 0 √ M2 equation of indices x = − 7/ 2 √3 8. Solve the equation: 1 + log2 (x – 2) = log2 x Answer: √ M1 x √ M1 log2 2 + log2 (x – 2) = log2 x OR 1 = log2 (x – 2) log2 2(x - 2) = log2 x √ M1 2x – 4 = x √M2 x 21 = √ M2 (x – 2) x = 4 √3 Converting log to indices Use the concept of the rules of equation of logarithms. 9. The first three positive terms of a geometric progression are 2, p and 18. Find the value of p and the common ratio of the progression. Answer: p 18 = √ M1 use the concept of common ratio 2 p p2 = 36 p = 6 √ M2 d=3 √3 10. It is given that 11, y + 4 and 3y – x are three consecutive terms of a arithmetic progression. (a) Express y in terms of x. (b) Find the common difference if x = 8 Answer: use the concept of common difference (a) (y + 4) – 11 = (3y – x) – (y + 4) √ M1 y = x – 3 √2 (b) x = 8 then y = 5 The three terms now are: 11, 9, 7 √ M1 d = –2 √2 11. In a geometric progression, the first term is a and the common ratio is r. Given that the third term of the progression exceeds the second term by 12a, find the values of r. Answer: T3 = ar2 use formula Tn = ar n - 1 T3 – T2 = 12a ar2 – ar = 12a a(r2 – r) = 12a √ M1 r 2 – r – 12 = 0 (r + 3)( r – 4) = 0 √ M2 r = – 3, r = 4 √3 12. The variables x and y are related by the equation p q =1– . Diagram 12 shows the straight line graph y x 2 1 1 obtained by plotting against 2 . Find the value of y x (a) p, (b) q. Answer: 1/y 1 1 q y = p − px2 √ M1 (5, 6) 2 1 (a) p =2 p=½ √2 1/x2 0 √ M1 (b) q 6–2 Diagram 12 = gradient = 5–0 q = 2/5 √ 2 p 13. Diagram 13 shows a straight line AB. Find (a) the midpoint of AB, (b) the equation of the perpendicular bisector of AB. Answer: (a) Midpoint AB = ( 15 -1, 3 +7 ) y 2 2 B(15, 7) = (7, 5) √ M1 7–3 (b) MAB = =¼ √ M1 A( –1, 3) 15 + 1 x Mbisector AB = – 4 0 Eqn: y – 5 = – 4(x – 7) √ M2 y = – 4x + 33 √3 Use the concept of m1m2 = – 1 14. Diagram 14 shows a straight line PQ with the equation x y + 10 2k = 1. Determine the value of (a) h (b) k Answer: (a) x - intercept: 5h = 10 h=2 √1 (b) y - intercept: 2k = – 8 k=–4 √1 15. Diagram 15 shows the vectors OA, OB and OP drawn on a grid of equal squares with sides of 1 unit. Determine (a) │OP│, (b) OP in terms of a and b. Answer: a b P B (a) │OP│ = √ + 32 32 A b = 3√2 √ 1 OR 4.243 a O Diagram 15 (b) OP = 2b – a √1 use triangle law 16. The following information refers to the vectors a and b. a = (m 6 4 ) , b = ( 2 ) – 5 It is given that a = kb, where a is parallel to b and k is a constant. Find the value of (a) k (b) m Answer: 6 ( ) =k(2 ) m–4 5 (a) 6 = 2k √ M1 (b) m – 4 = 5k k=3 √1 m = 19 √1 17. Solve the equation tan2θ – 3tanθ + 2 = 0 for 0 ≤ θ ≤ 360o. Answer: (tan θ – 1) (tan θ – 2) = 0 √ M1 tan θ = 1; tan θ = 2 θ = 45o , 135o; θ = 63.43o , 243.43 o √ M2 θ = 45o , 63.43o, 135o , 243.43 o √ 3 18. Diagram 18 shows sectors OAB and ODC with centre O. It is given that OA = 4 cm, the ratio of OA : OD = 2 : 3 and the area of the shaded region is 11.25 cm2. Find (a) the length, in cm, of OD, (b) θ, in radians. Answer: D OD 3 A (a) = √ M1 4 2 O θ OD = 6 √2 B C (b) ½ (6)2 θ − ½ (4)2 θ = 11.25 √ M1 Diagram 18 θ = 1.125 √2 19. Given the function h(x) = kx3 − 4x2 + 5x, find (a) h’(x) (b) the value of k if h”(1) = 4 , Answer: (a) h’(x) = 3kx2 – 8x + 5 √1 basic differentiation (b) h”(x) = 6kx – 8 √ M1 h”(1) = 6k(1) - 8 √ M2 6k – 8 = 4 apply 2nd derivative k=2 √3 20. The gradient of the tangent to the curve y = x2(2 + px) at x = – 2 is 7. Find the value of p Answer: y = 2x2 + px3 dy/dx = 4x + 3px2 √ M1 At x = – 2; 4(– 2) + 3p( – 2)2 = 7 √ M2 12p = 15 p = 5/ 4 √3 7 21. Given that ∫ 2 f(x)dx = 10, Find 2 (a) the value of ∫7 f(x)dx, 7 (b) the value of k if ∫ 2 [ f(x) – k ] dx = 25. Answer: (a) −10 √1 7 7 (b) ∫ 2 f(x) dx – k ∫ 2 dx = 25 √ M1 7 10 – k [ x ] = 25 2 – k [ 7 – 2 ] =15 k = –3 √2 22. The mass of a group of 6 students has a mean of 40 kg and a standard deviation of 3 kg. Find (a) the sum of the mass of the students, (b) the sum of the square of the mass of the students. Answer: manipulating the formula of mean N=6 (a) ∑x = x (N) σ= 3 = 40 (6) = 240 √1 x = 40 ∑x2 manipulating (b) 32 = − 402 √ M1 the formula 6 of variance ∑x2 = (32 + 402) (6) = 6954 √2 23. There are 10 different coloured marbles in the box. Find (a) the number of ways 3 marbles can be chosen from the box. (b) the number of ways at least 8 marbles can be chosen from the box. Answer: 10 Applying the concept (a) C 3 = 120 of combination √ M1 √2 (b) 10 C8 + 10 C9 + 10 C10 √ M1 = 56 √ 2 24. A box contains 20 chocolates. 5 of the chocolates are black chocolates flavour and the other 15 are white chocolates flavour. Two chocolates are taken at random from the box. Find the probability that (a) both chocolates are black chocolates, (b) the chocolates taken are of different flavour. Answer: P(black – 1st round) = 5/20 = ¼ , 5/19 – 2nd round P(white – 1st round) = 15/20 = ¾, 14/19 – 2nd round (a) (¼ x 4/19) √ M1 = 1/19 √2 (b) (¼ x 15/19) + (¾ x 5/19) √ M1 = 15/38 √2 25. In a test, 60% of the students has passed. A sample of 8 students is chosen at random. Find the probability that more the 6 students from the sample passed the test. Answer: Applying the principle of binomial distribution p = 0.6, q = 0.4, n = 8, r > 6 probability 8 8 8 P( r > 6) = C 7 (0.6) 7 (0.4)1 + C8 (0.6) (0.4)0 √ M1 √ M2 √ M1 = 0.1064 √3 THE END Those who have worked hard with the correct method of learning Additional Mathematics surely will receive their rewards

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