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 α3+β3=α+β()α2−αβ+β2()=−23α+β()2−3αβ⎡⎣⎢⎤⎦⎥=−23−23⎛⎝⎜⎞⎠⎟2−1⎡⎣⎢⎢⎤⎦⎥⎥=−23−59⎛⎝⎜⎞⎠⎟=1027 (ii) Find a quadratic equation whose roots are 1α3and 1β3. [3] 1α3+1β3=β3+α3α3β3=1027127=10 – for obtaining sum and product of roots α and β. – for re-expressing into α+β and αβ . Acceptable too: – for finding sum of roots

y -3 A (0.6, 3) x 1α31β3⎛⎝⎜⎞⎠⎟=1α3β3=113⎛⎝⎜⎞⎠⎟3=27 x2−10x+27=0 4 The diagram shows part of the graph y=r−q3−px, where p, q and r are positive constants. The graph has a vertex at A (0.6, 3) and y-intercept of -3. (i) Determine the values of p, q and r. [3] 3−px=0px=3x=3p3p=0.6p=5r=3At the y-intercept of -3, x = 0, O – for finding product of roots

y cm x cm 3 cm −3=3−q33q=6q=2 (ii) State the value or range of values of k such that k=r−q3−px has (a) 1 solution, [1] k = 3 (b) 2 solutions. [1] k < 3 5 A buoy is formed by two identical right circular cones of sheet iron joined by its bases with a radius of x cm. The buoy has a vertical height of y cm and a slant height of 3 cm. (i) Express y in terms of x. [1] y2+x2=32y=±9−x2y=9−x2(-ve rejected as vertical height is positive) y cm 3 cm – applying Pythagoras’ Theorem to obtain y in terms of x

(ii) Given that x can vary, find the exact value of x for which the volume, V, of the buoy is stationary. [4] V=23πx2y=23πx29−x2dVdx=23πx2129−x2()−12−2x()⎡⎣⎢⎤⎦⎥+9−x24π3x⎛⎝⎜⎞⎠⎟=−2πx339−x2+9−x24π3x⎛⎝⎜⎞⎠⎟=139−x2−2πx3+4πx(9−x2)⎡⎣⎤⎦=139−x236πx−6πx3⎡⎣⎤⎦ At stationary point, dVdx=0139−x236πx−6πx3⎡⎣⎤⎦=036πx−6πx3⎡⎣⎤⎦=0x(6−x2)=0x=0,x=±6 (iii) Determine with reasons whether this value of V is a maximum or minimum. [2] Using first derivative test, x 6− 6 6+ dVdx >0 0 <0 Volume is maximum. Using second derivative test, d2Vdx2=2π3(18−9x2)9−x2−(18x−3x3)129−x2()−12(−2x)9−x2⎡⎣⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥ When x=6d2Vdx2=−43.5<0 Volume is maximum -expressing volume of buoy in terms of x -applying product rule to find dV/dx -identifying that dV/dx = 0 at stationary point x=0 rejected; -ve Rejected as radius x is positive

(iv) Find the exact surface area of the buoy when V is stationary, leaving your answer in terms of π. [1] Surface area =2πx()3()=66πcm2 6 The equation of a polynomial is given byp(x)=2x3+2ax2−x2+ax−a where a is a constant. (i) Find the remainder when p(x)is divided by (x + 1). [1] p(−1)=2−1()3+2a−1()2−−1()2+a(−1)−a=−2+2a−1−a−a=−3 Remainder = -3 (ii) Show that (2x – 1) is a factor of p(x). [2] p12⎛⎝⎜⎞⎠⎟=212⎛⎝⎜⎞⎠⎟3+2a12⎛⎝⎜⎞⎠⎟2−12⎛⎝⎜⎞⎠⎟2+a12⎛⎝⎜⎞⎠⎟−a=14+a2−14+a2−a=0 Since remainder = 0, (2x – 1) is a factor of p(x). (iii) By showing clearly your working, factorise p(x). [2] 2 2a – 1 a -a 0.5 1 a a 2 2a 2a 0 Alternatively, by long division -either first or second derivative test – conclusion that volume is maximum – for finding p(1/2) or using long division/synthetic division -conclusion stating remainder = 0, therefore (2x-1) is a factor of p(x). – Using synthetic division/long division

x2+ax+a2x−12x3+(2a−1)x2+ax+a2x3−x2_____________________2ax2+ax2ax2−ax_____________________2ax+a2ax−a 2x3+2ax2−x2+ax−a=(2x−1)(x2+ax+a) (iv) Find the range of values of a for which the equation p(x)=0has only one real root. [3] (2x−1)(x2+ax+a)=0 For p(x) = 0 to have only one real root, a()2−4(1)(a)<0a2−4a<0a(a−4)<00<a<4 7 The table below shows the data obtained from an experiment on the vertical motion based on the oscillation of a spring with different masses attached to it. Mass, x kg 0.02 0.03 0.04 0.05 0.15 Frequency of oscillations, y 16 13 11.4 10 6 It is known that the mass, x kg, and the frequency of oscillations per second, y, are related by the equation xy2=k, where k is a constant. (a) Plot y2 against 1xand draw a straight line graph. y2=kx 1x 50 33.3 25 20 6.67 y2 256 169 129.96 100 36 [3] -apply discriminant < 0 – solve inequality by factorising