TPJC_H2_MATHS_P2_ANS

ANNEX B TPJC H2 Math JC2 Preliminary Examination Paper 2 QN Topic Set Answers 1 Complex numbers (i)Since the coefficients of 3 2 31 212 0 az z z b − + + = are all real , complex roots occur in conjugate pair . Since a cubic equation has three roots , the third root must be a real root. (ii) 25 a = , 190 b = , 19 25 − 2 Vectors (ii) 7 (iii) length of perpendicular from O to AN . 3 Maclaurin series (a) 2 3 3 3 6 ... 2 x x x + + + (b)(ii) 2 a = , 6 b = 4 Application of Integration (a) ( ) 1 e sin e cos 2 x x x x D − + (b)(iii) π 3 1 6 − − 5 DRV (i) 1 36 (ii) 161 36 (iii) 0.112 6 Binomial Expansion (i)0.161 (ii)60 (iii) 3 4 11 11 p ∴ < < 7 Correlation & Linear Regression S

(ii)(a)0.9809 (b)0.9960 (iii) The scatter diagram shows that S increases at an increasing rate as h increases, and for 2 S ch d = + , r ≈ 0.9960 which is closer to 1, so the model 2 S ch d = + is a better model. (iv) S = 0.000182 h 2 + 672 (v)1550 Estimate for when h = 2200 metres is not reliable since h = 2200 metres is outside the range of the g iven data and extrapolation is not a good practice. 8 Normal Distribution (i)0.309 (ii)0.214 (iii)0.303 (iv) 314 9 Hypothesis Testing (i) Every dustbin has an equal probability of being selected and the selections of each dustbin are made independently . (ii) Since 50 n = is large, by Central Limit Theorem , the mean mass of rubbish in dustbins will be approximately normally distributed. (iii) 18.49, 23.6 Since p -value = 0.013937 > 0.01, we do not reject H 0 and conclude that there is insufficient evidence at 1% level of significance to claim that there has been a reduction in the mass of rubbish in dustbins. (iv) 56, n n + ≥ ∈

0 200 400 600 800 1000 1200 1400 0 500 1000 1500 2000

10 P&C, Probability (i) 63 800 (ii) 28 61 (iii) 504 (iv) 3360

H2 Mathematics 2017 Preliminary Exam Paper 2 Solutions 1(i) Since the coefficients of 3 2 31 212 0 az z z b − + + = are all real , complex roots occur in conjugate pair . Since a cubic equation has three roots , the third root must be a real root. 1(ii) Since 1 3i − is a root of 3 2 31 212 0 az z z b − + + = , ( ) ( ) ( ) 3 2 1 3i 31 1 3i 212 1 3i 0 a b − − − + − + = ( ) ( ) ( ) 26 18i 31 8 6i 212 1 3i 0 a b − + − − − + − + = ( ) ( ) 26 460 18 450 i 0 a b a − + + + − = Comparing real and imaginary parts: 26 460 0 a b − + + = ----------- (1) 18 450 0 a − = -----------------(2) From (2), 25 a = , 190 b = ( ) ( ) ( ) ( ) 1 3i 1 3i z z − − − + 2 2 10 z z = − + ( ) ( ) 3 2 2 25 31 212 190 2 10 z z z z z cz d − + + = − + + Comparing coefficient of 3 z : 25 c = Comparing constant: 190 10 d = 19 d = The real root is 19 25 − . 2(i) , , OA OB OC = = + = a a c c uuu r uuu r uuu r