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 π πsin π sin coscos π24 2 2 242 4 xx x xx x xx x xx xxx x xx x xx x                      π 11 π00 0 084 4 8 π 4          

5 | Page Question 5 No. Suggested Solution Remarks for Student (a)                            2 22 11ln ln 1 2 ln ln 1 ln 1 2ln 2 ln 3 ln 2 2 ln 3 ln 4 ln 3 2 ln 4 ln 5 ln 4 2 ln 5 ln 6 ln 3 2 ln 2 ln 1 ln 2 2 ln 1 ln ln 1 2 ln ln 1 ln 1 2 ln 2 ln 2 ln 2 ln ln 1 ln 2 ln ln 1 ln nn rr rr rr rr nn n nn n nn n nn n nn n                                      1 ln 2 (Shown)n   (b) As n , 11 11n nn    , therefore    2 2 11 1ln ln ln 2 ln1 ln 2 ln 2 n r rr n rn        (a finite number) So, the corresponding infinite series in convergent and the sum to infinity is ln 2 . (c)         20 20 9 22 2 10 2 2 11 11 11ln ln ln 21 10ln ln 2 ln ln 220 9 21 9 189ln ln20 10 200 rr r rr rr rr rr r                           Therefore, a =189 and b = 200.

6 | Page Question 6 No. Suggested Solution Remarks for Student (a) Use double angle formula  2 1cos cos 2 12 , we have     242 2 2 2 cos cos 1 cos 2 12 1 cos 2 2cos 2 14 11 1 cos 4 1 2cos 2 1 since cos 2 cos 4 142 2 11 3 cos 4 2cos 242 2 1 cos 4 4cos 2 3 (Shown)8                      (b) Volume generated 3 2 1.5 =π dyx  33 2 2 1.5 =π 9 d x x d3sin 3cosd xx    When 1.5x , π3sin 1.5 sin 0.5 6    When 3x , π3sin 3 sin 1 2   Therefore, volume generated  π 3 22 2 π 6 =π 9 9sin 3cos d    π 3 3 22 2 2 π 6 π 32 π 6 π 9 1 sin 3cos d π 27 cos 3cos d          π 42 π 6 π 2 π 6 π 2 π 6 81π cos d 181π cos 4 4 cos 2 3 d8 81π sin 4 2sin2 384 81π 3π 3 π382 8 2 81π 93 8 1 ππ 8π 9388 6 4                      

7 | Page Question 7 No. Suggested Solution Remarks for Student (a) 24 3 xy x   (b) Range of f =  fR0 , (c) Since f3R but f3D . So, ffRD , therefore the function 2f does not exist. Since answer is known ( 2f does not exist), it is important to write down how come ffRD by giving a specific value that inside fR but not inside fD . (d) For the function 1f to exits, f has to be one-one, so the greatest value of a is 2 . (e) Let 24 24 24 , since 0 for 2 33 3 xx xyx xx x        24 3 32 4 34 2 xy x yx y x yx y       Therefore 1 34f( ) 2 xx x    . Its domain is  1 ffDR 0 , 2 

8 | Page Question 8 No. Su ggested Solution Remarks for Student (a)(i) 21 2π1 3 and arg = π tan 3 3 ( is in the 2nd quadrant) zz z    Therefore 2πi 313 i = 2 ez  (ii) 2 πi 2 π 2ππ 2 ππ3 ii11 33 2 3 6 2ππ ii 32 2e 2e = 2ei* e2 e n nnnn nnz z             For i* nz z to be purely imaginary,  2 ππ 2 ππ πcos 0 (2 1) where36 36 2 31 1 3122 2 nn kk nk k           [Observe the graph of cosyx to get the expression π(2 1) 2k ] It is clear that the required smallest positive n occurs when 1k  . Therefore, the required smallest positive 2n . Recall that i 2ie   Alternatively, 2 ππ π 35,,,36 2 2 2 15 7,1 , 2 , , ,4 ,22 2 n n           (b) 21 ( 1 ) 3i 3 4 i ( 2 ) vw vw        Let iva b and iwx y , where ,,,abxy   From (1), 121 1 2vw v w     Thus, we can say that va  (with b = 0) From (2), 3i (i ) 3 4 i (3 ) i 3 4i ax y ay x       Comparing imaginary parts, 4x Comparing real parts, 33ay  From (1),  22 2 22 22 2 12 5 21 5 21 21 6 ( 3 3 ) 1 16 9(1 ) 1 2 16 9(1 2 ) 1 4 4 52 2 2 4 0 2, then, 3, ax y aa aa aa a a aa a y                So, possible solutions are 2, 4 3i and 12 21,4 i55 vw vw       Note that in this case w does not mean w or w . You can only say this when w is a real number. Since w is a complex number, then w means the magnitude of the complex number w (i.e.   22 R e () I m ()ww  )

9 | Page Question 9 No. Suggested Solution Remarks for Student (a) 1 32 :1 1 , 2 l a               r  and 2 23 :1 2 , 51 l               r  Since 1l and 2l cross at the point B, 3223 11 12 , f o r s o m e , 25 1 231 ( 1 ) 20( 2 ) 3( 3 ) a a                                           Using (1) and (2), 2a n d 1  , hence using (3), 2a Using 1  , 231 12 3 51 6                        r The coordinates of B is  1, 3,