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 ( ) ( ) 5 2 5 2 1 5 3 2 OX OA AX OA AC = + = + = + − = − a c a c a uuu r uuu r uuu r uuu r uuu r By midpoint theorem: 2 OB OX ON + = uuu r uuu r uuu r ( ) ( ) 1 1 5 3 2 2 1 7 4 ON   = + + −     = − uuu r a c c a c a 2(ii) Area of triangle OAB = 1 2 OA OB × uuu r uuu r Alternatively: ( ) 2 3 5 5 3 2 1 5 3 2 OX OA OC OC OA OX OX + = − = = − uuu r uuu r uuu r uuu r uuu r uuu r uuu r By Ratio Theorem: c a

( ) ( ) 1 4 2 1 0 2 = × + = × × = Q % a a c a c a a 8 ⇒ × = a c Area of triangle OAN = 1 2 OA ON × uuu r uuu r ( ) ( ) ( ) square u 1 1 7 2 4 nit 7 0 8 7 8 8 s 7 = × − = × × = = = Q % a c a a c a a 2(iii) AN OA AN × uuu r uuu r uuu r is the length of perpendicular from O to AN . Alternative answer: AN OA AN × uuu r uuu r uuu r is the shortest distance from O to AN . AN OA AN × uuu r uuu r uuu r is the area of a parallelogram formed with vector OA uuu r and unit vector AN uuu r as its adjacent sides. (Not recommended here) Area of triangle 7 OAN = 1 7 2 AN OA AN AN × = uuu r uuu r uuu r uuu r ( ) 14 14 14 1 7 4 56 7 5 AN OA AN AN ON OA × = = − = − − = − uuu r uuu r uuu r uuu r uuu r uuu r c a a (shown) c a 3(a) ( ) 2 e ln 1 3 x x + ( ) ( ) ( ) 2 2 3 2 3 3 1 2 ... 3 where 1 3 1 2! 2 3 x x x x x x     = + + + − + − − < ≤             K ( ) 2 2 3 9 1 2 2 ... 3 9 ... 2 x x x x x   = + + + − + −    

2 3 2 3 3 9 3 9 6 9 6 ... 2 x x x x x x = − + + − + + 2 3 3 3 6 ... 2 x x x = + + + where 1 1 3 3 x − < ≤ 3(b)(i) 3 π 3 π sin sin π 2 4 4 QR PR θ =   − −     3 π π sin sin 2 4 4 QR PR θ =   −     3 π sin 4 π π sin cos 2 cos sin 2 4 4 QR θ θ = − 1 2 1 1 cos 2 sin 2 2 2 QR θ θ = − QR = 1 cos 2 sin 2 θ θ − (shown) 3(b)(ii) When θ is small, QR ( ) 2 1 2 1 2 2! θ θ ≈   − −       2 1 1 2 2 θ θ = − − ( ) ( ) 1 2 1 2 2 θ θ − = − + ( ) ( ) 2 2 2 1 2 2 2 2 ... θ θ θ θ = + + + + + 2 2 1 2 2 4 ... θ θ θ = + + + + 2 1 2 6 ... θ θ = + + + 2 a = , 6 b = 4(a) ( ) e sin e sin e cos e sin e cos e sin e sin e cos e sin e sin e sin e cos e sin e sin e sin d d d d Hence, e cos 1 e sin e sin e cos 2 d d 2 d d x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x C x x x x x x x x x C x x D −   − +     = − − = = + = − − = − + = − + ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫

4(b)(i) Area of first rectangle, : k x n = ( ) ( ) 2 2 2 1 2 2 2 2 2 2 3 2 3 2 3 2 k k k k n n A n n nk k n n nk k k k n n n = ⋅ = = − − − − − − 4(b)(ii) Area of second rectangle, 2 : k x n = ( ) ( ) 2 2 2 2 2 2 2 3 2 (2 ) (2 ) 2 2 3 2 k k k n A n n n n k k k k n n = ⋅ = − − − − Area of third rectangle, 3 : k x n = ( ) ( ) 2 3 2 2 2 3 3 3 2 (3 ) (3 ) 3 3 3 2 k k k n A n n n n k k k k n n = ⋅ = − − − − By observation, combined area of n rectangles: 2 2 2 2 1 2 , 3 n r rk A n n nrk r k = = − − ∑ where 2 1 a and b = = 4(b)(iii) ( ) 2 2 2 2 1 3 1 2 0 3 1 2 0 3 1 3 1 2 2 0 0 3 1 2 1 0 3 d 3 2 1 ( 2 2 ) 1 2 d 3 2 Area under curve from 0 2 2 1 d d 3 2 4 ( 1) 3 2 1 sin 1 2 2 to 3 1 1 2 1 2 r rk n n anrk br k x x x x x x x x x x x x x x x x x x x ∞ = − − − − − − √ − − − − − − − − = − − − − − − − − +     − − +     = − = = =     − = =      −  − ∑ ∫ ∫ ∫ ∫ 3 1 0 3 1 3 1 2 1 0 0 1 1 1 3 2 sin 2 3 1 3 sin sin 2 2 π π 3 3 6 π 3 1 (exact) 6 x x x − − − − − −     +     = − − − −                 = − 1− − −               = −1− + = − −

5(i) 6 1 P( ) 1 3 5 7 9 11 1 1 36 r X r k k k k k k k = = = + + + + + = = ∑ 5(ii) E( ) 1( ) 2(3 ) 3(5 ) 4(7 ) 5(9 ) 6(11 ) 161 161 36 X k k k k k k k = + + + + + = = 5(iii) Required Probability 2 2 P({6,6,4}) P({6,5,5}) 11 7 3! 11 9 3! 36 36 2! 36 36 2! 1738 869 0.112 (3 s.f.) Accept: 15552 7776 = +         = +                 = = 6(i) Let X be the number of rocks containing fossils out of 20 rocks. B(20, 0.07) X  P( 3) 1 P( 2) 0.161 (3 s.f.) X X ≥ = − ≤ = 6(ii) Let Y be the number of rocks containing fossils out of 20 rocks. B( , 0.07) Y n  P( 3) 0.8 Y ≥ ≥ Method 1a: Using GC Table n P( 3) Y ≥ 59 0.79085 < 0.8 60 0.80023 > 0.8 61 0.80925 > 0.8 Hence, least 60. n = Method 1b: Using GC Table P( 2) 0.2 Y ≤ ≤ n P( 2) Y ≤ 59 0.20915 > 0.2 60 0.19977 < 0.2 61 0.19075 < 0.2 Hence, least 60. n = Method 2: Using the binomial distribution function 1 2 2 P( 2) 0.2 P( 0) P( 1) P( 2) 0.2 ( 1) 0.93 (0.07)(0.93) (0.07 )(0.93) 0.2 2 n n n Y Y Y Y n n n − − ≤ ≤ = + = + = ≤ − + + ≤ Using GC to sketch the graph: Hence, least 60. n = 6(ii