June 2025 Paper 2 Q19
19 \(\quad \mathrm{f}(x) = 4x^3 - 7 \qquad\qquad \mathrm{g}(x) = 6x + 8\)
Work out the value of \(x\) for which \(\quad \mathrm{f}(2x) = \mathrm{gf}(x)\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(4 \times (2x)^3 - 7\) or \(32x^3 - 7\) | M1 | oe expression for \(\mathrm{f}(2x)\) may be seen in an equation |
| \(6 \times (4x^3 - 7) + 8\) or \(24x^3 - 34\) | M1 | oe expression for \(\mathrm{gf}(x)\) may be seen in an equation |
| \(8x^3 = -27\) or \(\sqrt[3]{\dfrac{-27}{8}}\) | M1dep | oe correct equation of form \(ax^3 = b\) eg \(x^3 = -3.375\) or oe cube root eg \(\sqrt[3]{-3.375}\) dep on M2 |
| \(-\dfrac{3}{2}\) or \(-1.5\) with no other solution | A1 | oe value do not accept unprocessed cube roots |
Additional guidance
| \(\mathrm{f}(2x)\) and \(2\mathrm{f}(x)\) is 1st M0 unless \(\mathrm{f}(2x)\) chosen or used | |
| \(\mathrm{gf}(x)\) and \(\mathrm{fg}(x)\) is 2nd M0 unless \(\mathrm{gf}(x)\) chosen or used | |
| Missing brackets must be recovered |