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 x                    Comparing numerators, ( 3)( 4) ( 2)( 4) ( 2)( 3)A x x B x x C x x x         When 2x , (1)(2) 2 1A A    When 3x , ( 1)(1) 3 3B B    When 4x , ( 2)( 1) 4 2C C     So,     1 3 2 2 3 4 2 3 4 x x x x x x x          (b)    1 1 1 3 2 2 3 4 2 3 4 1 3 2 3 4 5 1 3 2 4 5 6 1 3 2 5 6 7 1 3 2 6 7 8 1 3 2 1 2 1 3 2 1 2 3 1 3 2 2 3 4 1 3 1 2 3 2 3 4 4 3 3 4 1 1 2 6 3 4 n n r r r r r r r r r n n n n n n n n n n n n n n                                                           (c)    1 1 2 3 4 6r r r r r      Since the question says “state”, working is not required.

5 | P a g e Question 6 No. Suggested Solution Remarks for Student (a) Note that C is not defined when 24 0, i.e. 0 or 4ax x x a   . (b) Smallest possible value of y occurs when 24ax x is the largest. The largest value of 24ax x occurs at 2x a . Thus, the smallest possible value of y is 2 1 1 24 (2 ) (2 ) aa a a  . (c) 2 2 2 1 1 4 4 ( 2 )y ax x a x a     Thus, replace by 2 2 2 2 2 2 1 1 1 4 4 ( 2 ) 4 x x ay y ax x a x a a x         The transformation is a transaltion in the negative x-direction by 2a units. 0x 4x a y 0 4a

6 | P a g e Question 7 No. Suggested Solution Remarks for Student (a) 1sin 1e ln sinxy y x     Differentiate with respect to x,   2 2 2 2 2 1 d 1 d 1 d1 d d1 d y y x x yx yx yx y x          Differentiate with respect to x,     22 2 2 2 2 2 2 2 2 d d d d1 2 2 2d d d d d d1 d d d d1 (1)d d y y y yx x yx x x x y yx x yx x y yx y xx x                    Alternatively, 1 1 sin sin 2 2 e d 1 ed 1 d1 d x x y y x x yx yx        Differentiating wrt x         1 2 2 22 2 2 2 2 2 2 2 2 2 2 2 1 d d d1 2 12 d d d d d d1 1d d d d d1d d d d1 d d y y yx x x x x x y y yx x xx x x y yx x yx x y yx y xx x                   (b) Differentiate (1) with respect to x,    3 2 2 2 3 2 2 3 2 2 2 3 2 2 d d d d d1 2 d d d d d d d d d1 2 2d d d d y y y y yx x xx x x x x y y y yx x xx x x x          When 0x , 2 3 2 3 d d d1, 1, 1, 2.d d d y y yy x x x    So, 1sin 2 3 2 3 1 2 1 1e 1 1 2! 3! 2 3 x x x x x x x            

7 | P a g e Question 8 No. Suggested Solution Remarks for Student (a) The direction vectors of the plane are 2 3 1       and 3 1 2 2 3 5 0 1 1                             A normal vector to the plane 2 2 8 1 3 5 0 8 0 1 1 16 2                                          Equation of the plane 1 3 1 . 0 2 . 0 3 2 0 2                             r The Cartesian equation of the plane is 2 3x z  . (b) Two lines are perpendicular if their direction vectors are perpendicular. 2 1 3 . 2 2 6 4 0 1 4                     Thus, 1l is perpendicular to 2.l (c)(i) 1 2 3 2 4 1 7 2 1 2 3 1 2 1 3 2 0 1 3 4 3 1 4                                                                                      r r Since 1 2r r is perpendicular to 1l ,  1 2 2 . 3 0 1 7 2 1 2 1 3 2 . 3 0 3 1 4 1 14 14 0 1                                                             r r Likewise for the line 2l ,  1 2 1 . 2 0 4 7 2 1 1 1 3 2 . 2 0 3 1 4 4 21 21 0 1                                                           r r

8 | P a g e The position vectors are 3 2 1 2 3 5 0 1 1                             and 4 1 3 1 2 3 3 4 1                             Alternatively, 1 2 14 2 2 3 7 7 1 4 1 7 1                                         Since 1 2r r is perpendicular to both 1l and 2l , then 1 2 2 1 , for some 1 7 2 2 1 3 2 1 3 4 1 k k k                                    r r  From G.C., 1, =1  The position vectors are 3 2 1 2 3 5 0 1 1                             and 4 1 3 1 2 3 3 4 1                             Note that 1 2b b is a vector perpendicular to both lines 1l and 2l , where 1 2 and b b are direction vectors of 1l and 2l . (c)(ii) Length of common perpendicular is 1 3 4 5 3 2 24 2 6 1 1 2                              

9 | P a g e Question 9 No. Suggest