RI 2016 P2 A-Level H2 Math Solution

1 | P a g e Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student d 0.1d V t  1tan tan where depth of water2 r r h h hh      2 2 31 1 1 1 3 3 2 12V r h h h h          2 1 33 2 3 d 1 d d 4 d 1 36When 3, 12 1 36 d0.1 4 d d 0.0251d V h ht t V V h h h t h t                     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 | P a g e Question 2 No. Suggested Solution Rem arks for Stud ent (a)(i) 2 2 2 2 2 2 3 dcos d , cos d 1 2 d 1sin sin d 2 , sin d 1 2 1 1 dsin cos cos d , sin d 1 2 2 d 1sin cos sin 1, cos d vx nx x u x nx x ux nx x nx x x v nxn n x n vx nx x nx nx x u x nxn n n n x ux nx x nx nx c v nxn n n x n                         (ii)       22 2 2 2 3 2 2 2 2 2 2 2 2 2 2 2 1 2 2cos d sin cos sin 2 22 cos 2 cos 4 cos 2 2 cos 2If is even, cos d 4 2 2 6If is odd, cos d 4 2 6 x nx x x nx x nx nxn n n n nn n n nn n x nx x a n n n x nx x a n n                                       

3 | P a g e (b)    2 2 2 0 2 2 20 22 20 2 5 29 9 25 9 5 d d 1Volume d 9 2 d d 2 d 0 99 d 2 59 1 9 d2 1 9 1 d2 1 9 ln2 1 9 1 ln 9 ln 52 5 1 4 5ln2 5 9 u xy x u x x x u x x x x x ux x x x x ux u uu uu u uu                                                           Question 3 No. Suggested Solution Remarks for Student cos , 1 cos , for 0 2x t t y t t       (i)

4 | P a g e     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 x x                        cos , 1 cos , for 0 2 d d1 sin , sind d 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 x t t y t t x y t tt t y t tx t t D x t x y                                   (ii)     cos 1 0 0 0 0 Area d 1 cos 1 sin d 1 cos sin sin cos d 11 cos sin sin 2 d2 1sin cos cos 24 1 3sin cos cos 24 4 a a a a a a y x t t t t t t t t t t t t t t t t a a a a                                 (iii) 1 2 d 1 d 2t y x       2 1 1 , 12 2 Thus, equation of normal is 1 2 2 2 1 1 is ,02 is 0, 1 1 1 1area of triangle 1 12 2 4 t x y y x y x E F OEF                                   

5 | P a g e Question 4 No. Suggested Solution Remark s for Student (a)  3 i 1 3 i 1z z       2arg where tan 5z    (a)(i) (ii) Equation of circle:     2 23 1 1 ...(1)x y    3 1 1 Im Re 0 2 2 Z1 Z2  3 1 1 Im Re 0 2 2 Z1 Z2 

6 | P a g e Equation of line: 5 2y x c  Sub (3, 1) into line, we have 15 171 2 2c c  