A Level 2016 P2 Solutions for new syllabus (edited CSC)

2016 H2 Maths Paper 2 SECTION A Qn Solution 1 Let h m and V m3 be the depth and volume of water respectively at time t minutes. Given: d 0.1 (1)d V t   and 5.0tan  d d d (2)d d d h h V t V t   , so we find V in terms of h first. tan 0.5 0.5 rh    2311 3 12(0.5 )V h h h 2 21 4 d (3)d4 Vh hh    Sub (1) & (3) into (2): 22 d 4 0.4 0.1d h t h h   When 3V  ,   3 1 3361 12 3hh     , so d 0.0251 (3 s.f.)d h t  The rate of increase of the depth of water is 0.0251 m per minute. 2(a)(i) 2 2 sin 2cos d sin d x nxx nx x x nx x nn 2 cos cossin d (1) d cos 1 cos d cos 1 sin x nx nxx nx x x nn x nx nx xnn x nx nxnn           2 2 2 2 23 sin 2 cos 1cos d sin sin 2 cos 2sin where is an arbitrary constant x nx x nxx nx x nx n n n n x nx x nx nx ccn n n            h r  Integration by parts 2 d d d sin d cos 2 vux x u nx vx x n nx    d d d co sin d 1 s vux x u n nxvx x n    

(a)(ii) 2 3 2 3 22 22 2 2 22 23 224 sin2 4 cos2 2sin2 sin 2 cos 2sin 4 cos2 2 cos 4 (1) 2 (1) cos d sin 2 cos 2sin ( the other terms are 0) (when is even) n n n n n nnn n n n n nn nn nn x nx x x nx x nx nx n n n n                                      22 22 4 (1) 2 ( 1) 26 or (when is odd) or nn nn n    (b) 29 x xxy   22 2 2 )9( )( x xxy            2 2 0 32 20 2 5 29 5 29 5 12 9 51 9 2 2 9 52 5 4 529 Volume of solid d d 9 99 1. d 29 9 d2 9d ln 9 ln 5 ln 9 1 ln yx x x x uu uu u u uu u u u uu                               3 (i) ttx cos , ty cos1 , for 02 t  When D meets the x-axis, 0 cos 1 0 or 2y t t t       1x or 12  x d d d sin d d d 1 sin y y x t x t t t    At the maximum point, d 0 sin 0d y tx    0, , 2t  When 0t or 2 , we get the points on the x-axis. When t , we get the maximum point. cos 1x       , 1 cos 2y    Note:       sin 0 cos 2 1, cos 2 1 1 n nn       29 x xxy  Let 2299u x x u     d 2 2 9d 1d d 29 u xux xu u      When x = 2, u = 5; x = 0, u = 9.

(ii) 1 10 dArea required d d d xa xy x y t t     0 0 0 1 2 1 4 11 44 3 1 44 (1 cos )(1 sin ) d 1 cos sin sin 2 d sin cos cos 2 sin cos cos 2 1 sin cos cos 2 a a a t t t t t t t t t t t a a a a a a a a                                (iii) At P, where 2t  , 2x  , 1y and 2 2 sind1 d 1 sin 2 y x   . Equation of normal at P,  212yx    21yx    At E, 0y , 1 2xa  At F, 0x , 1yb     Area of triangle OEF is      21111 2 2 2 2 1ab      units2 4(b)(ii) 2 2iw 222 ( 2) 8w     , 1 2 24tan    i 48ew          1 ii i 44 44 * 8e 8e 8e n n n n ww  