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 1 112 2 2 10 1 1 42 r r rr r r                     

7 | P a g e Question 7 No. Suggested Solution Remarks for Student (i)   1 1 1 1 e d e ed d1, e , e e 0d Equation of tangent: e d1, e, e e 2ed Equation of tangent: e 2e 1 2e e x x x y x y xx yx y x y yx y x y x y x                           (ii) tan 2e 79.6      Note that one of the tangents is a horizontal line. So the gradient of the other line allows us to get the angle directly.

8 | P a g e Question 8 No. Suggested Solution Remarks for Student (a)        1 1 1 5 7 642 2 64 1 22 2 32 2 126 2 2 2 13 k k k a a a k           (b)    4 4 Note that 0. If 1, then sum of first 4 terms 4 0 Thus, 1. 1 01 1 1 or 1 (rejected) Thus, 1, \ 0 0, even , oddn f r f r f r r r r r r f nS f n                (iii) Let a be the first term.      4 2 3 142 2 3 7 ...(1) 2 3 0 0 or or 2 or 3 Given 0, Possible or or 2 3 Into (1), only gives 0 7, 7 11th term 10 63 a d a d a a d a d a d a d d d a a ad a d a a a d a d                       

9 | P a g e Question 9 No. Suggested Solution Remarks for Student (i) cos isinw    (a) 2 2 1 1cos isin cos isin cos isincos isin cos sin cos isin cos isin 2cos w w                             Alternatively,     i i i i 1 1e e e e cos isin cos isin 2cos w w                     (b)            2 2 2 1 1 2 1 1 21 1 cos isin 2 1 cos isin1 1 cos sin 2 1 cos i 2sin1 2 2cos 2 2 cos 2 1 cos i 2sin 2 2 cos i 2sin 2 2 cos i sin 1 cos i 2sin cos2 2 2cos 2 i tan2 w w w w                                                    Alternatively, i i i i i0 2 2 2 i i ii i0 2 2 2 1 e e e (e e ) 1 e e e (e +e ) w w              

10 | P a g e i i 2 2 i i 2 2 cos i sin cos isine e 2 2 2 2 cos i sin cos isine +e 2 2 2 2 2isin2 2cos2 i tan2                                              (ii) 2 2 Let i Since 1 1 z a b z a b         22 2 2 3i i 3i 3 6 9 10 6 z a b a b a b b b                2 2 2 2 1 3i 1 3i i 1 3 9 1 6 9 9 10 6 z a b b a b b a b              3i 11 3i z z   OR Since 1 cos i sinz z        22 3i cos isin 3i cos sin 3 10 6sin z                 2 2 1 3i 1 3i cos 3sin 1 3sin 9cos 10 6sin z               3i 11 3i z z   OR        1 * 1 3i 3i * 3i * 1 3i **1 3i 1 3i 1 3i 1 3i Note that 1 3i * 1* 3i * 1 3i * * 1 3i * 1 3i *3i 11 3i 1 3i z z z z zz z z zz z z z z z z z zz z z                       