NJC_JC2_H2_Maths_2012_Solutions_Paper_1

2012 NJC H2 Math Prelim P1 Solutions Page 1 of 16 Qn Suggested Solution 1   32f x ax bx cx d    where , , ,a b c d  R . Given  f 0 1 , 1d  .   2f ' 3 2x ax bx c   From the graph of  f 'yx ,    f ' 2 0 12 4 0 1 a b c          f ' 7 0 147 14 0 2a b c         f ' 2.5 9 18.75 5 9 3a b c      OR    f '' 2.5 0 15 2 0 4ab     Using GC to solve (1), (2) & (3): 4 10 56,,27 9 9a b c     324 10 56f1 27 9 9 x x xx     For  f x is concave downwards  f '' 0x , 2.5x . 2(i) 6 3 0 AB OB OA        1 2 1 2 2 6 12 3 1 3 32 2 0 0 AP                                                 Since 1 3AP AB for 1 and A is a common point, this shows that A, B and P are collinear.

2012 NJC H2 Math Prelim P1 Solutions Page 2 of 16 2(ii) Given that area of triangle OAP is 162 5 , 1 162 52 1 11 3 162 523 2 16 11 3 3 162 523 20 OA AP AB                                2 2 2 12 1 3 1 324 5 20 2 1 4 324 5 5 1 2 4 5 324 5 1 45 324 5 1 108 1 108 or 1 108 107 or 109                                            Since P is on BA produced, AP k AB for a negative value of k. Hence 109 . 3 1st part              2 2 3 2 2 2 3 2 3 2 3 23 1 1 1 1 3 1 2 2 2 2 21 1 (2 ) (2 ) (2 ) 2 2 ! 3! f ( ) e 1 2 1111 22 111 22 131 22 x x x x x xx x x x x x x x x x x x x                                   3(a) When 1 ,3x

2012 NJC H2 Math Prelim P1 Solutions Page 3 of 16   21 3 1 9 23 23 1 1 1 1 3 1e 1 2 1 3 3 2 3 2 3 5 1 1 1 3 1 13e1 3 3 2 3 2 3 9                                 11 99 5 13135e 9 e 9 1339   3(b)   29f ' 1 2x x x    Using   29f ' 1 , 2x x x      2 2 2 2 23 23 22 2 ef ( ) 2 e 1 2 12 ef ( ) 2 f ( ) 12 e f ( ) 2 f ( ) 12 d 1 31d 2 2 1321 22 91 2 2 2 51 2 x x x x x x x x x x x x x x x x x x xx x x x x x x x x xx                                Alternative method:      2 2 1 2 3 2 2 2 2 2 e e 1 2 1 2 12 131 1 2 422 11 2 4 2 2 51 2 x x xx x x x x x x x x x x x xx                       

2012 NJC H2 Math Prelim P1 Solutions Page 4 of 16 4(i)       2 2 22 2 2 2 4 2 4 2 dd 2dd d2 2 0 d d d d shownd x ut xu tu ttt ut tu t t ut ut t ut u tt u ut          4(ii) 2 2 2 1 d 1dt 1 ' ' where ' uu tCu t tCx tx C CtC         Since there was 0.2 milligrams of bacteria after 15 minutes, then   2 0.250.2 0.05 0.2 0.06250.25 1 16 CC C      216 16 1 tx t  When t = 4,     2 16 4 256 or 3.9416 4 1 65x    4(iii)