RI 9758/01 2022

1 | P a g e Question 1 No. Suggested Solution Remarks for Student       i 2 1 ...(1) 2 i i 6 ...(2) (2) 2i : 2i 2 i 2 12i ...(3) (1)+(3): i 4i 2 1 12i 1 12i 2 5i 1 12i 2 5i 58 29i 2 i29 29 Sub back into (1): 1 i 2 i1 i i2 2 z w z w z w z z z z z zw                              This is a non-calculator question. Detailed working needs to be shown. Question 2 No. Suggested Solution Remarks for Student (a)            1 12 2 22 22 f tan 2 1f 1 2 1 2 2 2f 1 2 2 2 1 2 x x x x x xx x x x                               (b)        1 2 2 2 2 f 0 tan 2 0.95532 1 1f 0 0.33333 31 2 2 2 2 2f 0 0.31427 91 2 0.31427f 0.955 0.333 ... 2 0.955 0.333 0.157 ... x x x x x                     Note that all calculated values are to be in radians – check that you set calculator to the correct mode. Note the degree of accuracy for this question is 3 s.f. Raffles Institution H2 Mathematics (9758) Solution for 2022 A-Level Paper 1

2 | P a g e Question 3 No. Suggested Solution Remarks for Student (a)    3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 33 3 1 1 d 3e 2e e e e 3e2 2 d 2 1 1 d 3e 2e e e e 3e2 2 d 2 3e 3ed e 2e23d e 2ee 3e2 1 d 2 1ln 2, 33 d 2 1 t t t t t t t t t t t t t t t t t tt t xx t yy t y x yt x                                Gradient of normal = 1 3 lne for 0x x x  (b) 3 3 e 2e t t x y x y      3 3e 2et tx y x y       2 2 2 2 2 22 1, 2 2 2 x yx y x      Since question says state the restriction for x, we can use GC to draw  3 31 e 2e2 t tx   to get the restriction on x. (x as Y1 and t as X in GC)

3 | P a g e Question 4 No. Suggested Solution Remarks for Student (a)   2 2 2 2 2 2 2 d d coscotd d sin sin cos (using quotient rule)sin 1 (since sin cos 1)sin cosec xxx x x x x x x xx x            (b) Note that sin 2 2sin cosx x x and sintan cos xx x , therefore  2sinsin 2 tan 2sin cos 2sincos xx x x x x x  This is a “show” question (c)   9 18 9 18 9 2 18 29 18 9 18 cosec6 cot 3 d 1 dsin 6 tan 3 1 d (from (b))2sin 3 1 cosec 3 d2 1 cot 3 (from (a))6 1 cot cot6 3 6 1 1 36 3 1 1 3 6 3 1 3 or 93 3 x x x xx x xx x x x                                     Note that you are expected to simplify the answer. You can also use GC to verify whether your answer is correct.

4 | P a g e Question 5 No. Suggested Solution Remarks for Student (a)          2 2 2 2 2 2 2 2 2 2 2 2 2 8 14 52 16 64 28 196 52 1 16 28 208 0...(1) line is tangent to curve means (1) has one repeated root, 16 28 4 1 208 0 256 896 784 832 832 0 48 896 576 0 3 56 36 0 x mx x x m x mx m x m x m m m m m m m m m                                (b) since tangents intersects at (0,0), equations of both tangents will be 1 2 and y m x y m x  , where 1 2 and m m satisfy the equation in (a), solving 23 56 36 0m m      2 2 3 56 36 0 56 56 4 3 36 6 56 52 6 218 or 3 m m m              2 2 2 Sub 18 into (1) in (a), 325 520 208 0 0.8 (from GC or equation can be reduced to 0.8 0) 18 0.8 14.4 2Sub into (1) in (a), 3 13 104 208 09 3 12 (from GC or equation can be reduced to m x x x x y m x x x                   2 12 0) 2 12 83 x y      Coordinates are 12,8 and 0.8,14.4  .

5 | P a g e Question 6 No. Suggested Solution Remarks for Student (a)  2 2 f a x a a kax k a kx a x a x a x a          Sequence of transformation: 1. Translate a units in the positive x direction 2. Scale the graph by a factor 2a k parallel to the y-axis 3. Translate a units in the positive y direction (b)    1f f ax ky x a yx ya ax k x y a ay k ay kx y a ax kx x x a              (c) 2 1f ff f fx x x x    (d)      2023 2022f 1 f f 1 f 1 1 a k a     From (c),     2022 2 2 2f f f fx x x  

6 | P a g e Question 7 No. Suggested Solution Remarks for Student (a) 3 3 2 2 6 4 1 3 31 1 13 3 1 13 lnln d 3 ln 1 3ln d d 0 1 3ln 0 ed 1e ln e e3 1Coordinates are e , e3 xy x x x y x x x x x x x y x xx y                         (b) 3 3 1 3 32 3 11 32 1 ln d 1 1ln d2 2 1 1ln 318 4 1 1 1ln 318 36 4 2 1 ln 39 18 x x x x x x x x                      

7 | P a g e Question 8 No. Suggested Solution Remarks for Student (a)     2 2 22 2 2 2 1 d2 1 2 2 3 d2 1 2 2 3 d d , 12 1 1 3ln 2 1 1 3ln 1 1 32ln 1 1 x xx x x xx x x x x xx x x x x c x x c x x c x                                 Idea is to apply   f df x xx    (b)    2 20 1 22 12 20 2 1 222 2 10 2 2 1 d2 1 2 1 2 1 d d2 1 2 1 3 3ln 2 1 ln 2 11 1 9 9ln 2 3 ln 9 1 ln 24 4 16ln 9 4= 2ln3 x xx x x x x xx x x x x x x x x x                                                  Note:   2 1 12 1, 2 2 12 1 , 0 2 x x x x x        

8 | P a g e Question 9 No. Suggested Solution Remarks for Student (a) 2 2 2 2 2 2 14 2 14 2 4 4 14 4 10 0 50 (rejected since 0) or 2 a d ar a d ar a d a d a a d a ad d a ad d ad ad d d                 (b)   2 sin 1 cos sin 1 cos 2sin cos2 2 2cos 2 1tan , 2 2 S k                  (c)  7 7 7 7 3, sin3 2 1cos 2 1 1 3 112 2 11 2 3 1 212 2 3 3 129 3 128 43 3 128 a r a rS r                                    

9 | P a g e Question 10 No. Suggested Solution Remarks for Student (a)        2 2 2 2 2 1 d 2 d 1 has no stationary points, d 0 1 2 0 has no solutiond 2 2 0 has no solution 4 4 2 0 2 0 1Since 0, 2 0 2 or . 2 a by ax b x y a bax x C y a x a bx ax ax b a a b a a b a a b a b b a                             (b) Given 2b a , 2 4b a  . Since 0a , 2a b From (a), C has no stationary point. (c) As shown in diagram for (b) (d) 1a From graphs, 2 or 1x x 

10 | P a g e Question 11 No. Suggested Solution Remarks for Student (a)    22 2 2 2 936 939 72 939 15 936 72 15 939 15 441 6 (rejected since 15) or 36 QP p p p p p p                    (b) 100 200 2 20 70 5 400 940 50 8 10 94 500 14 7000 1 50 30 1 3 28 2 5 400 5 1 600 1 2560 2 20 2 Cartesian equation of the pl r                                                                                                ane is 5 2 2560x y z   Question asked for cartesian equation of the plane (c)        200 312 : 20 24 15 7 Sub into 5 2 2560 5 200 312 20 24 2 15 7 2560 1 Coordinates are 512, 44, 22 PQl r x y z                                    Question asked for coordinates (d) Let  be angle of PQ made with vertical 312 0 24 0 7 1 7cos 313312 0 24 0 7 1 88.719                                     Thus, the angle of PQ made with horizontal is 90 1.3 (1 d.p.)   The horizontal refers to a plane perpendicular to the z-axes. 178.7o is also acceptable in this case as question did not specify acute angle.