RI 9758/01 2023 - new, pls ignore previous version due to NRIC in metadata

1 | Page Question 1 No. Suggested Solution Remarks for Student  2 ln 11 5y x Differentiate with respect to x,    1d 21 1 5 ( 5 ) 1 01 1 5d d 10 11 5d y xxyx y yxx       When 2x ,  2 ln 11 5(2) 1 eyy    d 10 11 5 10(e)(11 10) 10ed y yxx      The equation of tangent is e1 0 e ( 2 ) 10e 21e yx yx       Question 2 No. Suggested Solution Remarks for Student (a) 32 1 2 3 4 10 842 6 1 27 9 3 206 64 16 4 469 nua n a nc n d ua b c d ua b c d ua b c d ua b c d             Solving, 4, 23, 46, 29ab c d    So, 3242 34 6 2 9nun n n  (b) 324 23 46 29 25000nun n n   When n =16, 21565nu  < 25000 When n =17, 25546nu  > 25000 Range of values of n is  :1 7nn Those who gave the answer as 16.9n will not get the full marks for not recognizing that n has be an integer. Raffles Institution H2 Mathematics (9758) Solution for 2023 A-Level Paper 1

2 | Page Question 3 No. Su ggested Solution Remarks for Student (a)      ab a ab b         .0 .. . . 0           aba abb ab ab abbaab a b   2 1 0, since . . 0 and .       ab abb a ab ab 1 2 1 1, since 0    ab ab ab   .. 0 abb aab because ab is perpendicular to both a and b. The reason needs to be given. (b) Let  be the angle between the direction of a and the direction of b. Then, .c o s 1  ab a b ---- (1) sin 1 ab a b ---- (2) sin(2) (1), 1 cos tan 1       ab ab So, 135  

3 | Page Question 4 No. Suggested Solution Remarks for Student (a) 1cos cos d cos( ) cos( ) d2 11 1 sin( ) sin( )2 px qx x p q x p q x x p qx p qx Cpq pq          Refer to the list of factor formulae in MF26 for assistance. (b) 2 11cos d sin sin d 11 sin cos xn x x x n x n x xnn xn x n x Dnn    Integration by parts. d, cosd vux n x x ds i n1, d un x vxn (c) Using the results in part (b), 0 2 0 22 22 cos d 11 sin cos 11 1 1sin cos (0)sin (0) cos (0) 11cos , since sin 0 for and cos 0 =1 xn x x xn x n xnn nn n nnn n n nn nnn              When n is even, π 220 11 0cos π 1 and cos d 0nx n x x nn    When n is odd, π 220 11 2cos π 1a n d c o s dnx n x x nn      Therefore, π 20 cos d kxn x x n , where 0 when is even 2 when is odd nk n  (d) Since cos 2 0xx  when π0 4x and cos 2 0xx  when ππ 42 x , so in the interval π0 2x , πcos 2 , when 0 ; 4cos 2 ππcos 2 , when . 42 xx x xx xx x     Again, using your GC, sketch cos 2yx x .

4 | Page π 2 0 ππ 42 π0 4 ππ 42 π0 4 ππ 42 π0 4 cos 2 d cos 2 d cos 2 d cos 2 d cos 2 d sin 2 cos 2 sin 2 cos 2 24 24 ππ πsin cos 0c o s 042 2 24 2 4 ππ