YJC_H2_MATH_P1_ANS

1 YISHUN JUNIOR COLLEGE 2015 JC2 PRELIMINARY EXAM PAPER 1 H2 MATHEMATICS SOLUTION Qn Solution 1 Let x, y, and z be the number of trays of blueberry, strawberry and chocolate cupcakes respectively. Time: 8 7 6 17 60 1020x y z     Amt: 0.6 0.6 0.8 96x y z   Price: 12 (1) 12 (0.9) 12 (0.8) 1572x y z   Using GC, 50, 50, 45x y z   2(a) If , a b 0 and , then 0ab implies that the two vectors a and b are perpendicular to each other i.e. ab . 2(b) If a lies in axisx and m a 0 , then m is parallel a , and hence is parallel to i . Since 1m then mi or i (just one will do) 2(c) Method 1: Let the diagonals BD and AC intersect at E. Given DE=EB -------(1) and AE=EC-----------(2) (*) (*) [from (1) and (2)] = AB AE EB DC DE EC EB AE AB    AB DC i.e. AB DC and //AB DC Therefore ABCD is a parallelogram. (Proven) Method 2: Using Ratio Theorem,     11 22OE OA OC OD OB    OA OC OD OB   OB OA OC OD   AB DC i.e. AB DC and //AB DC Therefore ABCD is a parallelogram. (Proven) 2(d) OP p x , OQ q y and OR r sxy Since ,P ,Q and R are collinear, PQ kPR for some k .    q p k r s p k r p ks      y x x y x x y ()k r p p  ----- (1) ks q ----- (2)

2 (1) (2) : r p p sq   rq pq ps  Therefore ps rq pq (Shown) 3 External volume =   2( 1) 1rh  Internal volume= 2 1000rh  2 1000h r   2( 1) 1 1000V r h    2 2 1000( 1) 1 1000Vr r       (Shown) 2 23 d 1000 20002 ( 1) 1 ( 1)d V rrr r r                  OR 2 2 2 3 2000 1000 d 2000 20002 2 2 d VV r r r r r r r r              For stationary ,V d 0d V r  2 2 3 2000 2000 2000( 1) 2 0r r r r         OR 23 2000 20002 2 0 r rr    i.e. 3 2000( 1) 2 0r r     OR    4 3 3 1000 1000 0 1 1000 0r r r r r          Since 3 20001 0 2r r    OR 31 0 1000rr     3 3 1000 10r    and 2 3 2 3 1000 1000 10 100h r        2 2 3 4 d 2000 60002 ( 1)d V rr r r               = 21.6105 > 0 when r = 3 10  OR 2 2 3 4 d 4000 60002 0 , 0d V rr r r      minimum for r    R OR r 3 10     3 10  3 10     d d V r negative 0 Positive Slope Therefore V is minimum when 3 10rh   cm.

3 4 2 2 dd dd yy xx  dd ddd yy xxx d d y yCx   , C d1 lnd x x C y Dy C y     , D eex D xC y A      , eD A  e xC y B  , 0B e xy C B  When 0x , 0y and d 1d y x  0 CB   i.e. CB and 1 0 1 CC    Therefore f ( ) 1 e xyx    4

4 5i 2 f ( )yx 5ii 1 f ( )y x