RI 2019 P2 A-Level H2 Math Solution

1 | P a g e Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student (i)      3 3 2 2 3 5 2 2 2 21 1 d3 3 2 41 13 15 I x x x x x x x c           Don’t forget to include the arbitrary constant of integration. (ii)        2 2 2 2 2 4 2 5 3 5 3 2 2 d1 2 d 1 2 d 2 1 d 2 d 2 2 5 3 2 21 15 3 xx u u u I u u u u u u u u u u u u d x x d                         2 1 2 1 1 x u u x     Final answer in terms of x. (iii)                  5 3 3 5 2 2 2 2 3 3 5 5 2 2 2 2 3 5 2 2 3 5 2 2 5 5 2 2 2 2 2 41 1 1 15 3 3 15 2 2 4 21 1 1 13 3 15 5 2 101 1 13 15 2 21 1 13 3 2 21 13 3 x x d x x x c x x x x x d c x x x d c x x x d c x x d c d c                                              Show difference in answer in parts (i) and (ii) do not depend on x. Raffles Institution H2 Mathematics (9758) Solution for 2019 A-Level Paper 2

2 | P a g e Question 2 No. Suggested Solution Remarks for Student (i)    2 2 3 5 8 2 3 8 1 xy x x x x x       (ii) 8 or 1 23x x   (iii)      2 03 8 1 2 03 8 1 8 1 or 23 x x x x x x x x          

3 | P a g e Question 3 No. Suggested Solution Remarks for Student (i) 2 2 2 900 2 2 450 450 r rh r rh rh r            2 2 2 3 2 2 2 2 2 2 2 450 450 d 450 3 0 150d 150 ( 0) d 1506 0 for all 0, that is, gives max .d max 450150 450 150150 150 300 150 450 rV r h r r r r V r rr r r V r r r Vr V r h r r rh                                                        2 2 2 1 450 150 1 450 450 150 2 : 1: 2 r r h r r r h r r h            

4 | P a g e Question 4 No. Suggested Solution Remarks for Student (i)          2 2 3 2 f 2sec 2 tan 2 f 2 2sec 2 tan 2 tan 2 2 2sec 2 sec 2 4sec 2 tan 2 4sec 2 f 0 1 f 0 0 f 0 4 f 1 2 x x x x x x x x x x x x x x               (ii)  0.02 2 0 1 2 d 0.0200053333 0.02001 (5 d.p.)x x   (iii) 0.02 0 sec2 d 0.0200053355 0.02001 (5 d.p.)x x   (iv) The approximation is good for small values of x. Part (ii) uses Maclaurin series with polynomial of degree 2 to approximate the integral with x = 0.02 which only differ from actual value from 9th d.p. This already provides a good approximation. (v) We would need g and all derivatives of g to be defined in order to apply Macluarin series. However 1cosec2 sin 2x x is undefined at x = 0.

5 | P a g e Question 5 No. Suggested Solution Remarks for Student (i)        5 2 4 5 1 4 5 1 4 14 1 4 5 15 1 4 4 5 4 OX OB BX OX OA AX b BD a AC b OD OB a OC OA b b a b a a b a b a a b b a a b OX b                                                                                                           a Another way to understand the solution is that X is the point of intersection between the line BD and line AC. Equation of line BD is OB BD r     Likewise, the eqn of the line AX is OA AC r     (ii)           1 5 1 2 4 5 4 55 1 2 4 4 Since and are nonzero nonparallel vectors, then Coeff of : 4 1 3 4 ...(1) Coeff of OY OD OC OY OX b a a b b a b a a b b a a b b                                                                       5 5 : 5 2 1 3 2 ...(2)4 4 9(1) (2) : 64 24 8 9 3 24 5 8 10 9 4 3 3 : 3: 8 a OY OX b a b a OX OY                                      Similar to (i), Y is the point of intersection between the line CD and the line OX Eqn of line CD: OC CD r     Eqn of line OX: OXr  

6 | P a g e Section B: Statistics Question 6 No. Suggested Solution Remarks for Student (i) These 22 clubs form the population as they are ALL the clubs in Division One whose approaches to training she is interested to find out. (ii) How Assuming no special treatment with regard to facilities for supporters of different divisions, he could randomly pick a certain number of clubs out of the 100, say 10, to do a thorough investigation. Why This is to avoid bias and it would also be more cost effective and manageable. (iii) 22 24 26 28 18 5 5 5 5 7.24 10C C C C    

7 | P a g e Question 7 No. Suggested Solution Remarks for Student (i) The event that one mug is faulty is independent of each other. The probability of a mug being faulty remains constant at 0.08. (ii) F ~ B(50, 0.08) P 7 1 P 6 0.10187 0.102 F F      (iii) Let W denote number of days out of 5 with at least 7 faulty mugs. W ~ B(5, 0.10187) P 2 0.99098 0.991 W   (iv)     8 810 2 2 2 1 45 1C p p p p   (v) P(no fault) + P(one fault)                         22 2 2 2 2 2 2 2 0.92 1 P fault with one mug P fault with one saucer 0.8464 1 2 0.92 0.08 1 2 0.92 1 0.8464 1 0.1472 1 1.6928 1 0.9936 1 1.6928 1 0.9936 1 1.6928 1 0.97 0.0689 p p p p p p p p p p p p p p p p                              

8 | P a g e Question 8 No. Suggested Solution Remarks for Student Orange Yellow Green White Total Horse 1 1 3 4 9 Rider 1 1 7 5 14 Dog 3 7 1 6 17 Bird 4 5 6 1 16 Total 9 14 17 16 56 You can just create the “total” column on the question booklet itself in the A level. (i) (a) 9 14 23 56 56   Just count relevant cells from the table above (b) 17 16 7 13 56 28    (ii) (a) 8 7 1 56 55 55   (b) Case 1: Dog is yellow, the other item is a non-yellow Horse/Rider/Bird Probability = 7 32 256 55  Case 1: Dog is not yellow, the other item is a yellow Horse/Rider/Bird Probability = 10 7 256 55  Required probability = 7 32 10 7 212 256 55 56 55 110      (iii) 12 56 55 77 where , refer to the number of his first and second favourites a b a b    20 1 20 or 2 10 or 4 5 Reference to the table, only 4,5 is possible ab      Thus, possible combinations are as follow 4 5 White Horse White Rider White Horse Yellow Bird Orange Bird White Rider Orange Bird Yellow Bird

9 | P a g e Question 9 No. Suggested Solution Remarks for Student (i) Since the manager wants to check “whether the mean resistance is in fact 750 ohms”, significant variation of both more or less than 750 is not acceptable. Thus, he should carry out a 2-tail test to see if the mean resistance differs from 750 ohms. Null hypothesis, 0H : 750 Alternative hypothesis, 1H : 750 where  is the population mean resistance of resistors rated at 750 ohms. (ii) Let X denote the resistance (in ohms) of a resistor rated at 750 ohms. Using GC, 756x Perform an 2-tailed test at 5% significance level. Under H0, 100~ N 750,8X     -value 2P 756 0.0897 0.05p X     , hence we do not reject 0H : 750 . The manager does not have sufficient evidence at 5% level of significance to claim that the mean resistance is not 750 ohms. Since X follows a Normal distribution, then so does X . The population variance is given in the question so we use it. Answer in context with reference to the alternative hypothesis. (iii) As the distribution of the resistance of the population of resistors is unknown, we need a large sample (of at least 30) so that we can use te Central Limit Theorem to approximate the probability distribution of X to follow a normal distribution.

10 | P a g e Question 10 No. Suggested Solution (i) (a) (b) The residuals are the distances labelled 1 5, ,e e in the diagram. (c)   5 2 1 1 1