2020_RI_H2_Math_Prelims_P1_Solutions

2020 RI H2 Mathematics Prelim Paper 1 Solutions 1 [4] 2216 9 144xy When 144 16(5) 85: 93xy    Differentiate equation with respect to x: d32 18 0 d yxy x   d 32 16 d 18 9 y x x x y y    d d d d d d x x y t y t   9 216 y x = 9 8 y x For d 0d x t  , x and y have different parity, and so the particle increases with respect to x at 8( 5, ) 3 . [Alternative 1 : for position of particle : Since d ,0d y xt  , particle moves in anti - clockwise direction. Hence for d 0d x t  , y should be negative.] [Alternative 2: for position of particle : differentiate w.r.t t and get dd32 18 0dd xyxy tt . Since dd, , 0dd xy xtt  , y should be negative.] At 85, 3   , d 9 8 3 d8 3 5 5 x t    cms-1 At 85, 3   , its rate of increase is 3 5 cms-1 [Alternative for d d x t , d d d 8 d 332 18 0 32 5 18 (2) 0d d d 3 d 5 x y x xxy t t t t         ] 2 [4] Let x, y and z be the amounts he invested into the 2%, 3% and 5% accounts respectively. 2 30000 ------- (1) 0.02 (1.03 1) 0.05 1423.50 ------- (2) 1000 ------- (3) x y z x y z xz         From GC, solving the 3 equations, 8000, 15000, 7000x y z  

3 (i) [2] dd dd yuy ux x u xx    22d( ) 2 d yyy x y x xx     2 2 2d( ) 2 d uux x x u u u x xx       2 2 2d( ) 2 d uux x x u x xx     2d( 1) 2 0 d uu u x x      2 1d 12d uu ux        22 1d 12 2 d uu u u x          (ii) [3] 22 22 1d 12 2 d 1 d 1 d22 uu u u x u uxuu            2111ln( 2) tan2 22 ux u C        2 111ln 2 tan2 22 yyxC x x        4 (i) [2]     22 2 2 2 22 67 3 9 7 34 x y x xy xy           (ii) [4] 223 9 6(3) 7 16x y y        4y   22 2 2 2 2 2 6 7 3 4 3 4x y x x y x y           Since 223, 3 4x x y    . Volume of solid generated     4 22 4 d 3 2(4)xy       24 22 4 4 2 2 2 4 3 4 d 72 9 6 4 16 d 72 yy y y y                 2 43 4 (2) (425 (6 ) 73 4 2)yy             (7,0) (-1,0)

2 2 128200 48 723 256 483          5(i) [1] 22 1cos d sin 2x x x x c  (ii) [3] 1cos 2 d sin 2 sin 222 dv cos 2d d1 1 sin 2d2 cos 2cos 2 d sin 224 xx x x x x u x x x u vxx xxx x x x c         [3] 222 44 22 44 2 22 4 4 22 2 cos 2 1cos d d 2 11 cos 2 d d22 1 cos 2 1sin 22 2 4 2 2 1 1 1 1002 4 8 2 2 4 16 31 64 8 16 xx x x x x x x x x x x x x x                                           

(a) [2] (b) [3] (c) [3] Note: There is no way to label the y-intercept as there is no information on the gradient of the tangent when x = 0 y ( ,0) x

7 [2]     222f 2 f 2 rr rr rr      