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    