RI 9758/01 2018

1 | Page Question 1 No. Suggested Solution Remarks for Student (i) 2 ln d 1 ln=d xy x yx xx = − (ii) ee 2211 ee 2211 ee 11 ln ln 1 1dd 1 1 lndd 1 ln 111 ee 21 e xx xxxx xxxxx x xx −+= −= −  = −−   =−− = − ∫∫ ∫∫ Raffles Institution H2 Mathematics (9758) Solution for 2018 A-Level Paper 1

2 | Page Question 2 No. Suggested Solution Remarks for Student (i) ( )( ) 2 3Sub into 2 7 3 27 2 7 30 21 3 1,32 6, 1 y yxx xx xx xx xx yy = += ⇒+ = ⇒ − += − −= = = = = (ii) ( ) ( ) 3 2 1 2 2 33 3 11 22 9Required volume 7 2 d 19726 1 36 3 186 125 6 xx x x x π ππ π π = −−   = −− +     = −+ +− = ∫ Note that question said exact and did not mention answer in terms of π

3 | Page Question 3 No. Suggested Solution Remarks for Student (i) 2 2 22 3 3 d 2 6 ...(1)d ...(2) dd 2 ...(3)dd Sub (2) and (3) into (1): d 2 26d d 6d d6 d yxy x y ux yu x xuxx ux x xu uxx ux x u xx = − = = +  +=− =− =− (ii) 2 2 22 2 3 3 3 2 when 1: 2 3 1 3 uC x y C y Cxxx y x CC yx = + = +⇒=+ = = =+⇒= − ∴=−

4 | Page Question 4 No. Suggested Solution Remarks for Student (i) ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) 2 2 22 2 22 22 22 2 2 3 22 2 32 2 2 32 2 0 2 32 2 2 32 2 0 22 24 4 0 1 0 or 2 2 0 2 120, 1, 2 13 xx x xx x xx x xx x xx x xx xx xx x x xx x +−=− +− =− +− −− = + −+− + −−+ = + +−= += + −= −±= = −= = −± (ii) 1 3 1 or 0 1 3xx− − < <− < <− +

5 | Page Question 5 No. Suggested Solution Remarks for Student f : for , , 1 g : for x ax x x baxb xx x + ∈ ≠− ≠−+ ∈   Given ff = g, ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 22 2 11 1 11 1 Comparing coefficients of : 1 xa axb xxa bxb x a ax ab xx a bx b x aa b xx b ab x a a b x b xa b xb + ++ =+ ++ ++ + =++ + ++ + =+ ++ ++ += ++ + =− ( ) ( ) ( ) 1 ff ff 1 xx xaxx x − = += = − ff = g, we assume question is just referring to the rule.

6 | Page Question 6 No. Suggested Solution Remarks for Student Given 32a b ac×=×   (i) ( ) ( ) 32 3 2 22 0 // 3 2 Thus, 3 2 , where is a constant. a b c a ba c aca c a bc bc a λλ × − =× −× = ×−× = ∴− −=              (ii) ( ) ( ) 2 22 2 2 . cos 60 2 32 . 32 . 9 12 . 4 144 24 4 124 2 31 bc b c bc bc a a b bc c a λλ λ λλ = = − −= −+= − += ⇒= ± = ±            

7 | Page Question 7 No. Suggested Solution Remarks for Student (i) ( ) 22 22 2 22 2 2 2 2 41 2 28 Differentiate with respect to : dd4 16 2 2 dd d 2 16 2d d2 d 2 16 xy x xy x y x xy x yyx y x xy yxx y xy y x yx y xy x xy y − = + −= + −= ++ += − −= + (ii) ( ) 2 2 22 14 1When 1, 21 28 1 1 3 11Let and be 1, and 1, respectively, 33 121 d 179At 1, : 2 163 d 54 33 1 17Tangent: 13 54 17 1 54 54 121 d 179At 1, : 2 163 d 54 33 Ta yx y yy y PQ yP x yx xy yQ x −= = + −= + =±    −     − = =  + −= − −= − − −= = −  −− ( )1 17ngent: 13 54 17 1 54 54 1Solving coordinates of is ,0 17 yx xy N += − − += − −

8 | Page Question 8 No. Suggested Solution Remarks for Student (i) ( )12 32 5 and 15 15 2 5 5 2 2 40 uu A A uuA = = ⇒= + = =+= (ii) ( ) 1 2 3 2 5 : 2 5 ...(1) 15 : 4 2 15 ...(2) 40 : 8 3 40 ...(3) 15Using GC: , 5, 52 n nu a bn c u abc u a bc u a bc abc = ++ = ++= = + += = + += = = −= − (iii) ( ) ( ) ( ) ( ) ( ) 11 15 2 552 22 115 15 152 21 2 515 2 1 1 5 2 nn r r rr n n ur nn n nn n = = = −−  − = − +−  −  = −− +− ∑∑

9 | Page Question 9 No. Suggested Solution Remarks for Student 22 sin 2 , 2sin for 0xy θ θ θ θπ= − = ≤≤ (i) ( ) 2 2 dd 2 2cos 2 , 4sin cosdd d 4sin cos d 2 2cos 2 4sin cos 2 2 1 2sin 4sin cos 4sin cos sin cot xy y x θ θθθθ θθ θ θθ θ θθ θ θ θ θ = −= = − = −− = = = Ok to have 0 and π included though cot is not defined. (ii) ( ) ( ) ( ) 2 2 2 Point where is 2 sin 2 , 2sin Equation of normal: 2sin tan 2 sin 2 At point , 0, sin2sin 2 sin 2cos 2sin cos 2 sin 2 2 , that is, 2 yx Ay x x xk θα α α α α ααα αα αα α αα α α α = − − = − −+ = − = − −+ = −+ ∴= = (iii) ( ) ( ) ( ) ( ) [ ] ( ) 2 2 0 22 0 0 2 0 0 0 Total length of 2 2cos 2 4sin 2 d 4 8cos 2 4cos 2 4sin 2 d 8 8cos 2 d 8 8 1 2sin d 4sin d 4 cos 4 11 8 C π π π π π π θ θθ θ θ θθ θθ θθ θθ θ = −+ = −+ + = − = −− = =− = − −− = ∫ ∫ ∫ ∫ ∫

10 | Page Question 10 No. Suggested Solution Remarks for Student dd , where dd Iq qL RI V ItC t++= = (i) Differentiate with respect to t, 2 2 2 2 d d 1d d d d dd dd d d , since dd d d I I qVLR t t Ct t I II V qLR It tC t t ++ = + += = 2 2 Therefore, dd d 0, when 0 that is, is a constant.dd d I II VLR Vt tC t+ += = (ii) 2 22 2 2 2 22 2 22 22 2 22 2 2 22 2 22 2 e d ee e ed 22 d e eeeed2 2 4 4 ddSub into 0,dd e ee e 42 Rt L Rt Rt Rt Rt LL L L Rt Rt Rt Rt Rt L LL LL Rt Rt Rt Rt L LL I At I R ARA At A tt LL I AR AR AR AR AR tttL L L L L I IILR t tC AR ARAR t AR t LL − −− − − − −−−− − −− − = = + −= −  = −−+ = −+ + += −+ +− 2 22 22 2 22 2 2 e0 e e e e e0 42 14 04 Rt LL Rt Rt Rt Rt Rt LL L LL A tC AR AR AAR AR t t t L LC RL CLC R − −− − −− += −++ − += − +=⇒= (iii) Note that the given values satisfy (ii), when 2 434 and 3, from (ii), 4 LRL C R= = = = By right, we should sub R = 4, L = 3 and C = 0.75 into I and check that it still a solution of the DE.