RI 9758/02 2022

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                       Question 2 No. Suggested Solution Remarks for Student (a) Volume 72 0.9 1000k h   900 12.572 hk h   (b) 2 2 2 2 2 d d 1 25 2 4 0 when 0 0 25 4 25900 900 4 144 Since 0, 12 V ktt V kt C ht C 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 2 d 25 12.5 25d 6.25 25 0 when 0 0 6.25 25 10, 900 900 625 250 250 10m275 11 V kt htt V ht t C V t C 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

3 | P a g e (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                                 

4 | 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 2 1 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 0.004 27.1113 3 2 Least 28 n r r r r r r nn n                         Note the correct use of inequalities leading to the answer 28 which has to be an integer.

5 | P a g e Alternatively, n 2 2 1 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.

6 | P a g e Question 5 No. Suggested Solution Remarks for Student (a) Volume of spherical cap is      2 2 2 2 3 33 3 2 3 3 2 3 2 2 3 2 3 2 d d 1 3 1 1 3 3 2 1 1 3 3 3 1 3 1 33 r r h r r h r r h x y r y y r y y r r r r h r h r r r h r r h rh h rh h h r h                                              (b)         3 2 2 2 3 2 3 3 2 4 1 13402 15 3 15 3 3 15 33 3 3 13402 4500 15 135 93 28 450 3294 0 3 or 2.5158 (reject since 0) or 15.587 (reject since 15) 3 p p p p p p p p p p p p p p p p                        (c) Volume of 2nd ornament is      2 21 13 3 3 33 3 ( 3, 15) 846 p r p p r p p r          3 9

7 | P a g e Section B: Probability and Statistics Question 6 No. Suggested Solution Remarks for Student (a)     2 4 6 2 P A wins P A wins on throw 3 P A wins on throw 5 P A wins on throw 7 ... 5 1 5 5 5 1 5 5 5 5 5 11 1 1 ...6 6 6 6 6 6 6 6 6 6 6 6 5 5 5 51 ...36 6 6 6 5 1 36 51 6 5 11                                                   Exact answer is required. (b)    P B wins on her second throw | B wins P B wins on throw 4 1 P A wins 5 5 11 6 6 651 11 275 1296         Exact answer is required.

8 | P a g e Question 7 No. Suggested Solution Remarks for Student (a) Sample, as not all 75 employees responded (b) She can randomly select a certain number of employees from each department. The number selected may be proportional to the size of the department. Example, 10% of each department. This way is less time consuming and fair (as there is no biasness since employees are selected randomly from each department) instead of gathering views from every employee. (c) A (7) P (6) M (4) S (3) No. 5 1 1 1 7 6 4 3 5 1 1 1 1512C C C C    4 2 1 1 7 6 4 3 4 2 1 1 6300C C C C    4 1 2 1 7 6 4 3 4 1 2 1 3780C C C C    4 1 1 2 7 6 4 3 4 1 1 2 2520C C C C    3 2 2 1 7 6 4 3 3 2 2 1 9450C C C C    3 2 1 2 7 6 4 3 3 2 1 2 6300C C C C    3 1 2 2 7 6 4 3 3 1 2 2 3780C C C C    Total: 33642 Listing systematically is key in answering this question. Yes, there are 7 cases.

9 | P a g e Question 8 No. Suggested Solution Remarks for Student (a) 2 2 2 2~ N ,aX bY ap bs a q b t   (b) (i) & (ii) 2~ N 6, 2V P 10 0.0228V  Label 6 And also the 4 and 8 (c)   E 1.2Var 8 1.2 8 1 10 (not meaningful) or 6 W W p p p p p            1~ B 8,6 P 2 P 1 0.605 W W W        

10 | P a g e Question 9 No. Suggested Solution Remarks for Student (a) Note that it takes 5 moves from S to one of A,B,C,D,E or F. Let X = number of left moves after first move. X ~ B(4, p) (first move has probability ½ to move left or right)       4 3 4 4 3 4 3 4 P counter arrives at B P first move is L and 3 P first move is R and 4 1 1P 3 P 42 2 1 1 2 2 12 2 X X X X C p q C p p q p               Note the probability is ½ for first move from S to L or R. (b) 5 routes in total Both taking Right LeftLeft Left Left has probability 2 4 81 1 2 4p p     There are 4 routes where Both take Left on first move follow by another same 3 left and one right moves subsequently has probability 2 3 6 214 2p q p q     Sum of above      8 6 2 28 6 8 6 2 6 2 2 6 2 1 4 1 14 1 1 24 1 4 8 44 1 5 8 44 p p q p p p p p p p p p p p p p p                 Required probability      6 2 6 2 2 2 2 263 4 1 1 5 8 4 5 8 4 5 8 44 4 1 4 31 42 42 p p p p p p p p pp q pp q p              Answer must be simplified and in terms of p only