RI 2022 P1 A-Level H2 Math Solution

1 | Page 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 2i29 29 Sub back into (1): 1 i2 i1i i22 zw zw zw z zz 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 12 12 22 f 1 2 22 12 xx xx x x x xx x − − − = + ′ = = ++ ++ −+′′ = −+ + + = ++ (b) ( ) ( ) ( ) ( ) [ ] ( ) 1 2 2 2 2 f 0 tan 2 0.95532 11f 0 0.33333312 22 22f 0 0.31427912 0.31427f 0.955 0.333 ...2 0.955 0.333 0.157 ... x xx xx −= = ′ = = = + −−′′ = = =− + = +− + = +− + Note that all calculated values are to be in radians – check that you set calculator to the correct mode. As stated in the question, n ote the degree of accuracy for this question is 3 s.f. Raffles Institution H2 Mathematics (9758) Solution for 2022 A-Level Paper 1

2 | Page Question 3 No. Suggested Solution Remarks for Student (a) ( ) ( ) 3 3 33 3 3 3 3 33 3 3 33 33 33 33 1 1 d3e 2e e e e 3e2 2 d2 1 1 d3e2 e ee e3 e2 2 d2 3 e 3ed e 2e2 3d e 2ee 3e2 1 d 21ln 2, 33 d 21 t t tt t t t t tt t t tt tt tt tt xx t yy t y x yt x −− − −− − − − − − = + = +⇒= − = − = −⇒= + + += = −− += = = − Gradient of normal = 1 3− Recall that: lne for 0x xx= > (b) 3 3 e 2e t t xy xy − += −= ( )( ) ( ) 33e 2ettxyxy −+ −= ( ) ( ) 22 22 222 1, 2 22 xyxy x− = ⇒−= ≥ Since question says state the restriction for x , we can use GC to draw ( ) 331 e 2e2 ttx −= + to get the restriction on x. (x as Y1 and t as X in GC)

3 | Page Question 4 No. Suggested Solution Remarks for Student (a) ( ) 22 2 22 2 2 d d coscotd d sin sin cos (using quotient rule)sin 1 (since sin cos 1)sin cosec xxx xx xx x xxx x =  −−= −= += =− (b) Note that sin 2 2sin cosx xx= and sintan cos xx x= , Therefore ( ) 2sinsin2 tan 2sin cos 2sin cos xxx xx x x= = This is a “show” question, so detailed steps need to be shown. (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 cot63 6 11 36 3 11 3 6 3 13 or 933 x xx xxx xx xx x π π π π π π π π π π ππ = = = =− = −− = −− −=−   = ∫ ∫ ∫ ∫ Note that you are expected to simplify the answer. You can also use GC to verify whether your answer is correct.

4 | Page Question 5 No. Suggested Solution Remarks for Student (a) To find the intersection between the tangent line and the curve, ( ) ( ) ( ) ( ) ( ) ( ) ( ) 22 2 22 22 2 2 22 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, discriminant 16 28 4 1 208 0 256 896 784 832 832 0 48 896 576 0 3 x mx x x m x mx m x mx mm mm m mm m ++ − = + ++ − + = + +− + = = − −+ = − + − −= − − −= + 56 36 0m+= (b) since tangents intersects at (0,0), equations of both tangents will be 12 and ym x ym x= = , where 12 and mm satisfy the equation in (a), solving 23 56 36 0mm+ += ( )( ) 2 2 3 56 36 0 56 56 4 3 36 6 56 52 6 218 or 3 mm 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 13 104 208 093 12 (from GC or equation can be reduced to m xx xx y m xx x =− + += = − += = −− = =− + += =− ( ) ( ) 2 12 0) 2 12 83 x y += = −− = Coordinates are ( ) ( )12,8 and 0.8,14.4−− .

5 | Page Question 6 No. Suggested Solution Remarks for Student (a) ( ) ( ) 2 2 f ax a a kax k a kxa xa xa xa −++++= = = +−− − 2 22replace by r eplace by replace by 11 yyx xa y yaak ak aky y y yax xa xa xa −− + ++=  → =  → =  → = +−− − Sequence of transformation: 1. Translate a units in the positive x direction 2. Scale the graph by a factor ( ) 2ak+ parallel to the y-axis 3. Translate a units in the positive y direction (b) ( ) ( ) ( ) 1ff ax ky xa yx ya ax k x y a ay k ay kx ya ax kxx xa − += − −=+ −=+ += − += =− (c) Method 1 ( ) ( ) ( ) 21f ff f fx x xx −= = = Method 2 (NOT Recommended. Only 1 mark) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 ff ax kx xa ax kak xa ax k axa a ax k k x a ax k a x a a x kx ka xa k xka +=  − + + −= + − − ++ −= +− − += + + = = + Note that ( ) 22f() f ()xx≠ (d) ( ) ( ) ( ) 2023 2022f 1 f f1 f1 1 ak a += = = − From (c), ( ) ( )( )( ) 2022 2 2 2f ff f x xx= =

6 | Page Question 7 No. Suggested Solution Remarks for Student (a) 3 3 22 64 1 3 311 133 1 13 lnln d 3 ln 1 3ln d d 0 1 3ln 0 ed 1e ln e e 3 1Coordinates are e , e 3 xyx x x yx x x x xx x y xxx y − − − − = = −−= = =⇒− =⇒ = = =     (b) 3 3 1 3 323 1 1 32 1 ln d 11 ln d22 11ln 318 4 1 11ln 318 36 4 21 ln 39 18 x xx x x xx x − −− − = −+ = −−  = − −+ = − ∫ ∫ You can use the GC to check your answer. Recall that ln(1) = 0.

7 | Page Question 8 No. Suggested Solution Remarks for Student (a) Method 1 ( ) ( ) ( ) ( ) 22 2 2 21 21 21 1 1 1 1 1 xx xx x AB x x Ax B x −− =++ + = ++ + ++= + Comparing numerator, ( )21 1x Ax B−= + + Comparing coefficients, A = 2, 13 AB B−= + ⇒ = − Thus, ( ) 2 2 21 d21 23 d1 1 32ln 1 1 x xxx xx x xc x − ++ = − + + = ++ + + ∫ ∫ Method 2 ( ) ( ) ( ) 2 2 22 2 2 21 d21 2 23 d21 22 3 d d , 121 1 3ln 2 1 1 3ln 1 1 32ln 1 1 x xxx x xxx x x xxxx x xx c x xc x xc x − ++ +−= ++ += − ≠−++ + = + ++ + + = ++ + + = ++ + + ∫ ∫ ∫∫ Idea is to apply ( ) ( ) f df x xx ′ ′ ⌠ ⌡

8 | Page (b) ( ) ( ) 2 20 1 22 1220 2 1 2222 10 2 21 d21 21 21 d d21 21 33ln 2 1 ln 2 111 99ln 2 3 ln 9 1 ln 244 16ln 9 4= 2ln 3 x xxx xx xxxx xx xx xxxx − ++ −−=−+ ++ ++  = − + ++ + + ++ ++    = − +− + +− −     = ∫ ∫∫ Note: ( ) 21 12 1, 2 2 12 1, 0 2 x xx xx −  − ≤≤= − − ≤<

9 | Page Question 9 No. Suggested Solution Remarks for Student (a) 2 2 22 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 da d a ad a ad d a ad d ad ad dd += += ++ = + ++= + −= = ≠= (b) ( ) 2 sin 1 cos sin 1 cos 2sin cos22 2cos 2 1tan , 22 S k θ θ θ θ θθ θ θ ∞ = −− = + = = = (c) ( ) 7 7 7 7 3, sin32 1cos 2 1 1 31 122 11 2 3 1212 23 3 129 3 128 43 3 128 a r ar S r πθθ θ = = = = −= − − = −  −−  = −− = +× =  =

10 | Page Question 10 No. Suggested Solution Remarks for Student (a) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 1 d2 d 1 has no stationary points, d 0 1 2 0 has no solutiond 2 2 0 has no solution Discriminant 4 4 2 0 20 Since 0, then 2 0 aby ax b x y abax x C y ax a bx ax ax b a ab aa b a ab += ++ − += − − = ⇒ −−+ = − −= ∴ = −−< +< > +< (b) Given 2ba=− , 24ba−= . Since 0a> , 2ab<− From (a), C has no stationary point. Note that the question asked for the equations of the asymptotes and the coordinates of all intercepts. Do indicate them on the graph. (c) As shown in diagram for (b) (d) 321 1xx x−− ≤−− Comparing with 23 211 ab ay ax b ax a xx += ++ = − −−− , choose 1a= From graphs, 2 or 1xx≤− >