Higher November 2023 Paper 1 Q21
21 \(\mathrm{f}(x) = \dfrac{x - 9}{8} \qquad\qquad \mathrm{g}(x) = 2x^2 + 9 \qquad\qquad \mathrm{h}(x) = 4x\)
Solve \(\quad \mathrm{f}^{-1}(x) = \mathrm{gh}(x)\) [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(2(4x)^2 + 9\) | M1 | oe |
| \(32x^2 + 9\) | M1dep | dep on 2nd M1 may be implied by 4th mark |
| \(8x + 9\) | M1 | may be implied by 4th mark |
| \(32x^2 - 8x = 0\) or \(32x^2 = 8x\) | M1 | oe equation with brackets expanded rearranges their \(\mathrm{f}^{-1}(x) =\) their \(\mathrm{gh}(x)\) to correctly collect terms |
| 0 and \(\dfrac{1}{4}\) | A1 | oe eg 0 and \(\dfrac{8}{32}\) |
Additional guidance
| With no terms to collect in their equation the 4th mark cannot be awarded | |
| \(8x + 9 = 8x^2 + 36\) \(0 = 8x^2 - 8x + 27\) | M0M0M1 M1A0 |
| \(8x - 9 = 8x^2 + 36\) \(0 = 8x^2 - 8x + 45\) | M0M0M0 M1A0 |
| \(8x + 9 = 2(4x)^2 + 9\) \(8x + 9 = 16x^2 + 9\) \(8x = 16x^2\) | M1M0M1 M1A0 |
| \(8x + 9 = 4x(2x^2 + 9)\) \(0 = 8x^3 + 28x - 9\) | M0M0M1 M1A0 |