π₯2 ππ¦ 1. (a) Show that ππ₯ (π₯ − tan−1 π₯) = 1+π₯ 2 . 2 √3 [2] 1 (i) Show that ∫0 π₯ tan−1 π₯ ππ₯ = 3 π − 2 √3. [5] 9709/03/SP/20/# 5 1 1 2. The variables π₯ and π¦ satisfy the relation sin π¦ = tan π₯, where − 2 π < π¦ < 2 π. Show that ππ¦ 1 = ππ₯ cos π₯ √cos 2π₯ [5] 9709/32/F/M/19/# 5 1 3. (i) Differentiate sin2 π with respect to π. [2] −π ππ¨π¬ π½ −π π¬π’π§ π½ ππ¨π¬ π½ −π ππ¨π¬ππ π π½ ππ¨π π½ , ππ ππ π π¬π’π§ π½ π¬π’π§π π½ (ii) The variables π₯ and π satisfy the differential equation ππ₯ π₯ tan π + cosec 2 π = 0, ππ 1 1 for 0 < π < 2 π and π₯ > 0. It is given that π₯ = 4 when π = 6 π. Solve the differential equation, obtaining an expression for π₯ in terms of π. [6] π = √ππ¨π¬ππ π π½ + ππ 9709/31/M/J/19/# 5 1 4. The diagram shows the curve π¦ = sin 3π₯ cos π₯ for 0 ≤ π₯ ≤ 2 π and its minimum point π. The shaded region π is bounded by the curve and the π₯-axis. (i) By expanding sin(3π₯ + π₯) and sin(3π₯ − π₯) show that 1 sin 3π₯ cos π₯ = 2 (sin 4π₯ + sin 2π₯) [3] (ii) Using the result of part (i) and showing all necessary working, find the exact area of region π . [4] π ππ ππ¦ (iii)Using the result of part (i), express ππ₯ in terms of cos 2π₯ and hence find the π₯coordinate of π, giving your answer correct to 2 decimal places. [5] π. ππ 9709/32/M/J/19/#10 1 1 5. The curve π¦ = sin (π₯ + 3 π) cos π₯ has two stationary points in the interval 0 ≤ π₯ ≤ π. (i) (ii) (iii) 6. (i) (ii) ππ¦ Find ππ₯ . [2] π ππ¨π¬ (ππ + π ) π By considering the formula for cos(π΄ + π΅), show that, at the stationary points on 1 the curve, cos (2π₯ + 3 π) = 0. [2] Hence find the exact π₯-coordinate of the stationary points. [3] π π π ππ π ππ ππ 9709/33/M/J/19/#7 By differentiating 1 π 2 1 π 4 cos π₯ sin π₯ ππ¦ , show that if π¦ = cot π₯ then ππ₯ = − ππ¨π¬ππ π π. 1 Show that ∫ π₯ cosec 2 π₯ ππ₯ = 4 (π + ln 4). [2] [6] 9709/31/O/N/19/#6 7. The variable π₯ and π satisfy the differential equation 1 ππ₯ 1 sin π = (π₯ + 2) cos π 2 ππ 2 1 for 0 < π < π. It is given that π₯ = 1 when π = 3 π. Solve the differential equation and obtain an expression for π₯ in terms of cos π. [8] π = π − π πππ π½ 9709/32/O/N/19/#6 8. The diagram shows the graph of π¦ = π cos π₯ sin3 π₯ for 0 ≤ π₯ ≤ π, and its maximum point π. The shaded region π is bounded by the curve and the π₯-axis. (i) (ii) 9. (i) Find the π₯-coordinate of π. Show all necessary working and give your answer correct to 2 decimal places. [5] π = π. ππ By first using the substitution π’ = cos π₯, find the exact of the area of π . [7] π π 9709/33/O/N/19/#10 Using the expansions of cos(3π₯ + π₯) and cos(3π₯ − π₯), show that 1 (cos 4π₯ + cos 2π₯) ≡ cos 3π₯ cos π₯ 2 [3] (ii) 1 π 6 1 − π 6 Hence show that ∫ 3 cos 3π₯ cos π₯ ππ₯ = 8 √3. [3] 9709/32/F/M/18/#3 2 10. The variable π₯ and π satisfy the differential equation ππ₯ π₯ sin2 π = 2 tan π + 1, ππ 1 1 for 0 ≤ π ≤ 2 π and π₯ > 0. It is given that π₯ = 1 when π = 4 π. (i) (ii) π Show that ππ (tan2 π) = 2 tan π cos2 π . [1] 1 Solve the differential equation and calculate the value of π₯ when π = 3 π, giving your answer correct to 3 significant figures. [7] π = π. ππ 9709/32/F/M/18/#6 11. A curve has equation π¦ = π 3π₯ 1 2 tan π₯ . Find the π₯-coordinates of the stationary points of the curve in the interval 0 < π < π. Give your answer correct to 3 decimal places [6] π. πππ ππ π. πππ 9709/31/M/J/18/#3 3 4 1 4 π₯ 12. Let that πΌ = ∫ √1−π₯ ππ₯. 1 π 3 1 π 6 2 (i) Using the substitution that π₯ = cos π, show that that πΌ = ∫ 2 cos 2 π ππ. (ii) Hence find the exact value of πΌ. (iv) Find ππ₯ . [4] [4] π π π 9709/31/M/J/18/#5 1 13. The curve π¦ = sin (π₯ + 3 π) cos π₯ has two stationary points in the interval 0 ≤ π₯ ≤ π. (v) (vi) 14. (i) (iii) ππ¦ [2] π ππ¨π¬ (ππ + π ) π By considering the formula for cos(π΄ + π΅), show that, at the stationary points on 1 the curve, cos (2π₯ + 3 π) = 0. [2] Hence find the exact π₯-coordinate of the stationary points. [3] π π π ππ π ππ ππ 9709/33/M/J/19/#7 By differentiating 1 π 2 1 π 4 cos π₯ ππ¦ , show that if π¦ = cot π₯ then ππ₯ = − ππ¨π¬ππ π π. sin π₯ 1 Show that ∫ π₯ cosec 2 π₯ ππ₯ = 4 (π + ln 4). [2] [6] 9709/31/O/N/19/#6 15. The variable π₯ and π satisfy the differential equation 1 ππ₯ 1 sin π = (π₯ + 2) cos π 2 ππ 2 1 for 0 < π < π. It is given that π₯ = 1 when π = 3 π. Solve the differential equation and obtain an expression for π₯ in terms of cos π. [8] π = π − π πππ π½ 9709/32/O/N/19/#6 3 16. The diagram shows the graph of π¦ = π cos π₯ sin3 π₯ for 0 ≤ π₯ ≤ π, and its maximum point π. The shaded region π is bounded by the curve and the π₯-axis. (iii) Find the π₯-coordinate of π. Show all necessary working and give your answer correct to 2 decimal places. [5] π = π. ππ (iv) By first using the substitution π’ = cos π₯, find the exact of the area of π . [7] π π 9709/33/O/N/19/#10 4