RI 9740/01 2015

1 | Page Question 1 No. Suggested Solution Remarks for Student (i) 2 ay bx cx=++ 3 d2 d ya bxx= −+ At x = 1, d 2,d y x = we have 2 2 ...(1)ab− += At ( )1.6, 2.4 , − we have ( ) 2 1.6 2.4 ...(2) 1.6 a bc+ += − At ( )0.7,3.6 , − we have ( ) 2 0.7 3.6 ...(3) 0.7 a bc− += − Using GC to solve (1), (2) and (3): 3.59345 3.593, 5.18691 5.187, 7.30274 7.303 a b c =− ≈− =− ≈− = ≈ Can key in ( ) 2 1 1.6 (ii) Using GC, x = –0.589 (3 d.p.) (iii) 5.187 7.303 (3 d.p.)yx= −+ Raffles Institution H2 Mathematics Solution for 2015 A-Level Paper 1

2 | Page Question 2 No. Suggested Solution Remarks for Student (i) Label asymptotes: y = 1 and x = 1 (ii) From graphs and GC, the graphs intersect at 1.73, 0.414, 1.73x=− For the inequality to hold, we have 1.73 0.414 or 1.73xx− << > y = 1 x = 1 −1 O 1 2

3 | Page Question 3 No. Suggested Solution Remarks for Student (i) Note that 11 2 3f f f ... f n n nnn n    +++ +       is the sum of area of the n rectangles in the diagram. As n→∞ , the sum will approach the exact area under the curve. Hence ( ) 1 0 11 2 3lim f f f ... f f d n n xxn nnn n→∞    +++ + =       ∫ Since f is any continuous function, we can use y = x2 for convenience. (ii) 1433 3 1 3 3 3 0 0 1 1 2 ... 3 3lim d 44n n xx xn n→∞  + ++ = = =   ∫

4 | Page Question 4 No. Suggested Solution Remarks for Student Length of rectangle 22xy= + Length of semi-circle ( )22d xy= −+ But length of semi-circle also ( ) ( )1 22 2 22 x x xx xπππ= +=+=+ Thus, ( ) ( )22 2 22 2 11 222 d xy x x yd x x y d xx π π π −+= + ⇒+= −− ⇒= − − Total area, A 21 2xy x π= + 211 1 222 2xd xx x ππ= − −+ 21 22 xd x= − ( )1 42 xd x= − This is an quadratic expression with max value attained when 8 dx= , that is the mid-point of the 2 roots 0 and .4 d Thus, max value of A 211 42 8 8 32 dd dd = −=   m2 So, 1 32k =

5 | Page Question 5 No. Suggested Solution Remarks for Student (i) ( ) ( ) ( ) 2 2 1 4 2 replace by 3 3 replace by 1 34 yx xx yx yy yx = ↓− = − ↓ = − Translate 3 units in the positive x directions followed by scaling of factor 1 4 parallel to y-axis. (ii) (iii) 1 −1 O 1 3 4 x

6 | Page −1 O 2 4 x y 2 5/4 1

7 | Page Question 6 No. Suggested Solution Remarks for Student (i) ( ) ( ) ( ) 23 2322 8ln 1 2 2 ... 2 223 3 xxx x xx x+ = − + +≈ − + (ii) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) 23 23 23 4 1 121 1 ...2! 3! 1 12 ...26 c cc cc cax bx ax bcx bx bx ab c c ab c c cax abcx x x − −−+ = ++ +  − −−= ++ + Given that ( ) 2 2 3 231 82223 ab c cax abcx x x x x−+ + = −+ Thus, ( ) 2 1 8 851 3 33 3 5 a bc b c bc b b c = =− − − = ⇒− + = ⇒ = =− Coefficient of x4 is ( )( ) 3 3 5 33 32 1212 1043 55 5 6 6 27 ab c c c      − −− −−     −−      = =−

8 | Page Question 7 No. Suggested Solution Remarks for Student (i) 35 ,5 11OC a OD b= =      (ii) ( ) : , 3 5 3 15 BCl r b BC b ab ab λλ λ λλ = +∈ = +−  = +−         ( )5: 1 , 11 ADlr b a µ µµ= +− ∈   (iii) ( ) ( )35 115 11 33 1155 5511 11 11 3 11Using GC, , 4 20 a bb aλλµµ λ µ λµ λ µλ µ λµ +− = +− = −⇒ += −= ⇒+ = = =    Thus, 33 3 9 1 15 4 4 20 4OE a b a b  = +− = +          E 6 5 2 3 D B O A C

9 | Page 9 1 1 11 20 4 4 20 5 9 1 9 9 9 1 11 9 11 20 4 44 20 11 4 20 11 Thus, : 11: 9 AE a b a b a ED b a b b a b a AE AE ED = + −= −  =− +=−= − =   =              

10 | Page Question 8 No. Suggested Solution Remarks for Student 1.5h 1.5 60 60 5400 s 1.75h 1.75 60 60 6300 s =××= = ××= (i) ( )( )505400 2 50 1 2 63002 5400 50 2450 6300 59 77 T T T ≤ +− ≤ ≤+ ≤ ∴≤≤ Set of values of T is [ ]59,77 (ii) ( ) ( ) 50 50 1.02 1 5400 63001.02 1 108 1.02 1 126 63.845 74.486 t t t − ≤≤ − ≤ −≤ ∴ ≤≤ Set of values of t is [ ] ( )63.9,74.4 3 s.f. Examiner report does not accept [ ] ( )63.8,74.5 3 s.f. (iii) ( ) ( )( ) 50 1 63.845 1.02 59 50 1 2 11.475 11 − −+− = ≈