The function f is defined by f : x → 7 – 2x² – 12x for x ∈ R.

Mark scheme

Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.

1. [B1 B1] (i) To express 7 – 2x² – 12x in the form a – 2(x + b)², factor out -2 from the x terms: 7 - 2(x² + 6x). Complete the square for x² + 6x: x² + 6x = (x + 3)² - 9. So, 7 - 2((x + 3)² - 9) = 7 - 2(x + 3)² + 18 = 25 - 2(x + 3)². So, a = 25 and b = 3. Mark expression if present: B1 for 25 and B1 for – 2(x + 3)². If no expression award a = 25 B1 and b = 3 B1.
2. [B1FT] (ii) The stationary point for y = a - 2(x + b)² occurs at x = -b. From (i), x = -3. The y-coordinate is a = 25. So, the stationary point is (-3, 25). FT from answers to (i) or by calculus.
3. [B1FT] (iii) For g to have an inverse, it must be one-to-one. For a quadratic function, this means restricting the domain to one side of the vertex. The x-coordinate of the vertex is -3. So the smallest value of k for which g has an inverse is k = -3. FT from answer to (i) or (ii) NOT x = −3.
4. [*M1] (iv) EITHER: Let y = 25 - 2(x + 3)². Make x the subject: 2(x + 3)² = 25 - y, so (x + 3)² = (25 - y)/2. x + 3 = ±√((25 - y)/2). Since x ≥ -3, we take the positive square root: x + 3 = √((25 - y)/2). x = √((25 - y)/2) - 3. Makes their squared term containing x the subject or equivalent with x/y interchanged first. Condone errors with +/- signs.
5. [DM1] (iv) EITHER: Divide by ±2 and then square root allow ±.
6. [A1] (iv) EITHER: g⁻¹(x) = √((25 - x)/2) - 3. ± gets A0. Must now be a function of x. Allow y = isw if substituting x = - 3.
7. [*M1] (iv) OR: Rearranging equation of the curve: y = 7 - 2x² - 12x → 2x² + 12x + (y - 7) = 0.
8. [DM1] (iv) OR: Correct use of their 'a, b and c' in quadratic formula: x = (-12 ± √(12² - 8(y - 7))) / 4. Allow just + in place of ±.
9. [A1] (iv) OR: g⁻¹(x) = (-12 + √(144 - 8y + 56)) / 4 = (-12 + √(200 - 8y)) / 4 = -3 + √(200 - 8y) / 4. ± gets A0. Must now be a function of x. Allow y = isw if substituting x = - 3.