The diagram shows the curve y = 4 + 3x - x² intersecting the positive x-axis at the point A. The line y = mx + 8 is a tangent to the curve at the point B. [Figure 11.1]

Mark scheme

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

1. [M1] y = 0 →(x-4)(x+1) = 0; Solve
2. [A1] → A is (4,0) nfww; Indication somewhere that x = 4 when y = 0
3. [M1] 4+3x-x² = mx +8; Eliminate y.
4. x² + (m -3)x + 4 = 0
5. [M1] b²-4ac(= 0)→(m-3)² = 16; M1dep Use of discriminant
6. [A1] m=-1; Do not award if m = 7 is not discarded
7. [M1] Obtain quadratic x² + (m-3)x+4=0 using their m and attempt to solve.; Working must be seen for any marks to be awarded. Must not be awarded if m is not obtained correctly
8. [A1] Point B (2, 6)
9. [M1] Area under curve = ∫(from 2 to 4) (4+3x-x²)dx; Integrate powers increased in at least 2 terms
10. [A1] = [4x + (3/2)x² - (1/3)x³] (from 2 to 4)
11. [M1] =[16+24-64/3]-[8+6-8/3]; M1dep Insert limits of their 2 and 4 and subtract in correct order. May be implied by 18-2/3
12. [M1] Intercept is (8,0) so area of triangle = (6×6)/2 =18; Area of triangle using (their8-XB) * yB / 2 or Attempt to find other suitable areas to result in a complete method.
13. [A1] Shaded area = 18-7 1/3 = 10 2/3; Accept 10.7. Must not be awarded if point B is not obtained correctly.