The diagram shows part of the curve y = 1 - 4/(2x + 1)². The curve intersects the x-axis at A. The normal to the curve at A intersects the y-axis at B. [Figure 10]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.

1. [B2,1,0] dy/dx = [0] + [(2x+1)⁻³] × [+16] OE. Full marks for 3 correct components. Withhold one mark for each error or omission.
2. [B2,1,0] ∫ydx = [x] + [(2x+1)⁻¹] × [+2] (+c) OE. Full marks for 3 correct components. Withhold one mark for each error or omission.
3. [B1] At A, x = 1/2. Ignore extra answer x = -1.5
4. [*M1] dy/dx = 2 → Gradient of normal (=-½) With their positive value of x at A and their dy/dx uses m1m2 = -1
5. [DM1] Equation of normal: y - 0 = -½(x - ½) or y - 0 = -½(0 - ½) or 0 = -½ × ½ + c Use of their x at A and their normal gradient.
6. [A1] B (0, ¼)
7. [*M1] ∫₀^(1/2) ydx SOI with 0 and their positive x coordinate of A.
8. [DM1] [x + 2/(2x+1)] evaluated between 0 and ½. = (-½) Substitutes both 0 and their ½ into their ∫ydx and subtracts.
9. [B1] Area of triangle above x-axis = ½ × ½ × ¼ = 1/16
10. [A1] Total area of shaded region = 9/16 OE (including AWRT 0.563)
11. [*M1] ∫₀^(1/4) xdy SOI. Where x is of the form k(1-y)⁻¹/² + c with 0 and their negative y intercept of curve.
12. [DM1] [½(1-y)⁻¹/²] evaluated between 0 and ¼. Substitutes both 0 and their -3 into their ∫xdy and subtracts.
13. [B1] Area of triangle above x-axis = ½ × ½ × ¼ = 1/16
14. [A1] Total area of shaded region = 9/16 OE (including AWRT 0.563)
15. [*M1] ∫₀^(1/2) (their normal curve) with 0 and their positive x coordinate of A.
16. [DM1] [−½x² + ¼x] evaluated between 0 and ½. Substitutes both 0 and their ½ into their ∫ydx and subtracts.
17. [B1] ∫₀^(1/2) ydx = [x + 2/(2x+1)] evaluated between 0 and ½. Substitutes both 0 and ½ into the correct integral and subtracts.
18. [A1] Total area of shaded region = 9/16 OE (including AWRT 0.563)