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