The diagram shows part of the curve y = −x² + 8x – 10 which passes through the points A and B. The curve has a maximum point at A and the gradient of the line BA is 2. [Figure 9.1]

Mark scheme

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

1. [B1] dy/dx = -2x + 8; co.
2. [M1 A1] Sets dy/dx = 0 to find x-coordinate of A (x=4); A is (4, 6); Sets to 0 and attempt to solve for x. co.
3. [M1] Equation of line BA: y - 6 = 2(x - 4); Correct form of equation.
4. [M1] Simultaneous equations for line BA and curve: -x² + 8x - 10 = 2x - 2; Eliminates x or y completely.
5. [A1] Quadratic equation: x² - 6x + 8 = 0 or y² - 8y + 12 = 0; Method for quadratic eqn = 0.
6. [A1] Solution for B: B (2, 2); co (Must not be guessed from diagram).
7. [B2,1] Integral of -x² + 8x - 10 dx = -x³/3 + 4x² - 10x; 3 terms, loses 1 for each error.
8. [M1] Uses limits 2 to 4 for the integral; Uses x limits correctly – allow ±.
9. [A1] Area = 9 1/3; co – allow ± (2 must have been correctly found, not guessed).