Higher June 2024 Paper 1 Q15
15 The function f is defined as
\[\mathrm{f} : x \mapsto \frac{3x + 1}{x - 2}\](a) State the value of \(x\) that cannot be included in any domain of the function f (1)
(b) Express the inverse function \(\mathrm{f}^{-1}\) in the form \(\mathrm{f}^{-1}(x) = \ldots\) (3)
| Scheme | Marks |
|---|---|
| \((x =)\; 2\) | B1 |
| (1) |
Notes
B1: Accept \(x = 2\) and \(x \neq 2\)
\(x\) cannot be 2
Any response that contains 2 is also acceptable
DO NOT ACCEPT WHEN WRITTEN WITH INEQUALITY SIGNS
\(x \gt 2\) or \(x \lt 2\) or \(x \geqslant 2\) or \(x \leqslant 2\)
DO NOT ACCEPT
2 with another number eg 2 & 3
| Scheme | Marks |
|---|---|
| \(y(x - 2) = 3x + 1\) oe or \(yx - 2y = 3x + 1\) oe or \(x(y - 2) = 3y + 1\) oe or \(yx - 2x = 3y + 1\) oe | M1 |
| \(x(y - 3) = 1 + 2y\) oe or \(y(x - 3) = 1 + 2x\) oe | M1 |
Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{1 + 2x}{x - 3}\) | A1 |
| (3) | |
| (4 marks) |
Notes
M1: for factorising correctly
A1: oe eg \(\dfrac{-1 - 2x}{3 - x}\) (must be in terms of \(x\))