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                 