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 6 1 d 14 113 2 2 2057 18 xy t x x                      GC can be used to evaluate the area. (iii)(a) 2057 2057618 3  (why multiply by 6? Just imagine you are given area of a triangle. You scale the base by 2, and the height by 3, the resulting area will be 6 times the original area)

6 | P a g e (b)      2 256 1 6 1, and 03 12 After transformation, 226 13 3 418 3 3 6 4 16 3 6 4 1 xy t x y x y xy x y x                  Equivalent to 2 2554 6 225, and 06 x y y x y    

7 | P a g e Question 4 No. Suggested Solution Remarks for Student (i)     2 2 2 2 2 2 2 2 30 30 2 2 30 2 2 30 4 30 30 4 900 60 4 225 15 h a ah aH h a a a a a a a a a a                              (ii)   2 2 2 4 4 5 3 4 3 4 3 1 1 225 153 3 1 5225 15 259 3 d 252 100d 3 d 250 100 0d 3 25100 03 300 12 since 025 V a H a a V a a a a VV a aa V a aa a a a a                         21 12 225 15 12 48 45 144 53V     (iii)(a)   14 2 302 2 30 which is quadratic (factorised), so max occurs when 15 S ah aa a a S a              (b) Square (2 layers of the base)

8 | P a g e Section B: Statistics Question 5 No. Suggested Solution Remarks for Student (i) 0,4,10,25 (ii)                    1 3 1 1 3 1 2 2 232 3 1 2 2 431 1 3 1 2 132 3 1 2 3 2 2P 0 3 1 3 3 1 2 2 1 2 1P 4 3 1 3 3 1 2 2P 10 3 1 3 3 1 1 1P 25 3 1 3 3 1 r r r r r r r r r C C rS C r r r r r rCS C r r r rC C r rS C r r r r r rCS C r r r                                Good to check sum of prob =1 2 1 43 3 32 1 12 3 1 3 1 3 1 3 1 1 4 2 4 1 123 1 3 3 3 3 3 1 r r r r r r r r r rr                      E P 1 4 2 4 1 10 2 4 10 253 1 3 3 3 3 3 1 16 8 40 25 25 3 1 3 3 3 3 3 1 27 113 1 S s S s r r rr r r rr rr                                             2 2 2 2 2 E P 1 4 2 4 1 14 10 253 1 3 3 3 3 3 1 64 32 400 625 625 3 1 3 3 3 3 3 1 363 2193 1 S s S s r r rr r r rr rr                                                 22 2 2 2 var E E 1 1363 219 27 113 1 3 1 1 363 219 3 1 27 113 1 S S S r rr r r r rr                        2 2 1 360 300 3403 1 r rr      

9 | P a g e (iii)   2 2 var 38 1 360 300 340 383 1 : 3 S r rr GC r        

10 | Page Question 6 No. Suggested Solution Remarks for Student (i) Your sketch should show the symmetry about 8.05am (ii)  P 6 0.202T  (iii)    22~N 5 2 1 , 1 . 2 3 ~N 2 6 , 1 0 . 4 4 P3 0 0 . 1 0 8 TW TW TW      (iv)         ~ N 24, 37.44 P Day is fine | James is late P Day is fine AND James is late P James is late 0.7 P 30 0.7 P 30 0.3 P 30 0.606 TD TW TW T D          