Higher June 2019 Paper 2 Q27
27
\[\text{f}(x) = \frac{2x}{5} - 1\]Work out the value of \(\quad \text{f}^{-1}(3) + \text{f}(-0.5)\) [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(y + 1 = \dfrac{2x}{5}\) or \(5y = 2x - 5\) | M1 | \(x\) and \(y\) may be transposed oe 1st step eg \(\dfrac{y}{2} = \dfrac{x}{5} - \dfrac{1}{2}\) |
| \(5(y + 1) = 2x\) or \(5y + 5 = 2x\) | M1dep | \(x\) and \(y\) may be transposed oe 2nd step eg \(\dfrac{y}{2} + \dfrac{1}{2} = \dfrac{x}{5}\) implies M2 |
| \(\dfrac{5(y + 1)}{2}\) or \(\dfrac{5y + 5}{2}\) or \(\dfrac{5(3 + 1)}{2}\) or 10 | A1 | may use \(x\) instead of \(y\) oe expression or calculation eg \(\dfrac{5y}{2} + \dfrac{5}{2}\) or \(\dfrac{3 + 1}{\frac{2}{5}}\) |
| \(\dfrac{2 \times -0.5}{5} - 1\) or \(-1.2\) or \(-\dfrac{6}{5}\) or \(-1\dfrac{1}{5}\) | M1 | oe |
| 8.8 or \(\dfrac{44}{5}\) or \(8\dfrac{4}{5}\) | A1 | |
| Alternative method 2 | ||
| \(\dfrac{2x}{5} = 3 + 1\) or \(\dfrac{2x}{5} = 4\) | M1 | oe |
| \(2x =\) their \(4 \times 5\) | M1dep | oe implies M2 |
| 10 | A1 | |
| \(\dfrac{2 \times -0.5}{5} - 1\) or \(-1.2\) or \(-\dfrac{6}{5}\) or \(-1\dfrac{1}{5}\) | M1 | oe |
| 8.8 or \(\dfrac{44}{5}\) or \(8\dfrac{4}{5}\) | A1 | |
Additional guidance
| The 4th mark may be seen first and may be the only mark awarded | |
| f may be used for \(y\) | |
| Missing brackets must be recovered | |
| Answer 8.8 | M2A1M1A1 |
| First three marks in Alt 1 Can be gained using a reverse function machine for a full calculation (applied to 3) which may be seen in stages eg \(3 + 1 = 4\) and \(4 \times 5 = 20\) and \(20 \div 2\) Part marks are not possible for this approach | M1M1A1 |