2020_RI_H2_Math_Prelims_P2_Solutions

2020 RI H2 Mathematics Prelim Paper 2 Solutions Section A: Pure Mathematics [40 marks] 1 [5] Given that 1 3 ,2u  let the common difference of the arithmetic series be d and the common ratio of the geometric series be r. 2 2 3 3 3 21and (1)2 2 2 2 60 ( 3)( 2) 0 2, since r is non negative. rr rr rr r            333 322 dr -----------------(2) Substitute r = 2 into (2), 7 .2d  Sum of first 10 odd numbered terms of AP 10 32 9(7) 33022S    2 [5] 4 4 8 8 8 4 ( 1) (1)1 ( 1) (2)1 (2) 1 17 (1) 1 16 arS r arS r r r          44 4 4 12 11 17 1 16 1 1 1 ,16 2 2 rr r r r r        When 11 1 ,2 12 1 2 ar S a    When 22 12, 123 1 2 br S b    Ratio is a : 1 3 b or 3a : b

3(i) [4]  gR , 3   fD 0, Since gfRD , fg does not exist.  fR 1, 4  gD , 4  Since fgRD , gf exists.       gfR g 2 , g 4 1, 3   (ii) [4] Largest possible value of c is 1. For 1x , 11xx       g 1 3x x x          2 1 13 12 2 1 1 g 2 1 , , 0. x x y xy x y x x x x x                4 [2]     1 1 1 1 2 2 2 2 tan 2 tan 2 tan 2 tan 2 e d 1 1ed2 1 2 2e 4 d4 2e d d4 2 (Shown)d x x x x y y x x x yx x yxy x                   

4 (i) [5]         2 2 2 2 3 2 2 2 2 3 2 2 2 3 2 2 2 3 2 2 d4 2 ---------- (1)d d d d4 2 2 ---------- (2)d d d d d d d d4 2 2 2 2d d d d d d d d d4 4 2 2 ---------- (3)d d d d yxy x y y yxx x x x y y y y yx x x x x x x x y y y yxx x x x x              When 0, 1.xy From equations (1), (2), (3) we get 23 23 d 1 d 1 d 1, , .d 2 d 4 d 8 y y y x x x   By Maclaurin’s Theorem, 1 23 23 tan 2 1 1 1e1 2 4 2! 8 3! 1 1 11 ...2 8 48 x xxx x x x                       (ii) [3] 1 23tan 2 1 1 1e 1 ... 2 8 48 x x x x                 1 1 22 2 22 2 2 2 tan 2 tan 2 e 1 1 1 ... 1281 111 ... 1 2 3 ...28 111 2 1 3 ...28 e 3 17 1 ...281 x x x x x x x x x x xx xx x                                        

5(i) [2] 2 3OC a , 3OD b (ii) [4] E on BC :  1OE OB OC   E on AD :  1OE OA OD      21 1 3 3OE           b a a b 3 3 (1) 2 2 3 (2)     (1) + (2): 2 3 1    4 3OE ab (iii) [3] Area of CDE is 1 1 2 2 32 2 3 3 1 2 2 3 since 2 3 3 1 4 2 2 3 3 DC CE                            a b a b a b a b a a b b 0 a b a b (iv) [3] 4 3OE ab is the diagonal of the rhombus with sides 4 3 a and b . So 43 34     aab b , i.e. :4: 3OA OB  Alternatively,     22 4 cos cos 44 33 44 cos cos33 44 cos cos33 4 1 cos 1 cos , 1 cos 03 3 i.e. :4 3: AOE FOE OE OE OA OB                           