YJC_H2_MATH_P2_ANS

1 YISHUN JUNIOR COLLEGE 2015 JC2 PRELIMINARY EXAM PAPER 2 H2 MATHEMATICS SOLUTION Qn Solution 1a cosar  and sinbr  Therefore    i cos sin i cos isina b r r r         Let P be the point on the Argand diagram representing the complex number iab . r is the distance of the point P to the origin.  is the directed angle made by the line segment joining origin to point P, with the positive real axis. Cartesian equation of locus: 22 9xy 1b Given cos isinz  By De Moivre’s Theorem   111 cos isin cos isinzz          OR 1i1 e cos isinzz      OR 1 1 cos isin cos isin cos isinz        2 2 2 cos isin cos i sin     22 cos isin cos sin     cos isin Im( )z iP a b  a b r O Re( )z

2 OR *2* 1 cos i1 sinzzz z z     1 cos isin cos isinz z         2cos (Shown) 1 cos isin (cos isin )z z         2isin (Shown) 2 2 1 1 1 1 zzzz z z z z                Hence 2 2 1 2cos icos or icot1 2isin sin z z       2i 24yx and 28( )y k x 4 8( ) 2( )y y k y y k     2yk 2 4(2 ) 8x k k 22xk Diameter of the rim of the bowl  2 2 2 2kk   or 2 2 2k 42 k (Shown) 2ii Capacity of the bowl 2 8( ) d k k y k y 2 218 2 k k y ky   2 2 2 24182 22k k k k     24 k Volume of material used for the bowl 2 2 0 4 d 4 k y y k 222 0 24 k yk  228 0 4kk   24 k  capacity of the bowl 2iii Given 4k  Required area 22 22 0 112 4 d 84 k x x x  

3 42 2 0 12 4 d 8 xx 30.16988933 30.170 (3 d.p.) 3i C: 2 7 2 x qxy x   , 0q 232 2 qy x q x      or const2 2y x q x     Asymptotes: 2x and 2y x q   3ii If yx is an asymptote of C, then 2x q x   Thus 2 0 2qq    (Shown) 3iii   2 2 27 1 2 16 xx xx    ---- (*) 2 227 162 xx xx   yx y x O 2x Graph of C (0,3.5) ( 1.83,0) (3.83,0)

4 x-coordinates of points of intersection: 0.170, 3.98 Solving (*), 4 0.170x   or 2 3.98x 4i g( ) lnxx , 0x 1h( )x x , 0x Any line yk where k , cuts the graph of g at most once g is a 11 function and therefore function 1g exists. Let g( ) lny x x 1g ( ) e yyx  1g ( ) e xx , 1 ggDR  and 1 ggRD  4ii Since gh RD     , composite function hg does not exist. x g( )yx O (1,0) yk y 216yx yx y x O 2x Graph of C (0,3.5) ( 1.83,0) (3.83,0)

5 4iii 11gh( ) g(h( )) g ln lnx x x xx      gh hDD  gh : lnxx  , 0x ghR  4iv 2 11h ( ) h 1xx xx     9 2 2 2 2 1h ( ) h(h h h h ( )) h( )x x x x   4