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       22 ( 2) 2 ( 2) rr rr rr    4 .2 8.2 .2 ( 2) r r rrr rr   (3 8)2 (shown)( 2) rr rr   (i) [4]          33 3 8 2 f 2 f2 rnn rr r rrrr     f (5) f (3) +f (6) f (4) +f (7) f (5) f (8) f (6) ... + f ( ) f ( 2) + f ( 1) f ( 1) + f ( 2) f ( ) nn nn nn                12 f ( 1) f ( 2) f (3) f (4) 2 2 16 812 nn nn nn             1 2 .2 4 .2 4.2 16( 1) (2 4 4)2 16( 1) (3 2)2 16( 1) n n n n n nn nn nn nn n nn        (ii) [4]       22 13 3 2 2 3( 2) 2 2 2 ( 2) rrnn rr rr r r r r              22 3 2 3 3 8 2 2 81 4 32 2 rn r rn r r rr r rr            21 (3( 2) 2)2 ( 2)( 1) 1 164 n n nn     1(3 4)2 4( 2)( 1) nn nn   4, 2, 4A B C   

8(a) (i) [3] Let iz x y , ,xy  . Then 2 2 2 2 2 2 22 2 4 2 2 4i 3 ( i ) ( ) 2i 4i 3 4 3 3 4 ( 4)( 1) 0 1 3 24 xy xy z x y x y xy x x x x x x x                            When 1, 2xy . When 1, 2xy  Thus the roots are 1 2i and 1 2i . (a) (ii) [3]   42 22 2 2 6 25 0 (1) 3 16 0 3 4i 4i 3 or 4i 3 zz z z z              or 42 2 6 25 0 (1) 6 36 4(25) 2 8i3 3 4i2 zz z            For 2 4i 3z  , 1 2i, 1 2iz    Since (1) is an equation with real coefficients, the roots occur in conjugate pairs. Thus the roots of the equation 42 6 25 0zz   are 1 2i, 1 2iz    . (b) [4] 2 22 8 2i 8 2i 5 3i 5 3i 5 3i 5 3i 40 10i 24i 6i 53 34 34i 1i34 w               Thus 221 1 2w    and πarg 4w For nw to be real,  2 cos isin 44 nn nnw                 is real. Hence sin 0 , 44 nn kk      , and so 4,n k k  (since n > 0)

9 (i) [4] 2 d1 ln 1 ln , d d e 4 4e e , ln 4 e dee t tt t t t x x t t ttt y tt ty             2d e 4 d e (1 ln ) t t y xt   Now, 0 ln 0 1 ( 0)x t t t t      224 4 e d e 4e ande e d e yy x      Equation of normal at P 24e(0, ) e  : 22 22 4 e e e 4 e e e 4 4 e ey x y x     (ii) [3] 2 2d e 4 0 e 4 0 ln 2d e (1 ln ) t t t y txt        Min occurs at ln 2 ln 2 4ln 2(ln(ln 2)), 4x y e e    (iii) [4] Area   20 20.5ln 0.5 201 120.5ln 0.5 2 e 4 e d4 e e e 4 e 4 d e 1 ln d4 e e e 0.0943 (3s.f.) t t x y x x x t t                   y y x

10 (i) [3] 12 : 1 0 , 23 l                       r Let C be a point on l such that 1 1 2 OC   . 1 1 0 1 2 3 2 1 3 AC                                1 0 2 3 1 0 2 1 3 2                               n 1 3 1 3 3 : . 2 1 . 2 3 2 4 5 . 2 5 2 2 2 2                                              rr (ii) [3] 1 2 1 0 , for some 2 3 OF                       BF OF OB = 6.5 2 4 3       2 . 0 0 13 4 9 0 3 BF         1  6.5 2( 1) 4.5 4 4 3( 1) 3 BF                      (shown) (iii) [2] Shortest distance from B to 1 = length of projection of BF onto 1n = 1ˆBFn = 4.5 3 42 3 2 17                  = 23 2 17 units (iv) [3] Let  be the acute angle between 1 and 2

23 2 17sin 23 2 17 BF BF    1 2 2 2 23sin 24.5 (1d.p) 2 17 4.5 4 3      [Alternatively, 2 2 4.5 2 0 4 0 3 1 33 12 9.5 8 BF                                    n Let  be the acute angle between 1 and 23 2 17 11 3 12 2 . 19.5 2 8 91cos cos 24.5 (1d.p) 17 588.25 17 588.25                        ] 11 (i) [2] By trigo ratio, sin , cosEX a AX a 12 ( cos )( sin ) ( sin )2V a a a a a     3 sin (1 cos )a  [Alternatively, by area of trapezium  1 ( sin ) ( 2 cos )2V a a a a a    3 sin (1 cos )a  ] (ii) [5]       3 3 2 2 3 3 d cos (1 cos ) sin ( sin )d cos cos sin cos cos 2 32 cos cos Factor Formulae22 V a a a a                   d3 0 cos 0 or cos 0d 2 2 3 or2 2 2 or (NA)3 V             2 F B E A X a Alternatively:     3 3 2 2 3 2 2 32 3 d cos (1 cos ) sin ( sin )d cos cos sin cos cos 1 cos 2cos cos 1 (2cos 1)(cos 1) V a a a a a                              d1 0 cos or cos 1d2 or (NA)3 V       

    3 3 2 3 2 3 d cos cos2d dAt , s 33 in 2sin 23d 32 2 3