SST 2020 S4 AM Prelim P2 Solutions

SECONDARY 4 PRELIMINARY EXAMINATION ADDITIONAL MATHEMATICS Paper 2 4047/2 16 September 2020 (Wednesday) 2 hours 30 minutes CANDIDATE NAME Solutions CLASS INDEX NUMBER READ THESE INSTRUCTIONS FIRST Do not turn over the page until you are told to do so. Write your name, class and index number in the spaces above. Write in dark blue or black pen in the writing papers provided. You may use a pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. INFORMATION FOR CANDIDATES Answer all the questions in the space provided. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. The use of an approved scientific calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your answer scripts securely together. The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 100. For Examiner’s Use Q1 5 Q2 5 Q3 6 Q4 5 Q5 8 Q6 8 Q7 9 Q8 10 Q9 10 Q10 10 Q11 11 Q12 13 Total /100

Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation , Binomial Expansion where is a positive integer and 2. TRIGONOMETRY Identities Formulae for € ax2+bx+c=0 € x=−b±b2−4ac2a € a+b()n=an+n1" # $ % & ' an−1b+n2" # $ % & ' an−2b2+...+nr" # $ % & ' an−rbr+...+bn € n€ nr" # $ % & ' =n!r!n−r()!=nn−1()...n−r+1()r! € sin2A+cos2A=1sec2A=1+tan2Acosec2A=1+cot2A sinA±B()=sinAcosB±cosAsinB cosA±B()=cosAcosBsinAsinB tanA±B()=tanA±tanB1tanAtanB sin2A=2sinAcosAcos2A=cos2A−sin2A=2cos2A−1=1−2sin2A tan2A=2tanA1−tan2AABCΔ€ asinA=bsinB=csinCa2=b2+c2−2bccosAΔ=12absinC

1 (i) Differentiate x2ln3xwith respect to x. [2] dydx=x233x⎛⎝⎜⎞⎠⎟+ln3x(2x)=x+2xln3x (ii) Hence find xln3x∫dx. [3] ∫x+2xln3xdx=x2ln3x+C∫xdx+∫2xln3xdx=x2ln3x+C∫2xln3xdx=x2ln3x−x22+C∫xln3xdx=12x2ln3x−x24+D – apply product rule – apply anti-differentiation – separating terms to integrate and integration of x

2 Express x3+1x2+1()x−2()in partial fractions. [5] x3+1x3−2x2+x−2=1+2x2−x+3(x2+1)(x−2) Let 2x2−x+3(x2+1)(x−2)=Ax+Bx2+1+Cx−22x2−x+3=(Ax+B)(x−2)+C(x2+1) By Substitution: At x = 2, 2(2)2−2+3=C(5)C=95 At x = 0, 3=−2B+C2B=95−3B=−35 By comparing coefficient of x2, A+C=2A=2−95=15 x3+1(x2+1)(x−2)=1+x−35(x2+1)+95(x−2) – Use long division to convert improper to proper rational function – Decomposition of partial fractions into its correct form

3 The quadratic equation 3x2+2x+1=0 has roots α and β. (i) Show that the value of α3+β3 is 1027. [3] 3x2+2x+1=0x2+23x+13=0α+β=−23αβ=13