RI 2020 P2 A-Level H2 Math Solution

1 | P a g e Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student 2 2 Let d 2d 2 when 1 2 d 0 when 1 2 0d d 5 when 2 4 5d 5 1Using GC, , 5, 2 2 5 1 52 2 y ax bx c y ax bx y x a b c y x a bx y x a bx a b c y x x                             Alternative:     2 2 Let 2 1 for some constant d 2 1d d 55 when 2 2 5d 2 5 1 22 y k x k y k xx y x k kx y x                  Do ensure you get everything correct for such straight forward question Raffles Institution H2 Mathematics (9758) Solution for 2020 A-Level Paper 2

2 | P a g e Question 2 No. Suggested Solution Remarks for Student (a)(i) (A) Using GC, it can be observed that the sequence strictly increases without bound. While some may notice that the difference of the terms form a GP 2, 4, 8, … Key word is increasing here. (i) (B) Using GC, it is a constant sequence.

3 | P a g e (ii)  1 1 12 5 5 2n n n nu u u u      From GC, 5 11u  Alternative method (Without using GC):     1 2 1 3 2 4 3 5 4 2 5 2 5 2 5 2 2 5 5 4 15 2 5 2 4 15 5 8 35 2 5 2 8 35 5 16 75 101 11 u p u u p u u p p u u p p u u p p p                             (b)(i)    1 2 3 1 2 4 2 3 4 3 , 2 7 2 7 2 7 2 2 7 7 2 5 21 Since 2 , 2 5 21 2 2 7 7 v a v b v v v a b v v v b a b a b v v a b a b b                           Alternatively, 4 2 3 3 2 3 2 2 7 2 2 7 7 7 v v v v v v v b        

4 | P a g e (ii)   1 2 3 1 2 5 3 4 , 2 7 2 7 7 since 7 2 7 7 2 2 14 7 5 28 v a v b v v v a b a b v v v a a a                      (c)(i)                  1 3 23 2 3 23 2 3 3 2 2 2 2 2 1 2 For 2, term 11 4 1 11 1 4 1 1 11 1 4 1 3 3 1 11 2 1 4 3 3 1 22 11 4 3 25 16 1st term 1 11 4 6 3 25 16 term 3 2 th n n th n n S S n n n n n n n n n n n n n n n n n n n n n n n n S n n                                                                    5 16n Note that the 1st term, 6, satisfies the expression for the nth term. (ii)     3 23 2 3 3 2 11 4 3 11 3 4 3 60 11 4 60 0 2,3 or 10 10 since 3 mS S m m m m m m m m m                GC Plysmlt 2 can be used to solve the cubic equation.

5 | P a g e Question 3 No. Suggested Solution Remarks for Student (i)   d d6 , 6d d d 1 d d 111 6 1 11 2, d 2 Cartesian equation of : 11 2 14 2 39 x ytt t y x t yy t t x N y x x y                 Equation needs to be in the form ax by c  (ii) When y = 0, 1 6t ; When y = 11, 2t Method 1: Parametric Area =    2 1 6 d 1 39 d 14 11d 2 2 xy t t      =   2 1 6 6 1 6 d 30.25t t t  =2057 18 Method 2: Cartesian form   14 25 12 26 1 6 13 2 1 39Area