RI 2017 P1 A-Level H2 Math Solution

1 | P a g e Question 1 No. Suggested Solution Remarks for Student          2 2 3 2 3 2 3 2 22 2 3 2 2 3 e ln 1 2 21 2 ... ... , 1 12! 3! 2 3 22 ...2 3 2 4 3 6 ...2 3 x ax x x ax axx ax ax ax ax ax xax ax x ax a a a a aax x x                                           2No term in x , we have 24 0 0 (rejected since 0) or 4.a a a a a      Raffles Institution H2 Mathematics (9758) Solution for 2017 A-Level Paper 1

2 | P a g e Question 2 No. Suggested Solution Remarks for Student (i) (ii) To solve 1 b x ax a  , we consider the graphs 1y x a  and y b x a  drawn in (i), noting they intersect at a point where x > a.    2 1 , . 1 1 1 , reject since b x a x ax a x a x a x a x ab b b                1 b x ax a  1x a b  or x a You need to refer to (i) if you are using “Hence”. ab x=a y=b|x – a|

3 | P a g e Question 3 No. Suggested Solution Remarks for Student (i) 2 22 5 10 0y xy x    … (1) Differentiate (1) with respect to x: d d2 2 2 10 0d d y yy x y xx x    d d 5 0d d y yy x y xx x    … (2) When d 0d y x , 5y x Sub into (1): 2 2 2 2 1 125 10 5 10 0 20 10. or 2 2x x x x x         (ii) Differentiate (2) with respect to x: 22 2 2 2 d d d d d 5 0d d d d d y y y y yy xx x x x x          …(3) When 1 5 d, and 0 d2 2 yx y x   . Sub into (3): 2 2 2 2 2 2 5 d 1 d d 5 25 0 0d d d 42 2 y y y x x x      Thus, 1 5, 2 2     is a maximum turning point. Use 2nd Derivative Test here. 1st derivative test cannot be used as d d y x depends on both x and y. Note that exact values with working are required as no calculator is allowed for this question.

4 | P a g e Question 4 No. Suggested Solution Remarks for Student (i) 4 2 14 9 1 42 2 2 xxy x x x        Note that C is defined on all values of x except –2.   2 d 1 0 for all 2d 2 y xx x   Thus, gradient of C is negative for all points on C. (ii) 14 2y x   Asymptotes are 2, 4x y  (iii) A translation of 2 units in the positive x direction follow by a translation of 4 units in the negative y direction. Replace x by (x – 2) follow by replacing y by (y + 4) is not acceptable.

5 | P a g e Question 5 No. Suggested Solution Remarks for Student (i)     3 2f f 1 8 1 8 7 ...(1) f 2 12 8 4 2 12 4 2 4 ...(2) f 3 25 27 9 3 25 9 3 2 ...(3) x x ax bx c a b c a b c a b c a b c a b c a b c                                   Solving (1), (2) and (3), 3 3, , 72 2a b c   (ii)  3 23 3f 7 2 2x x x x    We know that in general a cubic function has 1, 2 or 3 x-intercepts.    2 2 2 2 2 3f ' 3 3 2 33 2 1 1 33 4 4 2 1 3 13 0 for all since 0 for all 2 4 2 x x x x x x x x x x x                              f as and f as . Since gradient of curve is always positive, curve is always increasing. x x x x    (in particular, f is 1-1) So, f 0 has only one root.x  Using GC, the root is –1.33 (3 s.f.) (iii)   2 2 1 3f ' 2 3 2 2 4 1 5 2 12 1 5 2 12 x x x x                

6 | P a g e Question 6 No. Suggested Solution Remarks for Student (i) It is a line passing through the point A with position vector a and is parallel to the vector .b (ii) It is a plane with normal vector n such that the dot product of the position vector of any point on the plane with the normal has the value d. Since n is a unit vector, the magnitude of d is the distance from the origin to the plane. (iii) . . . .. 0, . . . a tb n d a n tb n d d a nb n t b n d a nr a b b n                                 which is the position vector of the point of intersection of the line in (i) and the plane in (ii).

7 | P a g e Question 7 No. Suggested Solution Remarks for Student (i)       sin 2 sin 2 d 1 2sin 2 sin 2 d2 1 cos 2 2 cos 2 2 d2 1 1sin 2 2 sin 2 24 4 4 4 mx nx x mx nx x m n x m n x x m n x m n x Cm n m n                (ii)        2 0 2 0 2 2 0 0 0 0 0 f d sin 2 sin 2 d sin 2 sin 2 2sin 2 sin 2 d 1 cos 4 1 cos 4 2sin 2 sin 2 d2 2 1 1 1 1sin 4 sin 42 8 2 8 1 1sin 2 2 sin 2 22 2 2 2 sin x x mx nx x mx nx mx nx x mx nx mx nx x x mx x nxm n m n x m n xm n m n k                                                  0 for k  

8 | P a g e Question 8 No. Suggested Solution Remarks for Student (a)          2 2 1 i 2 5 5i 0 2 2 4 1 i 5 1 i 2 1 i 2 4 40 2 1 i 2 6i 2 1 i 2 6i 1 i 2 6i 1 ior2 1 i 1 i 2 1 i 1 i 2 2i 6i 6 2 2i 6i 6or 4 4 1 2i or z 2 i z z z z z z z z z z                                     You can also divide through by 1 i before applying the quadratic formula Note that detailed working is required as calculator is not allowed. (b) (i)      22 3 2 24 2 1 i 1 i 2i 2i 1 i 2 2i 2i 4 w w w w w                    4 3 2 39 58 0 4 2 2i 39 2i 1 i 58 0 Compare Real and Imaginary parts: 4 2 58 0 ...(1) 2 78 0 ...(2) (1) (2) : 4 24 0 6 Sub into (2) : 12 78 0 66 w pw w qw p q p q p q p p q q                                    (ii)         4 3 2 2 6 39 66 58 0 Note that 1 i is a root and coefficients are real, thus 1 i is also a root One quaratic factor is 1 i 1 i 2 2 w w w w w w w w               4 3 2 2 26 39 66 58 2 2w w w w w w w pw q          By comparing constant terms, 58 2 29q q   By comparing linear term, 66 2(29) 2 4p p     4 3 2 2 26 39 66 58 2 2 4 29w w w w w w w w         

9 | P a g e Alternatively     2 2 4 3 2 4 3 2 3 2 3 2 2 2 4 29 2 2 6 39 66 58 2 2 4 37 66 58 4 8 8 29 58 58 29 58 58 0 w w w w w w w w w w w w w w w w w w w w w                        

10 | P a g e Question 9 No. Suggested Solution Remarks for Student (a) 2 1 n n r r S u An Bn     (i)      1 22 1 1 2 1 n n nu S S An Bn A n B n A n B            (ii) 10 17 48 19 48 90 33 90 Solving, 3, 9 u A B u A B A B           (b)                                              2 2 2 22 2 2 2 2 3 2 23 2 2 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 22 2 22 2 22 1 1 1 1 1 1 1 1 2 2 4 1 1 14 1 1 2 0 14 2 3 1 2 3 4 2 3 ... 1 2 1 1 1 1 14 n n r r r r r r r r r r r r r r r r r r r r r r n n n n n n n n n n                                                  (c)   1 1 Let . ! ! 1 ! 1 n n n n n n xa n a x n x a n x n        1lim lim 0 1 1 n n n n xa a n      for each x. Therefore 0 ! r r x r    converges. 0 e! r x r x r    from MF26.