SST 2025 AMATH PRELIM P2 MS

SECONDARY 4 PRELIMINARY EXAMINATION ADDITIONAL MATHEMATICS Paper 2 4049/02 2 September 2025 (Tuesday) 2 hours 15 minutes CANDIDATE NAME Solutions CLASS 4 - INDEX NUMBER Candidates answer on the Question Paper. READ THESE INSTRUCTIONS FIRST Write your full name, class and index number in the spaces above. Write in dark blue or black pen in the space provided for each question. You may use a HB pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. The number of marks is given in brackets [ ] at the end of each question or part question. If working is needed for any question, it must be shown in the space below the question. Omission of essential working will result in loss of marks. The total of the marks for this paper is 90. The use of an approved scientific calculator is expected, where appropriate. If the degree of accuracy is not specified in the question and if the answer is not exact, give the answer to three significant figures. Give answers in degrees to one decimal place. For p, use either your calculator value or 3.142. This document consists of 21 printed pages including the cover page and 1 blank page. For Examiner’s Use Q1 8 Q2 10 Q3 9 Q4 9 Q5 5 Q6 6 Q7 11 Q8 8 Q9 8 Q10 6 Q11 10 Total 90

2 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation , Binomial Expansion where is a positive integer and 2. TRIGONOMETRY Identities sin2𝐴+cos2𝐴=1 sec2𝐴=1+tan2𝐴 cosec2𝐴=1+cot2𝐴 sin(𝐴±𝐵)=sin𝐴cos𝐵±cos𝐴sin𝐵 cos(𝐴±𝐵)=cos𝐴cos𝐵∓sin𝐴sin𝐵 tan(𝐴±𝐵)=tan𝐴±tan𝐵1∓tan𝐴tan𝐵 sin2𝐴=2sin𝐴cos𝐴 cos2𝐴=cos2𝐴−sin2𝐴=2cos2𝐴−1=1−2sin2𝐴 tan2𝐴=2tan𝐴1−tan2𝐴 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! ABCD€ asinA=bsinB=csinCa2=b2+c2−2bccosAΔ=12absinC

3 Answer all the questions. A4.2 Use of factor theorem, including factorising polynomials and solving cubic equations 1 (i) It is given that f(𝑥)=𝑥4−𝑝𝑥3+7𝑥2+𝑥−𝑞 has a quadratic factor 𝑥2−2𝑥−3. Show that 𝑝=5 and 𝑞=12. [5] 𝑥!−2𝑥−3=(𝑥−3)(𝑥+1) Given f(−1)=0 and f(3)=0 (−1)"−𝑝(−1)#+7(−1)!+(−1)−𝑞=0 (3)"−𝑝∙(3)#+7∙(3)!+3−𝑞=0 𝑞−𝑝=7 ---- (1) 27𝑝+𝑞=147 ---- (2) (2) − (1): 28𝑝=140 𝑝=5 and 𝑞=12 (shown) (ii) Hence, solve the equation 𝑥4−𝑝𝑥3+7𝑥2+𝑥−𝑞=0. [3] f(𝑥)=𝑥4−5𝑥3+7𝑥2+𝑥−12 =(𝑥!−2𝑥−3)(𝑥!+𝑏𝑥+4) [by inspection] By comparing coefficients of 𝑥!: 7=4−2𝑏−3 𝑏=−3 (𝑥!−2𝑥−3)(𝑥!−3𝑥+4)=0 (𝑥−3)(𝑥+1)(𝑥!−3𝑥+4)=0 𝑥−3=0 𝑜𝑟 𝑥+1=0 𝑜𝑟 𝑥!−3𝑥+4=0 Since discriminant of 𝑥!−3𝑥+4=0 is (−3)!−4(1)(4)=−7<0, there are no real solutions for 𝑥!−3𝑥+4=0. Solution: 𝑥=3 and 𝑥=−1

4 A6.2 Simplifying expressions involving exponential and logarithmic functions 2 (a) Without using a calculator, find the integer value of (5lg2) (2lg3) (5lg9) (2lg6). [3] 85$%!9 82$%#9 85$%&9 82$%'9=85$%!($%&9 82$%#($%'9 =85$%(!×&)9 82$%(#×')9 =85$%,-9 82$%,-9 = (5×2)$%,- =10$%,- =18 A6.1 Logarithmic functions including laws of logarithms and change of base (b) Given that log!(𝑦!)=4−log..0𝑥, express 𝑦 in terms of 𝑥. [3] log!(𝑦!)=4log!2−$1%!2$1%!!"# log!(𝑦!)=log!2"+log!𝑥1 log!(𝑦!)=log!2"+log!𝑥 log!(𝑦!)=log!(16𝑥) 𝑦!=16𝑥 𝑦=±4√𝑥 A6.2 Solving simple equations involving exponential and logarithmic functions (c) Solve the equation 2e𝑥=3−5√e𝑥. [4] 2e2=3−5e2! Let 𝑦=e$!. 2𝑦!=3−5𝑦 2𝑦!+5𝑦−3=0 (2𝑦−1)(𝑦+3)=0 Since e$!>0 for all x, 𝑦=0.5 e$!=0.5 2!=ln0.5 𝑥=2ln0.5=−1.39 (3 s.f.)

A6.3 Using exponential and logarithmic functions as models 3 A metal cube is heated to a temperature of 205℃ before being dropped into a liquid. As the cube cools, its temperature, 𝑇℃, t minutes after it enters the liquid is given by 𝑇=𝐾+175e345, where K and m are constants. (i) Show that the value of K is 30. [1] At t = 0, T = 2056𝐶. 𝐾+175e34(.)=205 𝐾=205−175 𝐾=30 (shown) When t = 3, the temperature of the cube reaches 128℃. (ii) Find the value of m. [3] When t = 3, T =128℃. 30+175e34(#)=128 175e3#4=98 e3#4=98175 −3𝑚=ln L98175M 𝑚=0.19327 𝑚=0.193 (3 s.f.) (iii) Find the rate at which the temperature of the cube is decreasing at the instant when t = 8. [3] 𝑇=30+175e3..,&#!75 d𝑇d𝑡=−33.82225e3..,&#!75 When t = 8, d𝑇d𝑡=−33.82225e3..,&#!7(-) =−7.2063 =−7.21 (3 s.f.) Temperature of the cube is decreasing at a rate of 7.21℃ per minute. (iv) Explain why the temperature of the cube can never fall below 30℃. [2] Since e3..,&#!75>0 for t , 175e3..,&#!75>0 , 30+175e3..,&#!75>30 Therefore, 𝑇>30 Hence the temperature of the cube can never fall below 30℃. 0³

6 A5.3 Use of the general term] 4 (a) In the expansion of (2+𝑥2)(12𝑥−𝑎𝑥)5, there is no term in 𝑥. Given that 𝑎≠0, find the value of the constant a. [5] General term for (12𝑥−𝑎𝑥)5= P5𝑟QP,!2Q038(−𝑎𝑥)8 =P5𝑟QP,!Q038(−𝑎)8𝑥!830 2𝑟−5=1 and 2𝑟−5=−1 𝑟=3 and 𝑟=2 Term with 𝑥3, =P52QP,!Q03!(−𝑎)!𝑥"30 =0"𝑎!P,2Q Term with x =P53QP,!Q03#(−𝑎)#𝑥'30 =−0!𝑎#𝑥 (2+𝑥2)(12𝑥−𝑎𝑥)5 =(2+𝑥2)(−52𝑎3𝑥+54𝑎2(1𝑥)+⋯) =L−5𝑎#+54𝑎!M𝑥+⋯ −5𝑎#+54𝑎!=0 0"𝑎!(−4𝑎+1)=0 𝑎=0 (rejected) or 𝑎=," A5.1 Use of Binomial Theorem for positive integer n (b) Write down and simplify the first three terms in the expansion of (2+𝑥22)5in ascending powers of 𝑥. Hence find the estimated value of (2.005)5, showing all your workings clearly. [4] (2+𝑥22)5=25+(51)24(𝑥22)+(52)23(𝑥22)2+⋯ =32+40𝑥!+20𝑥"+⋯ ---- (1) Solving (2+𝑥22)5= 2.0055 gives 𝑥=0.1 Substitute 𝑥=0.1 into (1): 2.0050=32+40(0.1)!+20(0.1)" =32+40(0.01)+20(0.0001) =32.402

G3.1 Use of tangent-chord theorem 5 The diagram shows a circle passing through the vertices of a triangle 𝐴𝐵𝐶. Points 𝐷 and 𝐸 are the midpoints of 𝐴𝐵 and 𝐴𝐶 respectively. The tangent to the circle at 𝐶 meets 𝐷𝐸 extended at the point 𝑇. Prove that points 𝐴,𝐷,𝐶 and 𝑇 lie on a circle. [5] By midpoint theorem, 𝐷𝐸 is parallel to 𝐵𝐶. Thus, ∡𝐴𝐷𝐸=∡𝐴𝐵𝐶. (corresponding angles) By tangent-chord theorem, ∡𝐴𝐵𝐶=∡𝐸𝐶𝑇 Since ∡𝐴𝐷𝐸=∡𝐸𝐶𝑇, they are angles in same segment, hence, points 𝐴,𝐷,𝐶 and 𝑇 lie on a circle. (shown) 𝐴 𝐵 𝐶 𝐷 𝐸 𝑇

8 G1.6 Use of the expressions of 𝑎cos𝜃+𝑏sin𝜃 in the form 𝑅cos(𝜃±𝛼) or 𝑅sin(𝜃±𝛼) 6 The expression 3cos𝜃+5sin𝜃 is defined for 0≤𝜃≤9! radians. (i) Express 3cos𝜃+5sin𝜃 in the form 𝑅cos(𝜃−𝛼), where 𝑅>0 and 𝛼 is an acute angle in radian. [2] 3cos𝜃+5sin𝜃=𝑅cos(𝜃−𝛼) 3cos𝜃+5sin𝜃=√3!+5!cos(𝜃−𝛼), where 𝛼=tan3,P0#Q =√34cos(𝜃−1.0303) (working value) =5.83cos(𝜃−1.03) (3 s.f.) Hence, (ii) solve the equation 3cos𝜃+5sin𝜃=4, [2] 3cos𝜃+5sin𝜃=4 √34cos(𝜃−1.0303)=4 cos(𝜃−1.0303)="√#" Basic angle =cos3,P"√#"Q=0.81482 𝜃−1.0303=0.81482,−0.81482 𝜃=1.84512,0.21548 As 𝜃 is acute, 𝜃=0.215 radians (3 s.f.) (accept 0.216 rad if exact form is used) (iii) find the minimum value of ,(#;1<=(0<>?=)!(,. [2] −√34≤3cos𝜃+5sin𝜃≤√34 0≤(3cos𝜃+5sin𝜃)!≤34 1≤(3cos𝜃+5sin𝜃)!+1≤35 135≤1(3cos𝜃+5sin𝜃)!+1≤1 Minimum value =,#0

9 G2.5 Transformation of given relationships, including 𝑦=𝑎𝑥@ and 𝑦=𝑘𝑏2, to linear form to determine the unknown constants from a straight line graph 7 (a) The mass, 𝑦 mg, of a radioactive substance decre