Higher June 2017 Paper 6 Q17
17 Show that \(\dfrac{\sqrt[3]{81}}{3}\) can be written as \(3^{\frac{1}{3}}\). [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\dfrac{\sqrt[3]{81}}{3} = \dfrac{\sqrt[3]{3^4}}{3}\) or \(\dfrac{\sqrt[3]{81}}{3} = \dfrac{\sqrt[3]{3^4}}{3}\) | M1 | \(\dfrac{\sqrt[3]{81}}{3} = \dfrac{\sqrt[3]{81}}{\sqrt[3]{3^3}}\) | In left-hand methods, M1M1 can be awarded if the denominator 3 is consistently omitted |
| \(= \dfrac{3^{\frac{4}{3}}}{3}\) or \(\dfrac{\sqrt[3]{3^3 \times 3}}{3} = \dfrac{3\sqrt[3]{3}}{3}\) | M1dep | \(= \sqrt[3]{\dfrac{81}{27}}\) | There may be other surd methods. M1 first productive step \(\sqrt[3]{81} = 81^{\frac{1}{3}}\) is not sufficiently productive as a first step M1dep second productive step from a correct first step |
| \(\left[= 3^{\frac{4}{3} - 1}\right] = 3^{\frac{1}{3}}\) or \(\sqrt[3]{3} = 3^{\frac{1}{3}}\) | A1 | \(= \sqrt[3]{3} = 3^{\frac{1}{3}}\) | Conversion to decimals scores 0 |