If $\mathrm{f}(x)=\frac{2 x-3}{3 x-4}, x \neq \frac{4}{3}$, then the value of $\mathrm{f}^{-1}(x)$ is
If $\mathrm{f}(x)=\frac{2 x-3}{3 x-4}, x \neq \frac{4}{3}$, then the value of $\mathrm{f}^{-1}(x)$ is
- $\frac{4 x-3}{3 x-2}$
- $\frac{3 x-2}{4 x+3}$
- $\frac{3 x-4}{4 x-2}$
- $\frac{2 x+3}{4 x-3}$
Solution
Let $\mathrm{f}(x)=y \Rightarrow x=\mathrm{f}^{-1}(y)$
$\begin{aligned}
& y=\frac{2 x-3}{3 x-4} \\
& \Rightarrow 3 x y-4 y=2 x-3 \\
& \Rightarrow x(3 y-2)=4 y-3 \\
& \Rightarrow x=\frac{4 y-3}{3 y-2} \\
& \Rightarrow \mathrm{f}^{-1}(y)=\frac{4 y-3}{3 y-2} \\
& \Rightarrow \mathrm{f}^{-1}(x)=\frac{4 x-3}{3 x-2}
\end{aligned}$
Asked in: MHT CET 2023 (14 May Shift 2)
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