A-Level Mathematics October/November 2024 Q5(a)(iii): Find an expression for f¯¹(x) and state the domain of the function f¯¹.

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  Let $y = f(x)$. To find the inverse function, we first write the expression for $f(x)$ and then interchange $x$ and $y$.

  Let $y = \frac{2x+1}{2x-1}$

  Interchange $x$ and $y$: $x = \frac{2y+1}{2y-1}$

  Now, we make $y$ the subject of the formula. $x(2y-1) = 2y+1$ $2xy - x = 2y + 1$ $2xy - 2y = x + 1$ $y(2x - 2) = x + 1$ $y = \frac{x+1}{2x-2}$

  So, $f^{-1}(x) = \frac{x+1}{2x-2}$.

  The domain of the inverse function $f^{-1}$ is the range of the original function $f$. The function $f(x) = \frac{2x+1}{2x-1}$ can be written as $f(x) = \frac{2x-1+2}{2x-1} = 1 + \frac{2}{2x-1}$. This is a hyperbola with a horizontal asymptote at $y=1$. Given the domain of $f$ is $x < \frac{1}{2}$, the denominator $2x-1$ is always negative. Therefore, the term $\frac{2}{2x-1}$ is always negative. This means that the value of $f(x) = 1 + (\text{a negative number})$ will always be less than 1.

  The range of $f$ is $y < 1$.

  Therefore, the domain of $f^{-1}$ is $x < 1$.

  Answer: $f^{-1}(x) = \frac{x+1}{2x-2}$ Domain of $f^{-1}$ is $x < 1$.

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  How the marks are awarded

  • M1* — Setting $y = f(x)$, interchanging variables to get $x = \frac{2y+1}{2y-1}$ and then multiplying by the denominator to get $x(2y-1) = 2y+1$.
  • DM1 — Correctly rearranging the equation to gather all terms involving $y$ on one side and factorising, as shown in the step $y(2x - 2) = x + 1$. This mark is dependent on the first M1.
  • A1 — Obtaining the correct final expression for the inverse function, $f^{-1}(x) = \frac{x+1}{2x-2}$.
  • B1 — Correctly stating the domain of the inverse function as $x < 1$, based on the principle that the domain of $f^{-1}$ is the range of $f$.

  Common mistakes

  • Making a sign error when rearranging the equation, for example getting $2xy - 2y = 1 - x$, which leads to an incorrect final expression for $f^{-1}(x)$.
  • Attempting to find the domain of $f^{-1}$ by looking at its own expression (e.g., from the denominator $2x-2 \neq 0$, giving $x \neq 1$) instead of correctly identifying it as the range of the original function $f$.
  • Leaving the final answer as $x = \frac{y+1}{2y-2}$ instead of expressing it as a function of $x$, i.e., $f^{-1}(x) = ...$ or $y = ...$.
  • Forgetting to state the domain of $f^{-1}$ entirely, thus losing the final mark.

  Examiner tip: To find the inverse of a function $y=f(x)$, first swap $x$ and $y$, then make $y$ the new subject; remember that the domain of the new inverse function is the range of the original function.

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