Higher November 2020 Paper 2 Q21
21 The function f is such that \(\mathrm{f}(x) = 5 + 6x - x^2 \quad\) for \(x \leqslant 3\)
(a) Express \(5 + 6x - x^2\) in the form \(p - (x - q)^2\) where \(p\) and \(q\) are constants. (2)
(b) Using your answer to part (a), find the range of values of \(x\) for which \(\mathrm{f}^{-1}(x)\) is positive. (5)
| Scheme | Marks |
|---|---|
| \(5 - (x \pm q)^2 + 9\) oe or \(p - (x - 3)^2\) oe or \(p - q^2 + 2qx - x^2\) and one of \(2q = 6\) or \(p - q^2 = 5\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: \(14 - (x - 3)^2\) | A1 |
| (2) |
Notes
M1: may be seen in working eg \(-[(x - 3)^2 - 9 - 5]\)
or
expanding \(p - (x - q)^2\) correctly and equating one of the coefficient of \(x\) or the constant term
A1: fully correct
SCB1 for \((x - 3)^2 - 14\)
| Scheme | Marks |
|---|---|
| e.g. \((x - 3)^2 = 14 - y\) [or \((y - 3)^2 = 14 - x\)] | M1 |
\(x = 3 \pm \sqrt{14 - y}\) [or \(y = 3 \pm \sqrt{14 - x}\)] | M1 |
| \((\mathrm{f}^{-1}(x) =)\; 3 - \sqrt{14 - x}\) | M1 |
| M1 | |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: \(5 \lt x \leqslant 14\) | A1 |
| (5) | |
| (7 marks) |
Notes
M1: correct steps to isolate their bracket
ft from (a) dep on expression in form \(\pm p \pm (x - q)^2\)
M1: complete method to find \(y\) in terms of \(x\) or \(x\) in terms of \(y\). Condone + for \(\pm\)
ft from (a) dep on expression in form \(\pm p \pm (x - q)^2\)
M1: for the correct inverse
M1: method to solve \(0 \lt 3 - \sqrt{14 - x}\) or a lower bound of 5 clearly shown, eg \(x \gt 5\) as part of the answer
A1: cao