RI 2022 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 2 3 3 2 2 2 d2 1d 2 d 2 d2 2 2 d 12 23 2 2 4 23 u x u x uu x x u x u u ux u u u u C x x C                       Remember to change back to x. Question 2 No. Suggested Solution Remarks for Student (a) Volume 72 0.9 1000k h   900 12.572 hk h   Be careful of the units! 2 3 72sec litres/sec = 72 litres 0.9m m 0.9 m 0.9 1000litres k k h h h      (b) Let V litres be the volume of water inside the container at time t. 2 2 d d d 1 2 25 , using (a)4 V ktt V kt t kt C ht C        2 2 2 0 when 0 0 25 4 25When full, 900 900 4 144 Since 0, 12 V t C V ht V h h ht t t t s             Raffles Institution H2 Mathematics (9758) Solution for 2022 A-Level Paper 2

2 | P a g e (c) 2 d 25 12.5 25d 12.5 25 d 6.25 25 V kt htt V ht t V ht t D           2 0 when 0 0 6.25 25 10, 900 900 625 250 250 10m275 11 V t D V ht t t V h h h h                Question 3 No. Suggested Solution Remarks for Student (a)   i61 2i 52 2 7i i5 6 303 1 2 3 3 3 i 3 12e 1e2 12e 3e2 73, arg 30 z z z z z z z                     Need to specify 3 3 and argz z explicitly. Answer left as 7i303e  is not acceptable. (b) Let A, B and C represent complex numbers z1, z2 and z3 respectively (c)     7 7i i30 303 1515 7 3 15 3 3e 3 e 7 3 5 7 9, , , ,30 2 2 2 2 2 7 15, 45,75,105,... smallest positive integer 15 3 3 3 2187 3 105 7arg 30 2 24 4 n nnnz n n n z z                                 

3 | P a g e Question 4 No. Suggested Solution Remarks for Student (a)     2 1 1 1 1 9 3 2 3 2 3 1 3 3 1 3 3 2r r r r r r        (b)    3 3 2 1 1 1 1 9 3 2 3 3 1 3 2 1 1 1 3 3 1 3 2 1 1 3 2 3 5 1 1 3 5 3 8 . . . 1 1 9 4 9 1 1 1 9 1 9 2 1 1 1 3 3 1 9 2 2 1 3 1 9 2 m m r m r mr r r r m m m m m m m m m m m m m m m                                               You are required to combine and simplify the answer as a single fraction. (c)    21 21 1 1 1 1 1 1 9 3 2 3 2 3 2 6 3 3 2 1 1 1 1lim9 3 2 6 3 3 2 6 n r nr r r n n r r n                                      (d)    2 2 1 1 1 1 0.0049 3 2 9 3 2 1 1 1 0.0046 3 3 2 6 1 0.004 27.1113 3 2 Least 28 n r r r r r r n nn n                            Alternatively, n 2 21 1 1 1 9 3 2 9 3 2 n r r r r r r                   27 0.004016 > 0.004 28 0.003876 < 0.004 Least n is 28. Note the correct use of inequalities leading to the answer 28 which has to be an integer.

4 | P a g e Question 5 No. Suggested Solution Remarks for Student (a) Volume of s