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= + ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) 1 2 1 22 1 22 32 2 32 2 1 1 1 11 1 1 2 1 224 1 1 24 884 1 1 5 10 84 1 1 2 344 k r k r rr rr k k k kk k kk k k kk kk k k kk k k k kk + = = + = +++ + + =++ + + ++ + = + +++ ++ = + +++ = + + ++ ∑ ∑ Therefore 1k kP P +⇒ is true. Since 1P is also true, nP is true for n +∈  by Mathematical Induction.

7 | Page (ii) 3 1 0 1 2 3 2 4 14 44 nnuu nn u u u u −= ++ = = = = (iii) ( ) 1 10 1 21 32 12 1 0 ... n rr r nn nn n uu uu uu uu uu uu uu − = −− − −= − +− +− +− +− = − ∑ ( ) ( ) ( ) ( ) ( ) 01 1 3 1 2 1 2 2 21 12 1 2 from (i)4 n n rr r n r n r u u uu rr rr nn n n − = = = ∴=+ − = ++ = ++ =+ + ++ ∑ ∑ ∑

8 | Page Question 7 No. Suggested Solution Remarks for Student (a) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 8i 17 7i 0 ...(1) 1 5i 1 8i 1 5i 17 7i 24 10i 41 3i 17 7i 24 41 17 10 3 7 i 0 Thus, 1 5i is a root of (1) ww+−− +− + = −+ +−− −+ +− + =− − + + +− + = −+− + −+ + = −+ ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) 2 Let i be the second root, then 1 8i 17 7i 1 5i i 17 7i 1 5i i 17 7i 5 5 i Comparing Re and Im parts, 5 17 57 Thus, 2, 3 Second root is 2 3i xy w w w w xy xy x y xy xy xy xy + +−− +− + = −−+ − + − + =−+ + − + = −− + − += −= = = + ( ) ( )( ) ( ) ( ) ( ) ( ) 2 22 22 OR Let i be the second root, then i 1 8i i 17 7i 0 2 i 8 8 i i 17 7i 0 Comparing Re and Im parts: 8 17 0 ...(2) 2 8 7 0 ...(3) Solving to get the same answers... xy xy xy x y xy x y x y xyxy xy x y + + +−− + +− + = − + +−+ − − +− + = − −+ − = − −+= (b) Since coefficients are real, both 1 + ai and 1 – ai are roots. ( )( ) ( )( ) ( ) ( )( )( ) ( ) ( ) 32 32 2 2 2 2 5 16 1 i 1 i 5 16 2 1 Comparing coefficients and constant term, 25 3 2 1 16 3 since 0 1 30 z z zk z a z a zb z z zk z z a zb bb ba a a k ab − + +=−+ −− − − + += − ++ − −−= −⇒ = ++ = ⇒= > = −+ = −

9 | Page Question 8 No. Suggested Solution Remarks for Student ( ) ( )f tany x ax b= = + (i) ( ) ( ) ( ) 2 22f sec 1 tanx a ax b a ax b a ay′ = += + + = +  ( ) ( ) ( ) 2 2 23f 2f 2 2 2x ay x ay a ay a y a y′′ ′ = = +=+ ( ) ( ) ( ) ( ) ( ) 2 22 22 2 22 3 32 32 34 3 32 34 f 2f 6 f 26 2 266 28 6 x a x ay x a a ay a y a ay a ay ay ay a ay ay ′′′ ′ ′= + = ++ + = +++ = ++ (ii) ( )f tan 4y x ax π= = +  ( )f 0 tan 1 4 π= =  ( ) ( ) 2 f0 1 2 aa a′ = += ( ) 222f0 2 2 4 aaa′′ =+= ( ) 3f 0 16 a′′′ = ( ) 3 22 3 8f 1 2 2 ...3 ax ax a x x= ++ + +

10 | Page (iii) Using part (i) with a = 2 and b = 0, ( ) ( )f tan 2yx x= = ( )f0 0 = ( ) ( ) 2 f 0 2 20 2′ = += ( )f0 0′′ = ( )f 0 16′′′ = ( ) 38f 2 ... 3x xx= ++ OR Using part (ii) 23 23 23 23 64tan 2 1 4 8 ...43 tan2 tan 644 1 4 8 ... 31 tan2 tan 4 tan2 1 641 4 8 ...1 tan2 3 tan2 1 tan2 1 2 2We can rewrite 11 tan2 1 tan2 1 tan2 2 642 4 8 ...1 tan2 3 1 1 tan2 x xx x x xx x x x xx xx xx xx x xx xx π π π  + = ++ + + + = ++ + + − + = ++ + +− + −+= =−+− − − = ++ + +− − 23 1 23 2 23 23 3 23 3 3 321 2 4 ... 3 321 tan2 1 2 4 ... 3 32 32124 24 33 322 4 ...3 81 2 ... 3 8tan2 2 ... 3 xx xx x xx x xx x xx x xx x xx xxx − = ++ + + − =++ + +  = −++ +++   −++ +  = −− + = ++