Higher November 2023 Paper 4 Q21
21 Solve.
\[x^{-\frac{1}{6}} = \frac{5x^{\frac{1}{3}}}{x^{\frac{3}{4}}}, \text{ where } x \neq 0\][3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 625 with no extras | 3 | M1 for \(\dfrac{x^{-\frac{1}{6}} \times x^{\frac{3}{4}}}{x^{\frac{1}{3}}} = 5\) or better M1 for \(-\frac{1}{6} + \frac{3}{4} - \frac{1}{3} = \frac{1}{4}\) or better e.g. \(x^{\frac{1}{4}} = 5\) | Alternative method : M1 for \(x^{\frac{1}{3} - \frac{3}{4}}\) or \(x^{\frac{-5}{12}}\) or \(x^{\frac{-1}{6} + \frac{3}{4}}\) or \(x^{\frac{7}{12}}\) and M1 for \(x^{\frac{-1}{6} - \textit{their} \frac{-5}{12}}\) or \(x^{\frac{7}{12} - \frac{1}{3}}\) or \(x^{\frac{3}{12}}\) or \(x^{\frac{1}{4}}\) could be \(x^{\frac{5}{12}}\) or \(x^{\frac{-1}{4}}\) depends on which side of the equation |