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,