RI 9758/01 2019

1 | Page Question 1 No. Suggested Solution Remarks for Student The coefficient of the cubic function are all real. Since 2i+ is a root of f (z) = 0, then ( )2 i* 2 i+= − is also a root. ( )( ) ( )( ) ( )( ) ( ) ( ) 2 32 32 3 2 f( ) 0 3 2i 2i 0 3 450 7 15 0 , 7 , 15 z zz z z zz zz z b cdaz bz cz d a z z z a aa b ac a d a = − − −+ −−= + −+= −−+= + ++= + + +  = −= − = Alternatively, With the given roots apply Factor Theorem. f( 3) 0 27 9 3 0 (1)a b cd− = ⇒− + − + = ( ) ( ) ( )f ( 2i ) 0 21 1 i 34 i 2i 0 2 3 2 0 (2) 11 4 0 (3) a b cd a b cd a bc +=⇒+ ++ ++ += + + += + += From (1), (2), (3) we form a system of linear equations with last column the coefficient of a. As before, , 7 , 15b ac a d a= −= − = Raffles Institution H2 Mathematics (9758) Solution for 2019 A-Level Paper 1

2 | Page Question 2 No. Suggested Solution Remarks for Student (i) 3 10xx+−= Using GC, Polysimult2 0.68233 0.682≈ Do note the GC skills required in this question. (ii) ( ) ( ) 033 1 1d 2 0 1d b a xx x xx x − +− = − +−∫∫ We can solve with GC, noting that b > a. 1.89243 1.892b= ≈ Alternatively, ( ) ( ) 033 1 42 4 2 42 1d 2 0 1d 0.68233 0.68233 0.68233 3.542 4 2 3.10465 042 b a xx x xx x xx x xx x − +− = − +−   +−− + − =     + −− = ∫∫ 1.89243 1.892b= ≈

3 | Page Question 3 No. Suggested Solution Remarks for Student (i) ( ) ( ) ( ) ( ) ( ){ } 3 32 32 32 3 3 Note that 1 3 3 1 f 2 6 6 12 2 3 3 1 10 2 1 10 2 15 2, 1, 5 x xxx x xxx xxx x x pq r −=− +− = − +− = − + −− = −− = −− = = −= − (ii) Sequence of 1. Translation of 1 unit parallel to the positive x-axis 2. Translation of 5 units parallel to the negative y-axis 3. Scale by factor 2 parallel to the y-axis.

4 | Page Question 4 No. Suggested Solution Remarks for Student (i) 00, 2 10 9 ln10 10, 2 10 or ln 2 lg 2 x xy yx = =−= = = ⇒= (ii) Solving 2 10 6 2 10 6 or 2 10 6 ln16 ln 4 2ln 2 ln 2 4 ln 2 ln 2 4 2 10 6 2 4 x xx x xx x −= −= −= − = = = = = − ≤⇒≤≤

5 | Page Question 5 No. Suggested Solution Remarks for Student (i) ( ) ( ) 2f e 4, g 2, xxx xx x = −∈ = +∈   ( ) ( ) ( ) ( ) ( ) 2 2 1 1 e4 e4 2 ln 4 1 ln 42 1f ln 4 2 Domain of f Range of f 4, x x y y xy xy xx− − = − = + = + = + = + = =−∞ Note that we use “round brackets” here. (ii) ( ) ( )( ) ( ) ( ) ( ) ( ) 11 1 fg 5 f fg f 5 g f5 12 ln 5 42 ln 3 2 x x x x x −− − = = = += + = −

6 | Page Question 6 No. Suggested Solution Remarks for Student (i) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 11 1 1 11 4 1 2 1 2 1 22 1 22 1 11 1 1 4 12 21 21 111 213 11 35 11 57 ... 11 23 21 11 21 21 1112 21 nn rr r rr r r r rr nn nn n = = = = −− −+ − + = −− −+  = − +− +− +− −− +− −+  = − + ∑∑ (ii) ( ) 10 222 11 1 1 111 41 41 41 11 1 12 2 2 10 1 1 42 r rr rrr ∞∞ = = = = −−−− = −−  + = ∑∑∑

7 | Page Question 7 No. Suggested Solution Remarks for Student (i) ( ) 1 11 1 e d eed d1 , e , e e 0 d Equation of tangent: e d1, e, e e 2ed Equation of tangent: e 2e 1 2e e x xx yx y xx yxy x y yxy x yx yx − −− − −− − = = − == =−= = = − = − =+= += + = + (ii) tan 2e 79.6 θ θ = =  Note that one of the tangents is a horizontal line. So the gradient of the other line allows us to get the angle directly.

8 | Page Question 8 No. Suggested Solution Remarks for Student (a) ( ) ( )( )( ) ( ) ( ) 1 1 1 57 642 2 64 1 22 2 32 2 126 2 22 13 k k k a aa k − − − = +− = + = = (b) ( ) { } 4 4 Note that 0. If 1, then sum of first 4 t erms 4 0 Thus, 1. 1 01 1 1 or 1 (rejected) Thus, 1, \ 0 0, even , odd n fr f r fr r rr r rf nS fn ≠= = ≠ ≠ − =− = ⇒= − = = −∈ =   (iii) Let a be the first term. ( ) ( )( )( ) 4 2 3 142 2 3 7 ...(1) 2 30 0 or or 2 or 3 Given 0, Possible or or 23 Into (1), only gives 0 7, 7 11th term 10 63 ad ad a ada da d a ddd a aada da a ad ad += += + + += =−− − < = −− − = −< = −= = +=

9 | Page Question 9 No. Suggested Solution Remarks for Student (i) cos isinw θθ= + (a) 22 11 cos isin cos isin cos isincos isin cos sin cos isin cos isin 2cos w w θθ θθ θθθθ θθ θθθθ θ += + + + −=++ + =++− = Alternatively, ( ) ( ) i i ii 11 e e ee cos isin cos isin 2cos w w θ θ θθ θθ θθ θ − +=+ = + =++− = (b) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 12 11 21 1 cos isin 2 1 cos isin1 1 cos sin 2 1 cos i 2sin1 2 2cos 2 2cos 2 1 cos i 2sin 2 2cos i 2sin 2 2cos i sin 1 cos i 2sin cos22 2cos 2 i tan 2 ww ww θθ θθ θθ θθ θ θ θθ θ θ θ θ θ θθ θ θ − +−=++ = − ++ +−= − ++ +−= − + + −+ += + = + = + = = Alternatively, ii i i i0 22 2 ii ii i0 22 2 1 e e e (e e ) 1e e e (e +e ) w w θθ θ θ θθ θθ − − −− −= =++

10 | Page ii 22 ii 22 cos isin cos isinee 22 22 cos isin cos isine +e 22 22 2i sin 2 cos 2 i tan 2 θθ θθ θθ θθ θθ θθ θ θ θ − −   + −−  −   = =   ++−     = = (ii) 22 Let i Since 1 1 za b z ab = + = ⇒+= ( ) 22 22 3i i 3i 3 69 10 6 z ab ab ab b b − =+− = +− = +−+ = − ( ) ( ) 2 2 22 1 3i 1 3i i 13 9 16 9 9 10 6 z ab ba bb a b += + + = −+ =−+ + = − 3i 11 3i z z −∴=+ OR Since 1 cos isinzz θθ=⇒= + ( ) 22 3i cos isin 3i cos sin 3 10 6sin z θθ θθ θ −= + − = +− = − ( ) 2 2 1 3i 1 3icos 3sin 1 3sin 9cos 10 6sin z θθ θθ θ += + − = −+ = − 3i 11 3i z z −∴=+ OR ( ) ( ) ( ) ( ) 1 *1 3i 3i * 3i * 1 3i **1 3i 1 3i 1 3i 1 3i Note that 1 3i * 1* 3i * 1 3i * * 1 3i * 1 3i *3i 11 3i 1 3i zz z z zz z z zzz z z z z zz zz zz = ⇒= −− − −= = =++ + + += + = + = − +−∴= =++