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 which passes 3 through ( 1, 2, 4) and has normal vector 5 . 2 r p q x y z x y z x y z R                                          

5 | P a g e Question 6 No. Suggested Solution Remarks for Student                      2 2 22 2 2 2 2 2 2 i 8i 0 i 2 i 8i i 0 1 i 2 i 8 i 1 i 0 2i 2 i 8 1 i 0 2 4i 8 1 i 0 2 8 4 8 i 0 Comparing imaginary part: 4 8 0 2 or 0(rejected z z t k k k k t k k t k k t k k t k k t k k k k k k                                                 2 2 as 0) Comparing real part: 8 16 0 8 2 i 8i 8 0 8i 8i 4 2 i 8 2 2 i 8i 32i 2 2 i 2 62 2i or i 5 5 k t t z z z                        Note that you can use GC for this question where appropriate such as evaluating   8i 32i 2 2 i   Note that coefficients of the equation are not real, so roots do not come in conjugate pair.

6 | P a g e Question 7 No. Suggested Solution Remarks for Student (i) 12 sin 4 d 2 cos 44x x x x c    (ii) Let u = x and d 2 sin 4d v xx  d 1d u x and 12 cos 44v x x  from part (i)  2 0 2 2 00 2 22 0 2 2 2 2 sin 4 d 1 12 cos 4 2 cos 4 d4 4 12 sin 44 8 16 2 8 4 4 8 x x x x x x x x x x x                                                    Clear working is required since exact value is required. Note that to check your answer, you can use GC to evaluate  2 0 2 sin 4 dx x x   and compare with the numerical value of your answer, in this case 2 4 8   (iii)       22 0 22 0 22 20 0 0 2 0 2 sin 4 d 4 4sin 4 sin 4 d 14 cos 4 1 cos8 d2 1 12 0 sin 82 8 92 4 4 x x x x x x x x x x x                             

7 | P a g e Question 8 No. Suggested Solution Remarks for Student (a) (i)   1 5 30 4 4 4 10 1.5 4 29 1.5 47.5 u u d d u          (ii)    50 21 22 50 21 ... 4 1.5 1 1672.5 r u u u r         Using GC (b) (i) 44 1.6384 0.8 4 201 0.8 r r S      (ii)   4 1 0.8 19.61 0.8 4 1 0.8 19.60.2 1 0.8 0.98 0.8 0.02 ln 0.8 ln 0.02 ln 0.0217.531ln 0.8 smallest possible 18 n n n n n n n            Note that ln 0.8<0

8 | P a g e Question 9 No. Suggested Solution Remarks for Student (i) Consider 12 and 2y x y x  They are perpendicular to each other since product of gradients is –1. Angles lines made with positive x-axis are 1 1 1tan 2 and tan 2       respectively. Thus 1 1 1tan 2 tan 2 2         (ii)     2 2 2 2 1 1 3 4 3 4 3 4 0...(1) Intersect (1) has real roots 0 and 9 4 4 0 4 16 9 0 2 1 2 9 0 1 9 2 2 90 2 9set of values of is 0,2 k x x x kx k kx x k k k k k k k k k k k                               Read question carefully. It is given k > 0. (iii) Good to label the intersection of the 2 curves if time permits: (–0.5, 0.8), (2, 0.2) C1 C2

9 | P a g e (iv)   2 1 2 2 2 21 11 22 1 1 1 2 d1 3 4 2tan ln 3 43 1 2 5tan 2 tan ln10 ln2 3 2 2ln 42 3 4ln 22 3 xx x x x                                         Recall from (i), 1 1 1tan 2 tan 2 2        

10 | P a g e Question 10 No. Suggested Solution Remarks for Student (i) d 0.03d P Pt  (ii) 0.03 1 d 0.03d ln 0.03 , 0 and is an arbitrary constant e , e 0t C P P t P t C P C P A A        Note that as 0.03, e 0 and thus 0tt P    So the number of sheep will approach 0 if this situation continues many years. Note that no further information were given in the question, so we can’t find value of A. Explain in context of the question (iii) d 0.03d P P nt   (iv)   0.03 0.03 0.03 1 d 10.03 d 100ln 0.03 , where is an arbitrary constant3 ln 0.03 0.03 0.03 0.03 e , e 100 e3 t b t P n P t n P t b b n P t b n P B B P n B                 (v) 0.03, e 0 and given 500 100 5003 15 tt P n n      