June 2018 Paper 1 Q9
9 \(\mathrm{f}(x) = 14 - x^2 \quad\) for all real values of \(x\).
Solve \(\quad \mathrm{f}(2x) = 5\)
You must show your working. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(14 - (2x)^2\) or \(14 - 4x^2\) | M1 | or \(14 - (2x)^2 = 5\) or \(14 - 4x^2 = 5\) |
| \(14 - 5 = (2x)^2\) or \(9 = 4x^2\) or \(9 - 4x^2 = 0\) or \(4x^2 - 9 = 0\) or \((2x + 3)(2x - 3) = 0\) | M1dep | |
| \((x =)\ \dfrac{3}{2}\) or 1.5 | A1 | |
| \((x =)\ -\dfrac{3}{2}\) or \(-1.5\) | A1 | |
| Alternative method 2 | ||
| \(14 - x^2 = 5\) and \(x = \pm 3\) | M1 | |
| \(2x = \pm 3\) | M1dep | |
| \((x =)\ \dfrac{3}{2}\) or 1.5 | A1 | |
| \((x =)\ -\dfrac{3}{2}\) or \(-1.5\) | A1 | |
Additional guidance
| A final answer of \(\sqrt{\dfrac{9}{4}}\) scores M1 M1 A0 A0 A final answer of \(\pm\sqrt{\dfrac{9}{4}}\) scores M1 M1 A1 A0 |