Higher June 2023 Paper 5 Q18
18 \(\sqrt[5]{p^2} = \left(\sqrt[3]{m}\right)^2\) and \(p = m^x\), where \(p \gt 0\), \(m \gt 0\) and \(p \ne m\).
Show that the value of \(x\) is \(\frac{5}{3}\). [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Correct method to establish \(x = \frac{5}{3}\) e.g. \(p^{\frac{2}{5}} = m^{\frac{2}{3}}\) or better or \(m^{\frac{2x}{5}} = m^{\frac{2}{3}}\) or \(\frac{2x}{5} = \frac{2}{3}\) or \(p = \left(\sqrt{\left(\sqrt[3]{m}\right)^2}\right)^5\) or better | M2 | M1 for \(p^{\frac{2}{5}}\) or \(m^{\frac{2}{3}}\) or [ \(\sqrt[5]{(m^x)^2}\) =] \(m^{\frac{2x}{5}}\) or for first step in making \(p\) the subject \(p^2 = \left(\sqrt[3]{m^2}\right)^5\) or \(\sqrt[5]{p} = \sqrt{\left(\sqrt[3]{m^2}\right)}\) or better If 0 scored, SC1 for [ \(p\) =] \(\sqrt[3]{m^5}\) written | or better e.g. \(\sqrt[5]{p} = \sqrt[6]{m^2}\) e.g. or better for M2 \(p = \sqrt[6]{m^{10}}\) Maximum mark is SC1 for those working backwards from \(x = \frac{5}{3}\) and this mark is for interpreting the index \(m^{\frac{5}{3}}\) as \(\sqrt[3]{m^5}\) |
| \(p^{\frac{1}{5}} = m^{\frac{1}{3}}\) leading to \(p = m^{\frac{5}{3}}\) or \(\frac{2}{3} \times \frac{5}{2} = \frac{5}{3}\) or \(6x = 10\) leading to \(\frac{5}{3}\) or \(p = \left(\sqrt[3 \times 2]{m}\right)^{(2 \times 5)}\) or better leading to \(p = m^{\frac{5}{3}}\) | A1 | After M2 earned and with no errors seen | |