NJC JC2 H2 Maths 2012 Solutions Paper 2

2012 NJC H2 Math Prelim Paper 2 Solutions Page 1 of 14 Qn Suggested Solutions 1(a) (i)    i 3 1 i 3 1 i 3 2        z w z z z z       arg( ) arg 1 i 3 arg 1 i 3 arg 3         z w z z Alternatively, Given that iez  ,where 0 and 0 2  , and i3wz , find iez  i 3 3wz  By vector addition, z w OA OB   By Pythagoras Theorem,   22 2 3 4 2 zw         1(a) (ii)     1 3arg arg tan 3 z w z          Real axis Im axis O A B z+w 

2012 NJC H2 Math Prelim Paper 2 Solutions Page 2 of 14 1(b) (i) 1 i 3 3z   is a circle with centre  1, 3 and radius 3 units.         arg 2 i4 3 3 arg 2 arg i2 3 3 arg i2 3 3 z z z             arg i2 3 3z  is a half-line with starting point at  0, 2 3 that makes angle of 3  with the positive real axis. Note that 1. the half line passes through the centre of circle. 2. The real axis is a tangent to the circle. 1(b) (ii) Let A denote the value of z that gives the greatest possible value of arg(z). iA x y 23 Re(z) Im(z) O 1 3 Locus of z 3  23 Re(z) Im(z) O 1 3 A 1x  3y 3 3 

2012 NJC H2 Math Prelim Paper 2 Solutions Page 3 of 14 3 3 3sin 32 33 3 32 yy y        1 1 1cos 32 33 3 12 xx x        33 1 i 322       z 2(a) x = distance of the car from the intersection y = distance between the car and truck By Cosine Rule, y2 = x2 + 482  2x(48)cos 120o y2 = x2 + 48x + 2304 (shown) Differentiating implicitly with respect to time, 2y d d y t = 2x d d x t + 48 d d x t The rate of change of the car is d d x t = −60 because it is traveling toward the intersection. When x = 15 y2 = x2 + 2304 + 48x y2 = 152 + 2304 + 48(15) = 3249  y = 57 2(57) d d y t = (2 (15) + 48) d d x t d d y t = – 41 1 19 = –41.1. (3sf) Hence, the distance between the car and the truck is decreasing at a rate of 41.1 km/h. 2(b) (1st part) Cost for roof = $P per unit area; Cost for curved surface area and base = $3P per unit area. Total cost = $C. Area = 2222 r r rh   (where h is the height of the cylinder)  27 ( ) 2 (3 )C r P rh P  C (7 6 )rP r h

2012 NJC H2 Math Prelim Paper 2 Solutions Page 4 of 14  766 C rhrP  7 66 Crh rP 322 3V r r h  3227  3 6 6      Crrr rP 3 327 3 6 6 Cr rr P    3 62 Cr r P  2(b) (2nd part) 2d3 0d 6 2 VC rrP    r 9 C P 2 2 d d V r = 3 r < 0 when r = 9 C P , the volume is maximum. Thus, cost of top 2 rP 9 C PP    = 9 C 3(a)       2 22 21 21 3 d49 2411 dd2 4 9 25 1 1 2ln 4 9 tan or 2 55 1 1 2ln 4 9 tan2 55                   x xxx x xxxx x xx x c xx x c

2012 NJC H2 Math Prelim Paper 2 Solutions Page 5 of 14 3(b)           2 2 d2 d 1 3 d 2 2 d ln 1 3d 1 3 3 Since the curve has gradient value of 2 at the point(0,1), then we have 22 ln1 23 d2 ln 1 3 2d3 2 ln 1 3 2 d3 23 ln 1 3 d 23 1 3 2 ln 1 3 13 y xx y x x cxx cc y xx y x x xx x x x D x xx                                             1 d 213 21 ln 1 3 ln 1 3 233 21 ln 1 3 ln 1 3 233 Since curve passes through the point 0,1 , we have 211 0 0 ln 1 0 133 21 ln 1 3 ln 1 333 x x Dx x x x x x D x x x x x D DD y x x x x                                      21 2 1 4 ln 1 3 13 3 3 x x x x       4(i) 24yx is the equation of a semicircle centred at (0,0) with radius 2. Hence 2 22 2 14 d Area of semicircle (2) 2 2      xx 4(ii) Solving   22 24xy   and 12 xy , we have Consider

2012 NJC H2 Math Prelim Paper 2 Solutions Page 6 of 14 2 2 2 2 2 2 1 4 2 2 1 442 5 304 5 4 12 0 xx xxx xx xx                 2 For 0, 5 4 12 0 62 or (rejected) 5 x xx x        Alternative by symmetrical properties, the other intersection is at 2. x 2 For 0 5 4 12 0 62 or (rejected)5 x xx x        4(ii) 1st part     2 22 2 2 22 22 2 2 22 2 Volume 1 2 4 d 2 2 1 4 4 4 4 d42 5 7 4 4 d (shown)4 x xx xx x x x x x x x                            4(ii) 2nd part   2 22 2 2 2 2 22 2 2 2 2 023 20 2 0222 2 20 2 23 5Volume 7 4 4 d 4 5 7 d d 4 4 d 4 5 7 d d +4 2 12 64 83 2 2 64 483 528 unit3 x x x x x x x x x x x x x x x x xx                                                           

2012 NJC H2 Math Prelim Paper 2 Solutions Page 7 of 14 5(i) Let n and N denote the sample size and population size respectively. 100% 2% 2 100 100 502 n N n N Nk n       Randomly select a starting point in the first 50 concert goers in the queue and then pick every 50th concert goer thereafter. 5(ii) There may not be sufficient information to categorize the audience members in relevant strata/find out the numbers of people in each strata. 6(i) r-value = −0.9728201266 = −0.973 (3 sig. fig.) 6(ii) (A) y ax b : r-value = −0.973 (B) 2y cx d : r-value = −0.999 (C) y e x f : r-value = −0.930 As x increases, y decreases at an increasing rate and (B) has r-value closest to −1, (B) is the best model. 6(iii) Regression line of y on x2 is 2 2 0.358974359 13.22512821 0.359 13.2 yx yx     When x = 3.8, y x

2012 NJC H2 Math Prelim Paper 2 Solutions Page 8 of 14   2 0.358974359 3.8 13.22512821 8.041538466 8.04 (3 sig. fig.) y    Since x = 3.8 is within the data range given, and r-value indicates a strong negative linear correlation between y and x2, the estimate is reliable. 7(a) (i) 8 letters 2 N's 2 A's 8!Number of ways 100802! 2! 7(a) (ii) Number of ways Number of ways without restrictions number of ways that each pair is together 8! 6!2!2! 9360     7(a) (iii) Number of ways 54! 4!4 2!2! 720    7(b) (i) Number of ways 4344 3222 432                7(b) (ii) Case 1: 3 trumpet players, 3 saxophone players 4 4 7Number of ways 560333           Case 2: 3 trumpet players, 4 saxophone players 4 4 7Number of ways 84 342           Case 3: 4 trumpet players, 3 saxophone players 4 4 7Number of ways 84 432          

2012 NJC H2 Math Prelim Paper 2 Solutions Page 9 of 14 Case 4: 4 trumpet players, 4 saxophone players 4 4 7Number of ways 7 4 4 1           Total number of ways = 560 + 84 + 84 + 7 = 735 8 8(i) P(has second CCA | first CCA is Music and Dance) = 0.25 8(ii) P(has second CCA) = 45 40 1 15 1 100 100 4 100 2p                = 9 1 3 9 7 20 10 40 20 40 pp    9 7 13 20 40 40 1 3 p p   8(iii) P(first CCA Clubs and Societies | no second CCA) =     P first CCA Clubs and Societies no secon d CCA P no second CCA  = 15 1 3 1100 2 40 13 27 91 40 40    Sports Music & Dance Clubs & Societies 2nd CCA No 2nd CCA 0.45 0.40 0.15 p 1 – p 2nd CCA No 2nd CCA 0.25 0.75 2nd CCA No 2nd CCA 0.50 0.50

2012 NJC H2 Math Prelim Paper 2 Solutions Page 10 of 14 8(iv)         2 P P P P 40 15 13 40 1 15 122100 100 40 100 4 100 2 3 169 3 25 1600 200 337 1600 A B A B A B                                       9 (i) Let μ be the mean distance Kelly can throw her javelin. H0: μ = 55 H1: μ > 55 Level of significance