RI 9758/02 2019

1 | Page   Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student (i)     33 22 35 22 22 11 d33 24 1131 5 Ix x x x xx xc          Don’t forget to include the arbitrary constant of integration. (ii)       2 22 22 42 53 53 22 d12 d 12 d 21 d 2 d 22 53 221153 yxu u x Iu u u u uu u uu u uu d xx d               2 1 2 1 1 xu ux   Final answer in terms of x. (iii)                 53 3 5 22 2 2 33 55 22 22 35 22 35 22 55 22 22 2 411 1 153 3 1 5 2 24211 1133 1 5 5 21 011 131 5 2211 133 221133 xx d x x x c xx x x xd c xx x d c xx x d c xx d c dc                                   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 | Page   Question 2 No. Suggested Solution Remarks for Student (i)    2 22 35 8 3 8 1 xxy xx x x     Label Asymptotes: 8 ,1 , 03xx y   Label axial intercepts: 10, , 2, 04   (ii) 8 or 1 23xx   (iii)       2 038 1 2 038 1 8 1 or 23 x xx x xx xx        

3 | Page   Question 3 No. Suggested Solution Remarks for Student (i) 2 2 2 900 2 2 450 450 rr h rr h rh r         2 22 3 22 2 2 2 2 2 450 450 d 450 3 0 150d 150 (0 ) d 150 6 0 for all 0, that is, gives max .d max 450150 450 150150 150 300 150 450 rVr h r r r r V rrr rr V rr r Vr Vr h r r rh                                         2 2 2 1 450 150 1 450 450 150 2 :1 : 2 r r hr rr hr rh          

4 | Page   Question 4 No. Suggested Solution Remarks for Student (i)           2 23 2 f2 s e c 2 t a n 2 f 2 2sec 2 tan 2 tan 2 2 2sec 2 sec 2 4sec2 tan 2 4sec 2 f0 1 f0 0 f0 4 f1 2 xx x x xx x x x xx x xx            (ii)  0.02 2 0 1 2 d 0.0200053333 0.02001 (5 d.p.)xx  (iii) 0.02 0 sec 2 d 0.0200053355 0.02001 (5 d.p.)xx  (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 | Page   Question 5 No. Suggested Solution Remarks for Student (i)       52 4 5 14 51 4 141 4 5151 44 5 4 OX OB BX OX OA AX bB D a A C bO D O B aO C O A bb a b a a b a ba ab ba ab OX b                                                                          a (ii)               1 51 2 4 5 4 551 2 4 4 4 1 3 4 ...(1) 55 52 1 32 . . . ( 2 )44 9(1) (2) : 6 4 24 8 93 24 5 94 OY OD OC OY OX ba ab ba ba ab b a OY OX b a                                                               81 0 33 :3 : 8 ba OX OY    

6 | Page   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 5555 7.24 10CCCC

7 | Page   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)    P7 1 P6 0.10187 0.102 FF     (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)   8810 2 2 2 14 5 1Cp p p p  (v) P(no fault) + P(one fault)                   22 22 2 22 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 pp p p pp p p pp p pp p p                    

8 | Page   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) 91 4 2 3 56 56   Just count relevant cells from the table above (b) 17 16 7 13 56 28   (ii) (a) 87 1 56 55 55   (b) Case 1: Dog is yellow, the other item is a non-yellow Horse/Rider/Bird Probability = 73 2 256 55 Case 1: Dog is not yellow, the other item is a yellow Horse/Rider/Bird Probability = 10 7 256 55 Required probability = 73 2 1 07 2 12256 55 56 55 110   (iii) 12 56 55 77 where , refer to the number of his first and second favourites ab ab   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 | Page   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 0.05pX    , 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) How Take a large random sample, say 50 instead of 8 in the previous test, of resistors rated at 1250 ohms, and take down the resistances. Calculate the sample mean and unbiased estimate of population variance using this new sample. Use Central Limit Theorem to approximate the distribution of X , to determine the p-value, and carry out the test. Why Distribution of the restistance of the population of resistors is unknown. (so we need a large sample to approximate the probability distribution of X ) The population variance is also unknown. (so we need to estimate it by 2s )

10 | Page   Question 10 No. Suggested Solution Remarks for Student (i) (a) Do on graph paper. For 135 3y x , it would be easier to plot 2 points and join them. Example join (42, 21) and (54, 17) to get the line 135 3y x (b) On same graph paper (c) 4 3 Using GC (d) Residual =  fba could be either positive and negative. Measuring vertical distance require the use of modulus, that is f ab and minimizing the sum of (absolute) residuals is difficult. Squar