AJC_H2Maths_2012Prelim_P2_soln

AJC Preliminary Examination 2012 H2 Mathematics Paper 2 (9740/02) Solution Section A: Pure Mathematics (40 marks) 1 DE = 2 2sin 2   = 4sin 2  Perimeter = 4x2 + 2 [ 4sin 2  ] = 8[1 4sin ]2  Unit cost = 2 28000 8[1 sin ] =   1 23500 1 sin    1 2 2 3500 1 3500(1 ...) 3500 1750        2 (i) f(4) = f(-2) = -f(2) = -1. (ii) (iii) 7 5 f ( ) dxx   =   3 2 1 14 4 1 32 x dx       = 33 1 (3 )42 3 x      = 48 [0 8]3 = 56 3 3 line l : 21 32 2 3 3 3 20 yx , z r                          (i) 3 2 1 1 3 3 3 2 2 0 CD k                                            Hence k = -1,  = 0 and  = 2 (ii) Equation of y-z plane is x = 0. At the point of intersection F of the line l and the y-z plane, 2-=0  =2. Hence F (0, 9, 2) (iii) By observation, 0 3 2 OE   and hence 2 '3 2 OC   E D /2 2 2 Alternative, use cosine rule - 4 4 3 6 9 1

Alternative method: Let E be the foot of  from C to y-z plane 1 00 0 r    Line CE: 2 1 2 3 0 3 2 0 2 r                              . When line CE meets y-z plane: 21 3 0 0 2 20 r                     . 2 1 0 3 ( 2) 0 3 2 0 2 OE                                By mid-point theorem, ' 2 OC OCOE   0 2 2 ' 2 2 3 3 3 2 2 2 OC OE OC                                 4 (a) 2 2 24( ) cos sindy x y x xdx   Using u x y , 11du dy dy du dx dx dx dx     2 2 21 4 cos sindu u x xdx   2 2 24 cos 1 sindu u x xdx      221 4 cosdu uxdx   2 11 1 cos 21 4 2 du x dxu   214 1 1 1 1 cos 242 du x dxu  11 1 12tan 2 sin 24 2 2 u x x c        1 1tan 2 sin 2 2x y x x c      (b)(i) 2dx R kxdt  , k is a positive constant At 2 , 0 dxxR dt 20 (2 )R k R 0 (1 4 )R kR   1 4k R 2 4 dx x Rdt R 22 4 14 R dx dtRx  2ln 2 Rx tCRx   22 ,22 C t t CR x R x e e Ae A eR x R x      

At 0, 0 1t x A    Hence 21 221 t t t R x e e x RR x e       (b)(ii) 11 2 1 1 t t exR e     , 1As , 0 2 tt x R e    [or use graphical method] The amount of drug in the body will not exceed 2R mg regardless of the period of treatment. 5 3 12 1i 212 z a    3 21arg( ) 1 4 2 1 2za a      (i)   i233 41 1 i e 2 k zz       , k = 0, +1, -1 2i 12 3e , 0, 1 k zk 