SST 2025 AMATH PRELIM P1 MS

[Turn over SECONDARY 4 PRELIMINARY EXAMINATION ADDITIONAL MATHEMATICS Paper 1 4049/01 29 August 2025 (Friday) 2 hours 15 minutes CANDIDATE NAME MARKING SCHEME 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 18 printed pages including the cover page. For Examiner’s Use Q1 3 Q2 4 Q3 5 Q4 5 Q5 6 Q6 7 Q7 8 Q8 8 Q9 8 Q10 8 Q11 8 Q12 10 Q13 10 Total 90

2 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation , Binomial Expansion , where n is a positive integer and 2. TRIGONOMETRY Identities sin(𝐴±𝐵)=sin𝐴cos𝐵±cos𝐴sin𝐵 cos(𝐴±𝐵)=cos𝐴cos𝐵∓sin𝐴sin𝐵 tan(𝐴±𝐵)=tan𝐴±tan𝐵1∓tan𝐴tan𝐵 cos2𝐴=cos2𝐴−sin2𝐴=2cos2𝐴−1=1−2sin2𝐴 tan2𝐴=2tan𝐴1−tan2𝐴 Formulae for ΔABC

3 [Turn over Answer all the questions. 1 A parallelogram of length AB (√6−1) cm has an area of (2√6+3) cm2. Without using a calculator, find the perpendicular height, h, in the form (𝑎+𝑏√6) cm, where a and b are integers. [3] 𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ℎ𝑒𝑖𝑔ℎ𝑡 ℎ =2√6+3√6−1 =2√6+3√6−1×√6+1√6+1 =12+3+5√65 =(3+√6) cm

4 2 The function f is given by f(𝑥)=sin22𝑥tan2𝑥. (i) Find f′(𝑥). [2] f′(𝑥)=(2sin2𝑥)(2cos2𝑥)tan2𝑥+sin22𝑥(2sec22𝑥) =4(sin2𝑥cos2𝑥)tan2𝑥+2sin22𝑥(sec22𝑥) =4sin22𝑥+2tan22𝑥 (ii) Explain why f is an increasing function for 0<𝑥<𝜋4. [2] Given 0<𝑥<𝜋4 sin22𝑥>0 and tan22𝑥>0, hence 4sin22𝑥+2tan22𝑥>0⇒ f′(𝑥)>0. Since f′(𝑥)>0, f is an increasing function.

5 [Turn over 3 The equation of a curve is 𝑦=(4+𝑝)𝑥2−2𝑥+𝑝, where p is a constant. The curve has a maximum point and intersects the line 𝑦=2𝑥−1. Find the range of values of p. [5] (4+𝑝)𝑥2−2𝑥+𝑝=2𝑥−1 (4+𝑝)𝑥2−4𝑥+𝑝+1=0 𝑎=4+𝑝,𝑏=−4,𝑐=𝑝+1 Since the curve intersects the line, 𝑏2−4𝑎𝑐≥0 16−4(4+𝑝)(𝑝+1)≥0 4−(𝑝2+5𝑝+4)≥0 −𝑝2−5𝑝≥0 𝑝2+5𝑝≤0 𝑝(𝑝+5)≤0 −5≤𝑝≤0 Since curve has a maximum point, 𝑝+4<0 𝑝<− 4 Hence − 5≤𝑝<− 4

6 4 Show that the curve 𝑦=−𝑥2+(𝑚+4)𝑥−(𝑚2+6) is always negative for all real values of m. [5] Discriminant=(𝑚+4)2−4(−1)(−𝑚2−6) =𝑚2+8𝑚+16−4𝑚2−24 =−3𝑚2+8𝑚−8 =−3(𝑚2−83𝑚)−8 =−3[(𝑚−43)2−169]−8 =−3(𝑚−43)2+163−243 =−3(𝑚−43)2−83 Since (𝑚−43)2≥0 for all real values of m, −3(𝑚−43)2≤0 −3(𝑚−43)2−83<0 ∴𝑏2−4𝑎𝑐<0 Since 𝑏2−4𝑎𝑐<0 and coefficient of 𝑥2 is negative, 𝑦=−𝑥2+(𝑚+4)𝑥−(𝑚2+6) is always negative for all real values of m. (shown)

7 [Turn over ALTERNATIVE 𝑑𝑦𝑑𝑥=−2𝑥+𝑚+4 =0 2𝑥=𝑚+4 𝑥=𝑚2+2 At 𝑥=𝑚2+2 𝑦=−(𝑚+42)2+(𝑚+4)(𝑚+42)−𝑚2−6 =14(𝑚2+8𝑚+16)−𝑚2−6 =−34𝑚2+2𝑚−2 =−34(𝑚2−83𝑚)−2 =−34[(𝑚−43)2+169]−2 =−34(𝑚−43)2−313 Since (𝑚−43)2≥0 for all real values of m, −34(𝑚−43)2≤0 −34(𝑚−43)2−313<0 ∴𝑦−coordinate<0 Since the y-coordinate of the turning point is negative and the graph has a maximum point, 𝑦=−𝑥2+(𝑚+4)𝑥−(𝑚2+6) is always negative for all real values of m. (shown)

8 ALTERNATIVE 𝑦=−𝑥2+(𝑚+4)𝑥−(𝑚2+6) =−[𝑥2−(𝑚+4)𝑥]−(𝑚2+6) =−[(𝑥−𝑚+42)2−(𝑚+42)2]−(𝑚2+6) =−(𝑥−𝑚+42)2+𝑚24+2𝑚+4−𝑚2−6 =−(𝑥−𝑚+42)2−3𝑚24+2𝑚−2 y-coordinate of maximum point is −3𝑚24+2𝑚−2 Discriminant of this function =22−4(−34)(−2) =4−6 =−2<0 Since discriminant of function is less than zero, and coefficient of 𝑚2 is negative, the y-coordinate of maximum point is always negative. Therefore, =−𝑥2+(𝑚+4)𝑥−(𝑚2+6) is always negative for all real values of m. (shown)

9 [Turn over 5 The line 4𝑦=5𝑥−6 cuts the curve 𝑥+4𝑥𝑦−3𝑦2=7 at the points P and Q. Find the length of the line PQ in the form 𝑎√𝑏, where a and b are integers. [6] 5𝑥=4𝑦+6 𝑥=4𝑦5+65−−−(1) 𝑥+4𝑥𝑦−3𝑦2=7−−−(2) Substitute (1) into (2) : (4𝑦5+65)(1+4𝑦)−3𝑦2=7 (4𝑦+6)(1+4𝑦)−15𝑦2=35 16𝑦2+28𝑦+6−15𝑦2−35=0 𝑦2+28𝑦−29=0 (𝑦−1)(𝑦+29)=0 𝑦=1 or−29 𝑥=2 or−22 Point P is (2,1) and Point Q is (−22,−29). Length of line PQ =√(2−(−22))2+(1+29)2 =√1476 =6√41 units

10 6 Express 𝑥3+𝑥2−10𝑥−22(𝑥−3)(𝑥+1)2 in partial fractions. [7] (𝑥−3)(𝑥+1)2=(𝑥−3)(𝑥2+2𝑥+1) =𝑥3−𝑥2−5𝑥−3 1 𝑥3 − 𝑥2 − 5𝑥 − 3 𝑥3 + 𝑥2 − 10𝑥 − 22 (−) 𝑥3 − 𝑥2 − 5𝑥 − 3 2𝑥2 − 5𝑥 − 19 𝑥3+𝑥2−10𝑥−22(𝑥−3)(𝑥+1)2=1+2𝑥2−5𝑥−19(𝑥−3)(𝑥+1)2 Let 2𝑥2−5𝑥−19(𝑥−3)(𝑥+1)2=𝐴𝑥−3+𝐵𝑥+1+𝐶(𝑥+1)2 =𝐴(𝑥+1)2+𝐵(𝑥−3)(𝑥+1)+𝐶(𝑥−3)(𝑥−3)(𝑥+1)2 2𝑥2−5𝑥−19=𝐴(𝑥+1)2+𝐵(𝑥−3)(𝑥+1)+𝐶(𝑥−3) Sub 𝑥=−1 2(−1)2−5(−1)−19=𝐶(−1−3) −12=−4𝐶 𝐶=3 Sub 𝑥=3 2(3)2−5(3)−19=𝐴(3+1)2 −16=16𝐴 𝐴=−1 Sub 𝑥=0 −19=𝐴+𝐵(−3)(1)+𝐶(−3) −19=−1−3𝐵+3(−3) 3𝐵=9 𝐵=3 Hence 𝑥3+𝑥2−10𝑥−22(𝑥−3)(𝑥+1)2=1−1𝑥−3+3𝑥+1+3(𝑥+1)2