AJC_H2_MATH_P1_soln

Anderson Junior College Preliminary Examination 2015 H2 Mathematics Paper 1 (9740/01) Qn Solution 1 2 2121 x xx     2 2 1 2 1 021 x x x x      243 021 xx x        22 1 4 3 0x x x       2 1 4 3 1 0x x x    31 42 x   or 1x (ans) 22 22 2 1 2 2 xx xx  2 2 2 12 2 12 1 x x x     2 3 1 1 42 x    or 2 1 1 x   2 110 2x  or 2 1 1 x   2 2x  or 2 1x   22xx  or or 1 1, 0xx    2 2 2 d 5 d 5dd y x y yxx y x     2 2 2 2 dd21 dd yyyy xx   3 2 2 3 2 2 2 3 d d d d d d2 2 2 0d d d d d d y y y y y yy y yx x x x x x        3 23 2 23 d d d d2 6 0d d d d y y y y yyx x x x      When x = 0, y = 5 d1 d5 y x 22 22 1 d d 310 25 125 d d 125 yy xx      3 33 33 1 1 3 d d 42 30 25 05 5 125 d d 625 yy xx                  23 23 1 3 45 5 125(2!) 625(3!) 1 3 2 5 5 250 1875 y x x x x x x            1 1/2 -3/4

3(i) 2arg( 4) 2 2arg( 4) 2 arg( 4)2              z z z 3(ii) From diagram, 8AB z AC   2 2 2 21 4 8 3 4 2z      17 8 7z   3(iii) maximum arg( 8)z 1132tan sin45 2.9096 2.91rad (3sf)       4 C: 12sinyx  L: 8 3yx   C & L intersect at 1 ,23   And y-intercept of L is -. Volume obtained when S is rotated 2 radians about the y-axis = 211 3 2 3 æ ö æ ö÷÷çç +÷÷çç÷÷ççè ø è ø - 2 3 0 sin 2 y dy   éù æö÷çêú ÷ò ç ÷çêú èøëû = 4 12 3 æö ÷ç ÷ç ÷çèø - 3 0 1 cos 2 y dy   -ò Re(z) Im(z) 2 (4,0) (4,3) 2 5 (8,0) A B C (8,0) 2 5 L C y x

= 2 9  - [ ]3 0sin2 yy   - = 2 9  - sin2 3 3   éùêú-êúëû = 2 9  - 2 3 6 2 2  éùêú+ êúêúëû = 23 4 18  - 5 By sine rule,     5 66 1 2 55 66 1 2 31 22 21 2 2 3 sin sin( ) 3 sin cos cos sin 3 cos sin 23 since is small 2 1 2 3 23 (shown) 2 2 3 AB AB                   Applying binomial expansion,       121 2 22211 22 221 2 2 3 1 3 3 1 3 3 3 1 3 3 7 3 7 33 3 3, 3, 22 AB a b c                              6(i) Given that 2sin xx   2 sin x xee  since xye is increasing  sin 2 110 xxe e   22 2 sin 00 x xe dx e dx    6(ii)    22 0 sinsin uxe dx e du     Let 1duux dx    =   2 sin 0 ue du    since sin (-u) = sin u C A B θ

6(iii) sin 0 xe dx    = 2 2 sin sin 0 xxe dx e dx     = 2 2 sin 0 xe dx    from the result in (ii) < 2 2 2 0 x e dx    from the result in (i) = 2 22 0 2 x e