The function f is such that f(x) = a²x² – ax + 3b for x ≤ 1/(2a), where a and b are constants.

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2016). The mark or mark type is shown in brackets.

1. [M1] Substitute in -2 and -3: f(-2) = a²(-2)² - a(-2) + 3b = 4a² + 2a + 3b. So, 4a² + 2a + 3b = 4a² - b + 8 => 2a + 4b = 8.
2. [A1] f(-3) = a²(-3)² - a(-3) + 3b = 9a² + 3a + 3b. So, 9a² + 3a + 3b = 7a² - b + 14 => 2a² + 3a + 4b = 14.
3. [M1] Substitute linear equation into quadratic & attempt solution: From 2a + 4b = 8, we get 4b = 8 - 2a. Substitute into 2a² + 3a + 4b = 14: 2a² + 3a + (8 - 2a) = 14 => 2a² + a + 8 = 14 => 2a² + a - 6 = 0 => (2a - 3)(a + 2) = 0.
4. [A1] a = -2 or a = 3/2
5. [A1] If a = -2, 4b = 8 - 2(-2) = 12, b = 3. If a = 3/2, 4b = 8 - 2(3/2) = 5, b = 5/4. (If A0A0 scored, allow SCA1 for either (-2,3) or (3/2, 5/4))
6. [M1A1] For a = 1, b = -1, f(x) = x² - x - 3. Complete the square: y = (x - 1/2)² - 1/4 - 3 = (x - 1/2)² - 13/4. (Allow with x/y transposed)
7. [DM1] Make x the subject: (x - 1/2)² = y + 13/4 => x - 1/2 = ±√(y + 13/4). (Allow with x/y transposed)
8. [A1] f⁻¹(x) = 1/2 - √(x + 13/4) (since original domain x ≤ 1/(2a) = 1/2, so choose negative root. Allow y =... Must be a function of x)
9. [B1^] Domain of f⁻¹ is x ≥ -13/4 (Allow >, [-13/4, ∞), etc)