Higher June 2023 Paper 2R Q17
17 Make \(x\) the subject of \(\;y = \sqrt[3]{\dfrac{6 + 5x}{x + 4}}\)
(4)
| Scheme | Marks |
|---|---|
| \(y^3 = \dfrac{6 + 5x}{x + 4}\) | M1 |
\(xy^3 + 4y^3 = 6 + 5x\) oe or \(x - \dfrac{5x}{y^3} = \dfrac{6}{y^3} - 4\) | M1 |
| \(xy^3 - 5x = 6 - 4y^3\) | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(x = \dfrac{6 - 4y^3}{y^3 - 5}\) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for removing cube root
M1: for multiplying by denominator and expanding in a correct equation
or
for gathering \(x\) terms on one side and the other terms on the other side in a correct equation in fractional form
M1: for gathering terms in \(x\) on one side and other terms the other side in a correct equation
or
for removing all fractions
A1: or \(x = \dfrac{4y^3 - 6}{5 - y^3}\)
SCB2 for \(x = \dfrac{6 - 4y^{\frac{1}{3}}}{y^{\frac{1}{3}} - 5}\) or \(x = \dfrac{4y^{\frac{1}{3}} - 6}{5 - y^{\frac{1}{3}}}\)
\(y^{\frac{1}{3}}\) can also be \(y^2\)