RI 2020 P1 A-Level H2 Math Solution

1 | P a g e Question 1 No. Suggested Solution Remarks for Student (i) 1 1 2 1 1 5 2 2 1 0 2 6 3                                          A vector normal to 1 is 1 1 3       . (ii) 1 4 1 5 3 6 19cos 11 77 11 77 Required angle, 49.2 (1dp)                       Question 2 No. Suggested Solution Remarks for Student            2 2 3 5 2 2 2 22 2 3 4 2 5 2 22 2 1 1 d d2 1 22 1 2 dd d 5 3 d1 1 At 1,1 , d d4 22 2 2 dd d 5 34 4 d 1 1 d d d 55 32 2 d d d 9 5Equation of tangent at (1,1) is 1 1 5 9 149 x y x yx y y yy y y yx x x x yx x x y x yxx y y y yx x x y y y x x x y x x y                          Note that question requires answer in the form ax + by = c, where a, b and c are integers Raffles Institution H2 Mathematics (9758) Solution for 2020 A-Level Paper 1

2 | P a g e Question 3 No. Suggested Solution Remarks for Student (i)     2 2 2 f ( ) ln(1 sin 3 ) 3cos 3f ( )1 sin 3 9sin 3 1 sin 3 9cos 3f ( ) 1 sin 3 9sin 3 9 1 sin 3 9 1 sin 3 x x xx x x x xx x x x x                Thus, k = –9 2 2sin cos 1x x  (ii) f (0) ln(1) 0 f (0) 3 f (0) 9           2f ( ) 9 1 sin 3 3cos3 f (0) 27 x x x      2 39 9f ( ) 3 ...2 2x x x x   

3 | P a g e Question 4 No. Suggested Solution Remarks for Student (i) i i i3 6 41 2 3 i 53 i i3 4 61 12 ii2 3 64 1 3i 2e , 1 i 2e , 2e 2e 1 1 1 5 5e e cos isin 12 122 2 22e 2e z z z z z z                                           Note that detailed working and exact values are required as the question states “Do not use a calculator in answering this question”. (ii) 1 4 2 3 1 4 1 4 2 3 2 3 5 5i i 2 31 12 12 2 3 1 5 5i i i i12 2 12 124 5 5 11i i i i12 2 12 124 i, where 1 1, that is i 1 e 2e2 i 2e e 2e 2e 2 cos isin 12 12 OR i 2e e 2e 2e 112 cos 12 z z a az z z z z z az z z z z zz z z z z z                                                  11isin 12          Note that there are 2 answers for z4.

4 | P a g e Question 5 No. Suggested Solution Remarks for Student (a)   Given But we know 0 , \ 0 a b b a a b b a a b a kb k                          That is, and are parallel.a b  If k = 0 , then 0 0a b    . But the question stated that a is a nonzero vector. Thus, 0k . (b)(i) 0 , , The set of all possible positions of point form a line which passes through point and parallel to vector . r p q r p kq k r p kq k R P q                      Clear description is expected. (ii) 0 1 3 2 5 0 4 2 3 3 5 10 2 8 0 3 5 2 5 represents all possible positions on plane