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 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 or 15.587 (reject since 0) (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 (with 3, 15) 846 p r p p r p p r          3 9

5 | 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                                                   Recognise that this is a geometric progression and the it is an infinite series The required answer is found by using the sum to infinity formula of a GP. So, knowledge of Pure Math topic is required. 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.

6 | P a g e Question 7 No. Suggested Solution Remarks for Student (a) Sample, as there were 75 employees in Staffing and the 53 respondents is a subset of the 75 employees. (b) A random sample should be chosen. This is to reduce bias. (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.

7 | 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        

8 | 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 denote 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

9 | P a g e (c)    4 2 2 4 3 2 3 2 2 3 P counter arrives at C 1 1P 2 P 32 2 1 1 2 2 3 2 X X C p q C p q p q p q             2 2 3 3 4 2 2 4 22 4 13 2 2 2 13 0 2 13 1 0 2 Since 0, 0.710 p q p q p q p p q p p p p p p           Can use GC, no need to find exact value of p which is 6 6 5 

10 | P a g e Question 10 No. Suggested Solution Remarks for Student (a) d and p may not be linearly correlated as –0.78 is not very near to –1. The scatter diagram also suggests a non-linear negative relationship. (b) No. Scaling of the values, including change of units, do not change the relationship. (c) As this data becomes a outlier, since the special car has different features from the other 6 cars, it will disrupt the relationship to the rest of the data as they were taken from cars of similar version Remember to elaborate in context. Merely stating that it is an outlier is NOT enough (d) (e) Squaring the distances so that the sum will not be zero or become negative, as the distances could be positive (above the line) or negative (below the line). This is referred to “method of least squares”