The diagram shows part of the curve y = √(5x – 1) and the normal to the curve at the point P (2, 3). This normal meets the x-axis at Q. [Figure X.X]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.

1. dy/dx = ½(5x-1)⁻½ × 5 (= 5/6) (B1 B1) B1 Without × 5 B1 × 5 of an attempt at differentiation
2. m of normal = -6/5 (M1) Uses m1m2 = -1 with their numeric value from their dy/dx
3. Equation of normal y - 3 = -6/5 (x - 2) OE (A1) Unsimplified. Can use y = mx + c to get c = 5.4 ISW
4. or 5y + 6x = 27 or y = -6/5 x + 27/5
5. EITHER: For the curve: (B1) Correct expression without ÷5
6. ∫√(5x-1) dx = (5x-1)^(3/2) / (3/2) ÷ 5 (B1) For dividing an attempt at integration of y by 5
7. Limits from 1/5 to 2 used → 3.6 or 18/5 OE (M1A1) Using 1/5 and 2 to evaluate an integrand (may be [y²])
8. Normal crosses x-axis when y = 0, → x = (4.5) (M1) Uses their equation of normal, NOT tangent
9. Area of triangle = 3.75 or 15/4 OE (A1) This can be obtained by integration
10. Total area = 3.6 + 3.75 = 7.35, 147/20 OE (A1)
11. OR: For the curve: (B2, 1, 0) –1 each error or omission.
12. ∫(x²+1) dy = ∫(y²/5 + 1) dy (M1A1) Using 0 and 3 to evaluate an integrand
13. Limits from 0 to 3 used → 2.4 or 12/5 OE (M1) Either to find side length for trapezium or attempt at integrating between 0 and 3
14. Uses their equation of normal, NOT tangent. (A1) This can be obtained by integration
15. Area of trapezium = ½(2+4½)×3 = 39/4 or 9¾ (A1)
16. Shaded area = 39/4 + 12/5 = 7.35, 147/20 OE (A1)