RI 9740/02 2016

1 | Page Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student d 0.1d V t = 1tan tan where depth of water2 r rh h hhαα=⇒= = = 2 231 11 1 3 3 2 12V rh h h hππ π = = = 2 1 33 2 3 d1d d4 d 1 36When 3, 12 1 36 d0.1 4d d 0.0251d Vh htt V V hh h t h t π π π π π = = = ⇒=  =   = Rate of increase of depth of water is 0.0251 m per minute. Raffles Institution H2 Mathematics Solution for 2016 A-Level Paper 2

2 | Page Question 2 No. Suggested Solution Rem arks for Stud ent (a)(i) 22 2 2 2 23 dcos d , cosd 12 d1sin sin d 2 , sind 1 21 1 dsin cos cos d , sind 1 22 d1sin cos sin 1, cosd vx nx x u x nx x ux nx x nx x x v nxn n xn vx nx x nx nx x u x nxn nn n x ux nx x nx nx c v nxn n n xn = = = −= = = −− + = = = + −+ = = − ∫ ∫ ∫ (ii) ( ) ( ) ( ) ( ) ( ) 2 2 22 23 22 2 2 2 22 2 2 22 1 22cos d sin cos sin 22 2 cos 2 cos 4cos 2 2cos 2If is even, cos d 4 2 2 6If is odd, cos d 4 2 6 x nx x x nx x nx nxnn n nnnn nnn n x nx x ann n x nx x ann ππ π π π π π π π πππ π ππ ππ ππ = +− = − = − = −= = = += = ∫ ∫ ∫

3 | Page (b) ( ) ( ) 2 22 0 2 2 20 22 20 2 5 29 9 25 9 5 d d1Volume d 9 2d d2 d 099 d 25 9 19 d2 1 91 d2 19 ln2 19 1 ln 9 ln 525 14 5 ln25 9 uxyx u x x x ux xx x xux xx x xu x u uu uuu uu π π π π π π π π = = − ⇒= − ⇒= − = =⇒= − = =⇒= − −=− = −  = −− = −− + + = +  ∫ ∫ ∫ ∫ ∫ Question 3 No. Suggested Solution Remarks for Student cos , 1 cos , for 0 2xt t y t t π=− =− ≤≤ (i)

4 | Page ( ) ( ) 0 1 cos 0 0 or 2 0 cos 0 1 or 2 cos 2 2 1 1, 0 and 2 1, 0 y t t xx π π ππ π = ⇒− = ⇒= ⇒=− = − = − = − −− ( ) cos , 1 cos , for 0 2 dd 1 sin , sindd d sin 0 sin 0d 1 sin 0, , 2 From above points where meets the -axis, max point is when cos 1, 1 cos 2 1, 2 xt t y t t xy tttt yt txt t Dx t xy π ππ π π ππ π π =− =− ≤≤ = += = = ⇒=+ ⇒= = ∴ = −= + = −= + (ii) ( )( ) ( ) cos 1 0 0 0 0 Area d 1c o s 1s i n d 1 cos sin sin cos d 11 cos sin sin 2 d2 1sin cos cos 2 4 13sin cos cos 2 44 aa a a a a yx t tt t t t tt t t tt tt t t aa a a − − = = −+ =−+− =−+− = −−+ = −−+ + ∫ ∫ ∫ ∫ (iii) 1 2 d1 d2 t y x π= = ( ) ( ) ( ) 2 11 , 122 Thus , equation of normal is 12 2 21 1 is ,0 2 is 0, 1 1 11area of triangle 1 12 24 t xy yx yx E F OEF ππ π π π π πππ = ⇒= = −= − − = −++ +  + +∴ = += +  

5 | Page Question 4 No. Suggested Solution Remark s for Student (a) ( )3i 1 3i 1zz−−=⇒ − + = 2arg where tan 5z αα= = (a)(i) (ii) Equation of circle: ( ) ( ) 22 3 1 1 ...(1)xy− +−= 3 1 1 Im Re 0 2 2 Z1 Z2 α 3 1 1 Im Re 0 2 2 Z1 Z2 α

6 | Page Equation of line: 5 2y xc= −+ Sub (3, 1) into line, we have 15 171 22 cc= − +⇒= 5 17 ...(2)22yx= −+ Solving (1) and (2), the two values are 3.37 + 0.0715i, 2.62 + 1.93i (b)(i) i 42 2i 8ew π−= −= 1 3 8133i i2 i3 44 4 22 811 i 122 37i ii 4 12 12 8 e