RI 9740/02 2014

1 | Page Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student 23xt= , 6yt= (i) d 6d x tt = , d 6d y t = d1 d y xt⇒= d1 0.4 0.4 2.5d y txt= ⇒= ⇒= (ii) Equation of tangent at point P(3p2, 6p) on C: ( ) 2163yp xp p−= − When x = 0, ( ) 216 3 36 3yp p y pppp− = − ⇒= − + = Coordinates of D are (0, 3p) Mid-point of PD is 239 ,22 pp  23 2 px= , 9 2 py= 2 232 2 2 9 27xy y ⇒= =  ∴Cartesian equation is 22 27xy= Question 2 No. Suggested Solution Remarks for Student ( ) ( ) 2 22 9 13 25 925 9 x x A Bx C xxxx +− += + −+−+ ( ) ( )( ) 229 13 9 2 5x x A x Bx C x+− = + + + − 22 5 55 5: 9 13 9 32 22 2x AA  = +−= + ⇒=    Raffles Institution H2 Mathematics Solution for 2014 A-Level Paper 2

2 | Page ( ) ( )0 : 13 3 9 5 8x CC= −= + −⇒= ( ) ( )( )1 :911 3 3 19 825 3x BB= + −= +++ −⇒= ( ) ( ) ( ) 222 2200 222 220 00 2221 0 0 1 1 9 13 3 3 8dd 25 925 9 3 38ln 2 5 d d2 99 33 8ln5 ln 9 tan22 33 33 382ln5 ln13 ln9 tan22 233 3 13 8 2ln tan2 45 3 3 xx xxx xxxx xx xx xx xx − − − +− += + −+−+ =  −+ + ++ = − + ++   = −+ −+ = + ∫∫ ∫∫ 31 382, ,,2 45 3 3ab cd∴= = = = You may use the GC to verify your answers. Question 3 No. Suggested Solution Remarks for Student (i) (a) Distance run by athelete who completes first 10 stages [ ] ( ) ( )( )( ) ( ) 2 4 8 12 ... 40 102 2 4 10 1 42 10 8 36 440m ++ ++ = +− = += (b) Distance run by athelete who completes first n stages ( ) ( )( )( ) ( ) 2 24 1 42 44 n n nn = +− = + ( )For 4 4 5000,nn +≥ consider the following n ( )44nn + 34 4760 < 5000 35 5040 > 5000 Least n is 35.

3 | Page (ii) OA1 = 4 A1A2 = 4 OA2 = 8=4(2) A2A3 = 8=4(2) OA3 = 16 = 4(2)2 A3A4 = 16= 4(2)2 OA4 = 32 = 4(2)3   An−1An = 4(2) n−2 OAn = 4(2) n−1 Required distance ( ) ( ) 12 4 8 16 ... 4(2) 212(4) 21 82 1 n n n −= ++ ++  −=  − = − Let B be the point where he completes 10km exactly. Need n such that ( )8 2 1 10000 1000021 8 10.289 n n n −≥ ⇒ ≥+ ⇒≥ ⇒Athlete ran a total of 10km after completing the 10th stage but not the 11th stage. Distance covered between O and B = 10 000 − ( ) 1082 1 − = 1816 OA11 = 4(2) 11−1 = 4096 Since OA11 > 1816, athlete is running away from O towards A11 when he reaches B and his distance from O is 1816 m.

4 | Page Question 4 No. Suggested Solution Remark s for Student (a)(i) ( )5i 4 5i 4zz+− = ⇒ −−+ = (ii) ( )6i 10 4i 6i 10 4izz zz−= ++⇒ −= − −− ( ) ( )1 5 4 c o s i 1 4 s i n 5 22 i 1 2244z ππ  =−− + + =−− + +     ( ) ( )2 5 4 c o s i 14 s i n 5 22 i 12244z ππ  =−+ + − =−+ + −     −5 −9 −1 1 −3 5 Im Re 0 4 −5 −9 −1 1 -3 5 Im Re 0 (0,6) (-10,-4) z1 z2 4

5 | Page (b)(i) i 63 i 2ew π−= −= ( )2 and arg 6ww π= =− ( ) ( ) 6