RI 9740/01 2016

1 | Page Question 1 No. Suggested Solution Remarks for Student ( ) ( )( ) ( ) ( )( ) 22 22 2 4 4 14 3 44 4 14 344 4 4 14 12 4 3 52 4 31 2 4 xx x xxx xxx x x xx x xx x xx x +−−+ −+− −+=−− + − − −− = − +−= − −+= − ( ) ( )( ) 24 4 14 34 31 2 04 xx xx xx x +− <+− −+ <− ( )( )( )31 2 40xx x− + −< 12 or 43xx<− < < Detailed working needed. We can use GC to check our answers, though question states “without using a calculator”. Raffles Institution H2 Mathematics Solution for 2016 A-Level Paper 1

2 | Page Question 2 No. Suggested Solution Remarks for Student (i) cos2 xy= 0 d 0d x y x = = 2 d 0.6931471 0.693 (3 s.f.)d x y x π= = −= − Can obtain answer from GC directly. No working needed. (ii) Tangent at x = 0: y = 2 Tangent at :2x π= 2 1 0.69315y x π − =−− 0.693 2.09(3sf)yx⇒= − + 2 21When 2, 0.69315 0.12811yx x π −= = − ⇒=− Required coordinates (0.128, 2).

3 | Page Question 3 No. Suggested Solution Remarks for Student ( ) ( ) ( ) ( ) ( ) ( ) 4 3 3 f f4 f 04 0 since 0 x kx l m x kx l a ka l la k = −+ ′ = − ′ =⇒ −= ⇒= ≠ ( ) ( ) 4 Thus f x kx a m= −+ ( ) ( ) 4 f a b ka a m b mb = ⇒ − += ⇒= ( ) ( ) 4 Thus f x kx a b= −+ ( ) ( ) 4 4 f0 0 c k a bc cbk a = ⇒ − += −⇒= (a,1/b) (0,1/c) y = 0 0

4 | Page Question 4 No. Suggested Solution Remarks for Student 4 7 14 3 ...(1) 8 ...(2) 11 ...(3) a d br a d br a d br += += += (i) (3) – (1): ( ) 14 4 4 108 1 ...(4)d br br br r=−= − (2) – (1): ( ) 7 4 435 1 ...(5)d br br br r=−= − 10 10 3 3 10 3 (4) 8 1: 5 58 8(5) 5 1 5 8 30 Since 1, using GC, 0.74045 0.74(2 d.p.) r rrr rr rr −= ⇒ −= −− ⇒ − += <= ≈ (ii) ( )3.85 0.741 n nbr br =−

5 | Page Question 5 No. Suggested Solution Remarks for Student 2 1, 0 2 a uv b      = −=        (i) ( ) ( ) ( ) 2 2 20 1 2 2 22 2 44 2 uv uv uu vuuvvv vu a b b ba b i ba ja k a +×− =×+×−×−× = ×    = ×−     = −=+− − −         Note that 0uu vv uv vu ×=×= −×=×     (ii) ( ) ( ) ( ) ( ) Given that 1 24 1 11 2 18 1 2 18 ba uv uv a uv uv a a =− − +×−= − − +×− =⇒ =⇒= ±     (iii) ( ) ( ) ( ) 22 222 . 0 . .0 0 2 1 23 uv uv u uv v uv vu + −= ⇒ −= ⇒−= ⇒ = = +− + =       

6 | Page Question 6 No. Suggested Solution Remarks for Student (i) Let nP be the statement ( ) ( ) ( ) 22 1 11 12 4 n r rr nn n n = += + ++∑ for .n +∈  For 1,n= ( ) ( )( ) ( )( ) ( ) 1 22 1 2 1 11 1 2 181111 12 244 r LHS r r RHS LHS = = += += = + ++ = = = ∑ Therefore, P1 is true. Assume kP is true for some k +∈  , i.e. ( ) ( ) ( ) 22 1 11 12 4 k r rr kk k k = += + ++∑ We need to show 1kP+ is true, i.e. ( ) ( )( ) ( ) ( ) ( )( ) ( ) 1 22 1 2 11 1 11 1 124 1 1 2 344 k r rr k k k k k k kk + = + = + ++ + +++ = + + ++ ∑ For 1,nk= + ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )