RI 9740/02 2015

1 | Page Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student (i) 32 m Note that we need 1 216 0 h−≥ (ii) ( ) ( ) 1 2 1 2 1 2 1 21 2 1 2 16d 10 16 d dd 10 20 16 40 16 where hh hh tt h tc tb h b c −−= ⇒− = ⇒ −= +− ∴=− − =− ∫∫ 1 2 0 when 0 160 Thus, 160 40 16 h tb th = =⇒= = −− ( )1 2 Half its maximum height, that is, 16 we have 160 40 16 16 160 40 8 46.9 years h t = = − −= −≈ Raffles Institution H2 Mathematics Solution for 2015 A-Level Paper 2

2 | Page Question 2 No. Suggested Solution Remarks for Student 12 : 2 3 , 46 Lr λλ    = −+ ∈  −−   (i) 11 21 30 60 2Acute angle = cos cos 73.4721 30 60 −−           −   = =           −     (ii) 2 5 6 OP  =−   Any point, , on has position vector, 12 2 3 for some . 46 RL OR λ λλ λ + =−+ ∈−−    ( ) ( ) ( ) ( )( ) 222 22 2 2 2 12 2 33 2 3 5 = 33 46 6 2 1 3 7 2 6 33 4 4 1 9 42 49 4 24 36 33 49 70 21 0 7 10 3 0 73 10 3 or 17 PR λ λ λ λλ λ λλ λ λ λλ λλ λλ λλ λλ +− = ⇒ −+ −−− + ⇒ −+ − +− = ⇒− + +−+ + −+ = ⇒ − += ⇒ − += ⇒ − −= ⇒= =   

3 | Page 312 7 12 3 323 o r 23 17 4 6 10 346 7 13 7 5 7 46 7 OR OR  +   +      =−+ = −+ =        −− −   −−     = −  −    13 17 773 1 51'1 2 77 10 46 58 77 OR                = +− =   −      −−        (iii) ( ) ( ) ( ) 2 1 2 1 2 36 52 3 7 3 2 6 46 26 1 1 Required plane is 36 2 11 36 2 2 5 11 6 4.xy z −−             + ×=×=            −+ − − − −       − + − = − + − −=

4 | Page Question 3 No. Suggested Solution Remarks for Student 2 1f : , , 1 1x xx x ∈>− (a)(i) Method 1: ( ) 22 2f '( ) 0 fo r 1 1 xxx x = >> − Thus, f is an increasing function for x > 1. Thus f has an inverse. Method 2: Sketch graph of f and explain using horizontal line. (ii) ( ) ( )1 2 2 2 1 ff 11Let 1 1 11 111 1f1 ,0 yx xy x y xx y x x DR− − = ⇒= −− ⇒= − ⇒= − > = − = = −∞  (b) 2 2g: , , 1 1 xx xx x + ∈ ≠±− ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2Let 1 21 20 For the equation to be defined, discriminate 0, thus, 1 4 2 0 4 8 10 4 2 130 4 1 30 223 2230 23 23 or 22 xy yx x x yx x y yy yy yy y yy yy += ⇒−= +− ⇒ ++ − = ≥ − −≥ − +≥ − +−≥ − −≥ −− −+ ≥ +−≥≤

5 | Page Question 4 No. Suggested Solution Remarks for Student (a) Let nP be the statement ( )( ) ( ) ( ) 2 1 12 5 1 3 31 7412 n r rr r nn n n = + += + + +∑ for .n +∈  For 1,n= ( )( ) ( )( )( ) 1 1 2 5 1 3 6 18 1 2161 1 1 3 31 74 1812 12 r LHS r r r RHS LHS = = + + =××= = + ++ = == ∑ Therefore, P1 is true. Assume kP is true for some k +∈  , i.e. ( )( ) ( ) ( ) 2 1 12 5 1 3 31 7412 k r rr r kk k k = + += + + +∑ We need to show 1kP+ is true, i.e. ( )( )