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 Stu