RI 2021 P1 A-Level H2 Math Solution

1 | P a g e Question 1 No. Suggested Solution Remarks for Student f (1) 5 5 (1)a b c d       f ( 1) 3 3 (2) a b c d         2f ( ) 3 2x ax bx c    f (1) 0 3 2 0 (3)a b c       1 1 3 2 0 0 1 4 3 2 0 f ( ) d 6 d 6 1 1 1 64 3 2 1 1 1 6 (4)4 3 2 x x ax bx cx d x ax bx cx dx a b c d                      Using GC to solve for a, b, c and d, we get 4, 6, 0, 7a b c d    Question 2 No. Suggested Solution Remarks for Student (a) (b) From the above graph, we can see that the 2 graphs intersect at two points. The left-most intersection point can be found using The question asks for exact solutions, so full algebraic working is required Raffles Institution H2 Mathematics (9758) Solution for 2021 A-Level Paper 1 15 5 5 2 5y x  2 1y x  1 2

2 | P a g e 2 2 5 2 1 2 6 0 2 4 24 1 72 x x x x x            From the graph, 0 5x  , so 7 1x  The right-most intersection point can be found using 2 2 5 2 1 2 4 0 2 4 16 1 52 x x x x x           From the graph, 5x , so 5 1x  The solutions are 7 1 and 5 1  . Question 3 No. Suggested Solution Remarks for Student (a) 1 1 2 2 3x y  Differentiate with respect with x, 1 1 2 2 1 1 2 2 1 1 2 2 1 2 1 1 d 02 2 d 1 d 1 d d d yx y x y xy x y y y x x x            (b) When x = 1, 1 21 3 4y y    Gradient of the curve at x = 1 is 1 24 21      Equation of normal at x = 1 is  14 12 1 7 2 2 y x y x     

3 | P a g e Question 4 No. Suggested Solution Remarks for Student (a)       1 116 8 14 1 18 8 2i2 i1 116 16 i i i1 18 8 ecos isin e ecos isin e ez                 4 2 2i i2 e e iz      Alternatively,      21 116 16 1 18 8 cos i sin cos isinz            21 1 216 16 1 18 8 cos isin 1 11cos isinz                        21 1 1 1 16 16 8 8 1 1 1 1 16 16 8 8 1 1 1 16 8 4 arg arg cos isin arg cos isin 2arg cos i sin arg cos isin 2 z                          So, z can be written as 4iez   . Note that calculator is not allowed in answering this question, thus all steps need to be clearly shown.    8 1 1 8 8 1 1 8 8 i cos isin cos +isin e            (b)(i) i i i cos isin 1 cos i sin e 1 e e 1 cos isin 1                    (b)(ii) The equation in (b)(i) holds for all values of , in particular we let 4  . Then,   44 4 (1 ) 1 (1 ) 1 (1) z z z z z z           But, 2 4i 1z z   Then, (1) becomes    44 4 4 (1 ) 1 1 (1 ) 0 z z z z          

4 | P a g e Question 5 No. Suggested Solution Remarks for Student (a)        2 3 4 2 3 4 ( 3)( 4) ( 2)( 4) ( 2)( 3) 2 3 4 x x x x A B C x x x A x x B x x C x x x x