Higher November 2020 Paper 3 Q30
30 \(\mathrm{f}(x) = \dfrac{1}{2}x \qquad \mathrm{g}(x) = x - x^2\)
Solve \(\quad \mathrm{f}^{-1}(x) = \mathrm{gf}(x)\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(2x\) | M1 | oe |
| \(\dfrac{1}{2}x - \left(\dfrac{1}{2}x\right)^2\) or \(\dfrac{1}{2}x - \dfrac{1}{4}x^2\) | M1 | oe \(\dfrac{1}{4}x^2 + \dfrac{3}{2}x = 0\) oe equation implies M2 |
| \(x\left(\dfrac{1}{4}x + \dfrac{3}{2}\right) = 0\) or \(x(x + 6) = 0\) | M1dep | dep on M2 oe method for correct quadratic equation eg \(\dfrac{-6 \pm \sqrt{6^2 - 4 \times 1 \times 0}}{2 \times 1}\) |
| \(x = 0\) and \(x = -6\) | A1 |
Additional guidance
| \(\dfrac{1}{2}x - \dfrac{1}{4}x^2 = 2x\) | M2 |
| \(2x - x^2 = 8x\) | M2 |
| \(x^2 + 6x = 0\) | M2 |