June 2024 Paper 2 Q18
18 Rearrange \(\quad m = \dfrac{2k^3 - 7}{5m - 3k^3} \quad\) to make \(k\) the subject. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(m(5m - 3k^3) = 2k^3 - 7\) | M1 | oe elimination of fraction eg \(5m^2 - 3mk^3 = 2k^3 - 7\) |
| \(5m^2 + 7 = 2k^3 + 3mk^3\) | M1dep | oe expansion of brackets and collection of terms in \(k^3\) eg \(-3mk^3 - 2k^3 = -7 - 5m^2\) |
| \(5m^2 + 7 = k^3(2 + 3m)\) | M1dep | oe factorisation eg \(k^3(-3m - 2) = -7 - 5m^2\) may be implied eg \(\dfrac{5m^2 + 7}{2 + 3m} = k^3\) or \(\sqrt[3]{\dfrac{5m^2 + 7}{2 + 3m}}\) |
| \(k = \sqrt[3]{\dfrac{5m^2 + 7}{2 + 3m}}\) | A1 | oe with \(k\) the subject eg \(\left(\dfrac{5m^2 + 7}{2 + 3m}\right)^{\frac{1}{3}} = k\) or \(k = \dfrac{\sqrt[3]{-7 - 5m^2}}{\sqrt[3]{-2 - 3m}}\) |
Additional guidance
| Up to M1 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| Condone one transcription error for A1 after fully correct answer seen | |
| Condone eg \(k = \sqrt[3]{\dfrac{5m^2 + 7}{2 + 3m}}\) in working lines with \(\sqrt[3]{\dfrac{5m^2 + 7}{2 + 3m}}\) on answer line | M3A1 |
| Correct answer followed by incorrect further work eg \(k = \sqrt[3]{\dfrac{5m^2 + 7}{2 + 3m}} = \sqrt[3]{\dfrac{5m^2 + 7}{5m}}\) | M3A0 |