Answer only one of the following two alternatives. OR The diagram shows part of the curve y = 2sin 3x . The normal to the curve y = 2sin 3x at the point where x = π/9 meets the y-axis at the point P. [Figure 11.2]

Mark scheme

Marking points from the official mark scheme (May/June 2012). The mark or mark type is shown in brackets.

1. [B1] (i) for dy/dx = 6cos3x
2. [B1] (i) For y and dy/dx values at x = π/9 (y = √3, dy/dx = 3)
3. [M1] (i) Use of m₁m₂ = -1 (gradient of normal = -1/3)
4. [M1] (i) for equation of normal and attempt to solve when x = 0 (y - √3 = -1/3(x - π/9))
5. [A1] (i) y = 1.85 (P(0, 1.85))
6. [B1] (ii) for Area of trapezium = 1/2 (y_P + y_point_on_curve) * x_point_on_curve (0.6251, allow unsimplified 1/2(√3+1.85)(π/9))
7. [M1] (ii) for attempt to integrate (∫ 2sin3x dx = -2/3 cos3x)
8. [A1] (ii) for correct integration (-2/3 cos3x)
9. [M1] (ii) for correct application of limits ([-2/3 cos3x] from 0 to π/9 = -2/3 cos(π/3) - (-2/3 cos(0)))
10. [A1] (ii) for final answer 0.292 (3 s.f.)
11. OR
12. [M1] (ii) for ∫ (equation of normal - equation of curve) dx (∫ ((1/3)x + 1.85 - 2sin3x) dx)
13. [A2,1,0] (ii) for correct integration ((1/6)x² + 1.85x + (2/3)cos3x)
14. [DM1] (ii) for correct application of limits (from 0 to π/9)
15. [A1] (ii) for final answer 0.292