P1 AM QP (three musketeers)

Preliminary Examination Paper 1 Three Musketeers Exam (final) Time: 2 hours 15 minutes Name: ……………………………. Marks: 90(Paper 1 Question Paper)Topics: the triple threat(trigonometry, differentiation, integration) Pages: 30 90 READ THE INSTRUCTIONS FIRST 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.

Formula List:………………………………..…………………………………………………………………………………………….O Level Additional Mathematics 4047/1 To all viewers (and holy grail moderators):setter of this paper is a sec 4 student who has taken a deep and fond interest in my beloved, amath 1) thank you holy grail mods for removing the previous copy!2) this paper, if you are using it as practice, is much more difficult from the regular question types. in my opinion, amath questions has gotten more vanilla and less interesting once you start to try more papers, hence this paper showcases how some amath questions can truly be very difficultthis amath paper solely consists of the three big chapters, and it encourages critical thinking 3) note that the heading should be 4049/1 instead of 4047/14) similarly, marks might not be given as fairly in this paper as in normal papers5) if you really want to do it timed, I suggest giving more time (about 15 minutes) to balance the fairness of this paperinspiration for some qns:nchs, tkss, bbss, blss, nhhs

………………………………..…………………………………………………………………………………………….O Level Additional Mathematics 4047/1 1(a)Find the value of . [2] (b) Answer all questions :) [1] Do not use a calculator for the whole of this question. It is given that cot (–B) < 0 and sec (–B) < 0, where 0 ≤ B ≤ 2π. Explain why sin — > 0.B2 –145cot [ 2 cos ( – —) ] (i) (ii)Given that cot (–B) = – –– , find the exact value of cos ( ––––– ). 125 B – π2 [2]

………………………………..…………………………………………………………………………………………….O Level Additional Mathematics 4047/1 (c)Find the exact value of cosec ( ––– ).7π122 [2](i) (ii)Hence, find sec ( ––– ).27π12 [1]

(a)Show, using the aid of a diagram, that ∫ 3√k – x dx = —– , for k > 0.k –k –––––22 3πk2 2 [2] (b)Hence, find in terms of k and/or π, the value of: (i) (ii) ∫ √4k – 4x –2π cos (–––) dx.0 k –––––––22 πx2k [2] ∫ √k – x + π tan (–––) dx.–k 0 –––––22 πx3k2 [2] ………………………………..…………………………………………………………………………………………….O Level Additional Mathematics 4047/1 2

(a) Prove that [1] (b) [3] ………………………………..…………………………………………………………………………………………….O Level Additional Mathematics 4047/1 3*(i) (ii) [1] Prove that –––––––––––––––– = tan –.1 – cos θ + sin θ1 + cos θ + sin θ θ2 Hence, solve ——————— = sin 3θ + sin 3θ2 sec 3θ + tan 3θ–––––––––––––––––– , for –1 ≤ θ ≤ 1.1 + cos 12θ + sin 12θ1 – cos 12θ + sin 12θ

Show that — x ln x = ax ln x + bx , where x > 0, and a and b are constants to be found.64(i) ………………………………..…………………………………………………………………………………………….O Level Additional Mathematics 4047/1 [1]ddx 53 5 (ii)Only by expressing ∫ ax ln x + bx dx = x ln x + c , where c is an arbitrary constant,53 5 46 evaluate ∫ x ln x dx.53 [2] 1 1 4

………………………………..……………………………………………………………………………………………. O Level Additional Mathematics 4047/1 (iv) [3] O x y y = x ln x 53 ––e2 Diagram 4 (iii)Look at Diagram 4 below. The shaded area can be expressed in two different ways.One way in expressing the shaded area is first making x the subject, then expressing it as a bound integral below: ∫m n x dy, where m and n are constants. Find the values of m and n, and the shaded area below. [2] It is given that –– = –––––– . At x = – – , the equation of the normal is parallel to 8y – 9x = 24, and at x = 2, y = k. Form an equation for y. k4x – 92dydx 12 M ↑&

………………………………..……………………………………………………………………………………………. O Level Additional Mathematics 4047/1 It is given that f ′′(x) = 10 sin 3x – 6 tan 6x – 11. Given that the normal of the line y = f(x) at x = 0 is perpendicular to the x-axis, find and simplify an expression for f ′(x). 2 2 (i) [3] (ii)Find an expression for –– ln (2– 2 sin nx), where n is a constant.ddx 2 [1] 5

………………………………..……………………………………………………………………………………………. O Level Additional Mathematics 4047/1 (iii)Hence, given that y = f(x) intersects the point ( — , — ), find an expression for f(x).π312 [3] (iv)In the case where y = f’(x), the gradient of y at x = 0 is equal to the minimum gradient of y = ax + 3x – 5x + 2, a ≠ 0. Find a.23 [1]