RI 2019 P1 A-Level H2 Math Solution

1 | P a g e Question 1 No. Suggested Solution Remarks for Student The coefficient of the cubic function are all real. Since 2 i is a root of f (z) = 0, then  2 i * 2 i   is also a root.         2 3 2 3 2 3 2 f ( ) 0 3 2 i 2 i 0 3 4 5 0 7 15 0 , 7 , 15 z z z z z z z z z z b c daz bz cz d a z z za a a b a c a d a                              Alternatively, With the given roots apply Factor Theorem. f ( 3) 0 27 9 3 0 (1)a b c d            f (2 i) 0 2 11i 3 4i 2 i 0 2 3 2 0 (2) 11 4 0 (3) a b c d a b c d a b c                  From (1), (2), (3) we form a system of linear equations with last column the coefficient of a. As before, , 7 , 15b a c a d a   Raffles Institution H2 Mathematics (9758) Solution for 2019 A-Level Paper 1

2 | P a g e Question 2 No. Suggested Solution Remarks for Student (i) 3 1 0x x   Using GC, Polysimult2 0.68233 0.682 Do note the GC skills required in this question. (ii)     03 3 11 d 2 0 1 d b a x x x x x x       We can solve with GC, noting that b > a. 1.89243 1.892b  Alternatively,     03 3 1 4 2 4 2 4 2 1 d 2 0 1 d 0.68233 0.682330.68233 3.54 2 4 2 3.10465 04 2 b a x x x x x x x x x x x x                            1.89243 1.892b 

3 | P a g e Question 3 No. Suggested Solution Remarks for Student (i)       3 3 2 3 2 3 2 3 3 Note that 1 3 3 1 f 2 6 6 12 2 3 3 1 10 2 1 10 2 1 5 2, 1, 5 x x x x x x x x x x x x x p q r                        (ii) Sequence of 1. Translation of 1 unit in the positive x-direction 2. Translation of 5 units in the negative y-direction 3. Scale by factor 2 parallel to the y-axis.

4 | P a g e Question 4 No. Suggested Solution Remarks for Student (i) 00, 2 10 9 ln10 10, 2 10 or ln 2 lg 2 x x y y x         (ii) Solving 2 10 6 2 10 6 or 2 10 6 ln16 ln 4 2ln 2 ln 2 4ln 2 ln 2 4 2 10 6 2 4 x x x x x x x                

5 | P a g e Question 5 No. Suggested Solution Remarks for Student (i)   2f e 4, g 2, xx x x x x               2 2 1 1 e 4 e 4 2 ln 4 1ln 42 1f ln 42 Domain of f Range of f 4, x x y y x y x y x x                Note that we use “round brackets” here. (ii)         1 1 1 fg 5 f fg f 5 g f 5 12 ln 5 42 ln 3 2 x x x x x           

6 | P a g e Question 6 No. Suggested Solution Remarks for Student (i)            2 2 1 1 1 1 1 1 4 1 2 1 2 1 2 2 1 2 2 1 1 1 1 1 4 1 2 2 1 2 1 1 1 1 2 1 3 1 1 3 5 1 1 5 7 ... 1 1 2 3 2 1 1 1 2 1 2 1 1 112 2 1 n n r r r r r r r r r r n n n n n                                         (ii)   10 2 2 211 1 1 1 1 1 4 1 4 1 4 1 1 1