RI 2015 P2 A-Level H2 Math Solution

1 | P a g e Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student (i) 32 m Note that we need 1 216 0h  (ii)    12 12 1 2 1 212 1 2 16d 10 16 d dd 10 20 16 40 16 where hh h h tt h t c t b h b c                1 2 0 when 0 160 Thus, 160 40 16 h t b t h         1 2 Half its maximum height, that is, 16 we have 160 40 16 16 160 40 8 46.9 years h t        Raffles Institution H2 Mathematics Solution for 2015 A-Level Paper 2

2 | P a g e Question 2 No. Suggested Solution Remarks for Student 1 2 : 2 3 , 4 6 L r                        (i) 1 1 2 1 3 0 6 0 2Acute angle = cos cos 73.4 72 1 3 0 6 0                                     (ii) 2 5 6 OP         Any point, , on has position vector, 1 2 2 3 for some . 4 6 R L OR                          2 2 2 2 2 2 2 2 1 2 2 33 2 3 5 = 33 4 6 6 2 1 3 7 2 6 33 4 4 1 9 42 49 4 24 36 33 49 70 21 0 7 10 3 0 7 3 1 0 3 or 17 PR                                                                   

3 | P a g e 31 2 7 1 2 332 3 or 2 3 17 4 6 1034 6 7 13 7 5 7 46 7 OR OR                                                                       13 17 7 731 5 1' 12 7 710 46 58 7 7 OR                                                          (iii)       2 1 2 1 2 36 5 2 3 7 3 2 6 4 6 2 6 11 Required plane is 36 2 11 36 2 2 5 11 6 4.x y z                                                                

4 | P a g e Question 3 No. Suggested Solution Remarks for Student 2 1f : , , 11x x xx    (a)(i) Method 1:   22 2f '( ) 0 for 11 xx x x   Thus, f is an increasing function for x > 1. Thus f has an inverse. Method 2: Sketch graph of f and explain using horizontal line. (ii)   1 2 2 2 1 ff 1 1Let 1 1 11 11 1 1f 1 ,0 y x x y x y x x y x x D R                   (b) 2 2g : , , 11 xx x xx             2 2 2 2 2 2 2Let 1 21 2 0 For the equation to be defined, discriminate 0, thus, 1 4 2 0 4 8 1 0 4 2 1 3 0 4 1 3 0 2 2 3 2 2 3 0 2 3 2 3 or 2 2 xy y x xx yx x y y y y y y y y y y y y                                

5 | P a g e Question 4 No. Suggested Solution Remarks for Student (a) Let nP be the statement     2 1 12 5 1 3 31 7412 n r r r r n n n n        for .n  For 1,n       1 1 2 5 1 3 6 18 1 2161 1 1 3 31 74 1812 12 r LHS r r r RHS LHS                 Therefore, P1 is true. Assume kP is true for some k  , i.e.     2 1 12 5 1 3 31 7412 k r r r r k k k k        We need to show1kP is true, i.e.           1 2 1 12 5 1 1 1 3 1 31 1 7412 k r r r r k k k k               For 1,n k 

6 | P a g e                              1 1 1 2 2 3 2 2 3 2 2 2 2 5 2 5 1 1 2 1 5 1 1 3 31 74 1 3 612 1 1 3 31 74 12 3 612 1 1 3 31 74 12 108 21612 1 1 3 43 182 21612 1 1 2 3 37 10812 1 1 2 3 2 1 31 1 712 k r k r r r r r r r k k k k k k k k k k k k k k k k k k k k k k k k k k k k k k k k k k k                                                                 2 4 1 1 1 1 3 1 31 1 7412 k k k k          Therefore 1kkP P is true. Since 1P is also true, nP is true for n  by Mathematical Induction. (b)(i)   2 2 2 1 1 4 8 3 2 1 2 3 2 1 2 3r r r r r r        (ii) 21 1 2 1 1 4 8 3 2 1 2 3 1 1 3 5 1 1 5 7 1 1 7 9 ... 1 1 2 1 2 1 1 1 2 1 2 3 1 1 3 2 3 n n r r r r r r n n n n n                           

7 | P a g e (iii) 3 31 1 1 1 10 103 2 3 3 2 3 498.5 smallest 499 n n n n          

8 | P a g e Section B: Statistics Question 5 No. Suggested Solution Remarks for Student (i) Not possible to obtain the sampling frame, that is, not possible to get the list of all customers, with their age, who patronize the supermarket. (ii) Assume manager wants to get opinions of 10 customers from each of the following groups (by ages) Children: Below 12 years old, Teenagers: 12 to 18 years old, Young Adult: 18 to 35 years old, Mature Adult: 35 to 50 years old, Above 50 years old. Approach the first 10 customers who are willing to provide their opinions using a questionnaire. (iii) As mentioned in (ii), we are approaching the first 10 customers who are willing to provide their opinions. We will not obtain opinions of those who are not as vocal.

9 | P a g e Question 6 No. Suggested Solution Remarks for Student (i) Let X be the number of red sweets in a small packet of 10 sweets. ~ B 10,0.25X P 4 1 P 3 1 0.775875 0.224X X       (ii) Let X be the number of red sweets in a large packet of 100 sweets. ~ B 100,0.25Y Since 100n is large, and 25 5np  and 1 75 5,n p   25,~ N 18.75Y approximately. P 30 P 29.5 by continuity corrections 0.14935 0.149 Y Y     Be sure to check the conditions and state the approximate distribution used. (iii) For 9740 syllabus: Let W be the number large packets, out of 15, containing at least 30 red sweets. ~ B 15,0.14935W P 3 0.824655 0.825W   For 9758 syllabus: Without using approximation for P 30Y , ~ B 15, P 30 , that is, ~ B 15,0.14954W Y W  P 3 0.82407 0.824W   For 9740 syllabus: Exam report suggests that students need to use answer from (ii), as the only answer given is 0.825.

10 | P a g e Question 7 No. Suggested Solution Remarks for Student (i) Number of errors per page are assumed to remain constant uniformly. Errors occure independently between pages of the newspaper. Let X be the number of errors on one page. ~ Po 1.3X (ii) 1 2 3 4 5 6 ~ Po 7.8Y X X X X X X      P 10 1 P 10 1 0.83523 0.165Y Y       (iii) 1 2 ... ~ Po 1.3nW X X X n         1.3 1.3 1.3 P 2 0.05 P 0 P 1 0.05 e 1.3 e 0.05 e 1 1.3 0.05 0 ...(1) n n n W W W n n               Solving (1) using GC graphically or table of values, least n = 4 To avoid careless mistake, we should also use the original inequality, P 2 0.05,W  to check the answer.