HCI_JC_2_H2_Maths_2012_Paper 1 Solution - for Sharing

1 2012 HCI H2 Math Prelim Paper 1 Marking Scheme Qn. Solutions 1 2 2 ( 2)(2 5) ( 5 3)(1)f ( ) ( 2) ' x kx kx xx x       22 2 2 5 4 10 5 3 ( 2) kx x kx kx x x        2 2 47 ( 2) kx kx x   Since f is an increasing function, 2 2 47f ( ) ( 2) ' 0kx kxx x    2 4 7 0kx kx    for all values of x Hence we have 0k  (linear) or for 0k  , 2 ( 4 ) 4 (7) 0kk   (quadratic) 2 16 28 0kk 4 (4 7) 0kk  Hence 70 4 k . 2 (i) and (ii) 2iz OQ a 3 ln lny xy x Differentiate implicitly w.r.t x, M O Re Im ia iaa Locus of z 4  B A a C Q

2 1 d d ln dd 1yy xy x x y y x x x      1 d d ln ln dd yy y x x y x y x x   1d ln ln d yx x y y x yx   22 d (1 ln ) (1 ln ) d 1 ln 1 ln y y x y x x xy x y    For the tangent to be parallel to the y–axis, d 0d x y  2 1 ln 0(1 ln ) y yx   1 ln 0y ey Hence from xyyx , we have ee xx e e0xx   Hence x = 1.32, and since the tangent parallel to the y–axis takes the form xc , hence equation of tangent is 1.32x 4(a)       3i i 1 2i 3i i i i 1 2i 3i 3 i i 2 3i 3 (1 )i (2 ) Comparing Re and Im parts, 3 2 1 131 4 1 i4 zz x y x y x y x y x y x y y y y x x x z                              

3 (b) 12 1 1 4 4 1 1 i2 2 2 4 1 3 1 ppz z z p p            5(i)               1 2 1 2 2 1 2 2 222 22 2 1 ln 1 2 21 2 ... 2 1 2 2 13 1 221 2 2 2 2 ...22 31 4 ... 8 51 ... 2 yx xx xx x x x x x x x xx                                       (ii)     1 1 1 2 1 and ln 1 2 1 1 1 e 1 e 1and2 2 2 2 e 1 1 OR 0.316 0.522 xx xx xx                     (iii) 2 2 2 0 0 5 20 d 1 d 6.67 OR 23y x x x x              Since the integration in  0,2 does not fall into the valid range of 0.316 0.5x   , the approximation is not good. 6(a) Assume the x th day to be the day with maximum amount of goods delivered. For the first x days: ( a = 1000, d = 100) [2(1000) ( 1)(100)]2 (1900 100 )2 x xSx x x     For the remaining (15- x) days: ( a = 1000+ (x - 1)100 – 100 = 800+100x, d = -100)

4 15 (15 ) [2(800 100 ) (15 1)( 100)]2 15 (200 300 )2 x xS x x x x         Since total goods delivered is 21300 tons, (1900 100 )2 x x + 15 (200 300 )2 x x  = 21300 2 31 198 0xx   ( 9)( 22) 0xx   x = 9 or x = 22 (NA, since x  15) Therefore, 9th June was the day with max goods delivered. Goods delivered = 1000 + (9  1)(100) = 1800 tons (b) (i) Series H is a GP with common ratio r. (1 ) 1 narH r   or ( 1) 1 narH r   C is a GP with common ratio 1/r. 1 11(1 ) 11 1 11 nn n rarC ar r r     1 1 1 ( 1) ( 1) 11 nn n n n H a r ar r C r r a ar uu     