RI 9758/02 2023 - new, pls ignore previous version due to NRIC in metadata

1 | Page Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student (a)            22 22 32 5 2 92 5 2 92 5 2 0 36 5 236 5 2 0 15 1 1 0 111 5 xx xx xx xx xx xx x                 The required set is 11:1 5xx   Alternatively, From the graphs, 111 5x The required set is 11:1 5xx   (b)        2 2 22 2 25 3 4 5 35 225 3 11 10345 45 45 5 1 xx x xxxx x xx xx xx x x            Raffles Institution H2 Mathematics (9758) Solution for 2023 A-Level Paper 2 5 2

2 | Page           2 2 25 345 25 3045 35 2 051 13 5 2 5 0 51o r < 2o r 53 x xx x xx xx xx xx x x xx x               Question 2 No. Suggested Solution Remarks for Student (a)  2 2 2 3 3 2 2 2 ds e c t a nln(sec ) tan ds e c d secd d 2sec sec tand 2sec tan dd2( S h o w n )dd yx xyx x xx y xx y xx xx xx yy xx           (b) 232 43 2 32 43 2 dd d dd dd22 2dd d dd dd yy y yy y y xx x xx xx            When 0,x ln sec 0 0,y d tan 0 0,d y x  2 2 2 d sec 0 1,d y x    3 3 d 21 0 0 ,d y x     4 2 4 d 20 0 21 2d y x   Therefore, 24 2 4 3 20 0 0 ... ...2! 4! 2 12 xx x xyx x     (c) When 1 π,4x   24 24 π 1ln sec ln 2 ln 242 ππ 44 ...21 2 ππ 32 3072 y        Therefore,  24ππln 2 16 1536 (d)  11 24ππ10 10 00 ln sec d d 21 2 0.0052187161 0.005219 (to 4 s.f.) xxxx x    Use GC to evaluate 5 3 1 2 5 + + + - -

3 | Page Question 3 No. Suggested Solution Remarks for Student (a)  2 2 32 11 5 32 2 2 1 0 85 1 4 BC AB OC OB OB OA OC OB OA                                      (b) 1 2OD d     , 11 2 1 1 0 22 0 , 2 2 4 55 88 AD BD dd dd                                      2222 22 20 04 58 25 48 41 0 2 5 1 61 6 6 4 65 1 17 2 AD BD dd dd dd dd d d                        (c) Let angle ADB =  1 20 0.4 71 .7 22cos 6549 141 644 7cos 83.81769576 83.8 (to 1 d.p.)65 AD BD AD BD                               Angle ADB 83.8 (to 1 d.p.) 

4 | Page (d) Since AD BD     , so ABD is an isosceles triangle. Let M be the midpoint of of AB. Then, 11 22 085 02 6.5 OM            Equation of the line of DM is 01 0 02 0 6.5 8.5 6.5 01 02 , 6.5 2                           r  Since P lies on line DM, then 01 022 , f o r s o m e 6.5 2 6.5 2 OP                                    22 222 2 2 2 22 11 22 2 2 6.5 2 8.5 6.5 2 8 11 22 22 22 1 . 52 12 2 2 2 1 2 2 1 . 5 2 81 6 8 48 4 2 . 2 56 DP BP                                                                           24 718 4 7 72       7 7201 77Therefore 0 2 72 366.5 2 241 36 OP                      Since P is the centre of the circle, then DP = BP = radius of the ticle

5 | Page Question 4 No. Suggested Solution Remarks for Student (a) 2 d23 4 d xx tt t  When 1 ,5t  2 17 723 52 5x   When 21,x 2 1823 2 1 3 2tt    since 1 .5t  Exact area between the curve C , the x-axis and the line 21.x    21 77 25 3 1 5 3 2 1 5 d 51 4 d 20 4 d yx tt t tt t       33 2 1 5 2 20 23 12152 units75 t t  Note that the question asks for exact answer (b) Cartesian equation of D is 42 0 (1) 5 y x x   Substitute 223xt and 51yt into (1),   2 32 32 2 2051 23 10 2 15 3 20 10 2 15 23 0 11 0 8 2 3 0 t t tt t tt t tt t         21o r 1 0 8 2 3 0tt t   Since discriminant   28 4 10 23 856 0    , the equation 210 8 23 0tt   , has no solution. Therefore, there is only one solution at 1.t  When 1t  , 5x  and 4y The curves C and D intersect at A 5, 4 and there are no other points of intersections. It is not enough to just say that there is just 1 answer for the value of t because this can be inferred from the question itself. You will need to show that t = 1 is the ONLY answer. (c) 2 d23 4 d xx tt t  and d51 5 d yyt t   Therefore d5 d4 y x t

6 | Page At A where 1,t  d5 d4 y x  Equation of the tangent of the curve at A is 545 4 1 6 5 2 54 59 44 yx y x yx         When the tangent to the curve C at the point A meets the curve D,    2 20 5 9 44 59 8 0 0 51 6 50 16 or 55 xx xx xx xx       When 16 5x , 20 25 16 4 5 y   The coordinates of the point where the tangent to the curve C at the point A meets the curve D for a second time is 16 25,54  .

7 | Page Section B: Probability and Statistics Question 5 No. Suggested Solution Remarks for Student (a)(i) A and B are mutually exclusive. A and D are mutually exclusive. (ii) { 1, 3, 5, 7, ..., 35} ( ) 18An A  {3, 6, 9, 12, ..., 36} ( ) 12Cn C  {3, 9, 15, 21, 27, 33} ( ) 6AC n AC     18 1 36 2PA  and  12 1 36 3PC    11 1 23 6PA PC    61 36 6PA C Since      P AC P A P C  , A and C are independent. Since A and B are complement, B and C are independent. (b)(i) After the ball has become stuck in the slot labelled 36, { 1, 3, 5, 7, ..., 35} {2, 4, 6, 8, ..., 34} A B   {3, 6, 9, 12, ..., 33} {6, 12, 18, 24, ..., 30} C D   A and B are still mutually exclusive. A and D are still mutually exclusive. (b)(ii) { 1, 3, 5, 7, ..., 35} ( ) 18An A  {3, 6, 9, 12, ..., 33} ( ) 11Cn C   18 35PA  and  11 35PC    18 11 198 35 35 1225PA PC   {3, 9, 15, 21, 27, 33} ( ) 6AC n AC     6 35PA C Since     PA C PA PC  , A and C are no longer independent.

8 | Page Question 6 No. Suggested Solution Remarks for Student (a)  48choosing 4 red counters 12 rb P rb         39choosing 3 red counters 12 rb P rb        Since the above two probabilities are equal (given in the question), 48 39 12 12 48 39 !! !! 4!( 4)! 8!( 8)! 3!( 3)! 9!( 9)! 4!( 4)! 8!( 8)! 3!( 3)! 9!( 9)! 4( 8) 9( 3) 43 2 92 7 95 4 ( rb rb rb rb rb rb rb rb rb rb rb rb br br rb                                  1) (Shown) It is not surprising that the equation can be reduced to 48 39 rb rb       because the 2 probabilitie being equal is equivalent to saying the number of ways of choosing 4 red counters is equal to the number of ways of choosing 3 red counters (b)  21 0choosing 2 red counters 12 rb P rb            39choosing 3 red counters 12 rb P rb        Therefore ,

9 | Page  39 21 0 5 3 12 12 5 39 21 0 3 !! 5 ! ! 3!( 3)! 9!( 9)! 3 2!( 2)! 10!( 10)! 5 3!( 3)! 9!( 9)! 2!( 2)! 10!( 10)!3 5 3(3 rb r b rb rb rb r b rb r b rb r b rb r b b                                            9) 10( 2) 54 5 1 02 0 10 25 5 (2) r br rb       Solving (1) and (2), 15r  and 35b The required probability  1 red counter 15 35 11 1 50 12 0.051551948 0.0516 (to