MHT CET · Maths · Functions
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
- A \(\frac{4 x-3}{3 x-2}\)
- B \(\frac{3 x-2}{4 x+3}\)
- C \(\frac{3 x-4}{4 x-2}\)
- D \(\frac{2 x+3}{4 x-3}\)
Answer & Solution
Correct Answer
(A) \(\frac{4 x-3}{3 x-2}\)
Step-by-step Solution
Detailed explanation
Let \(\mathrm{f}(x)=y \Rightarrow x=\mathrm{f}^{-1}(y)\)
\(\begin{aligned}
& y=\frac{2 x-3}{3 x-4} \\
& \Rightarrow 3 x y-4 y=2 x-3 \\
& \Rightarrow x(3 y-2)=4 y-3 \\
& \Rightarrow x=\frac{4 y-3}{3 y-2} \\
& \Rightarrow \mathrm{f}^{-1}(y)=\frac{4 y-3}{3 y-2} \\
& \Rightarrow \mathrm{f}^{-1}(x)=\frac{4 x-3}{3 x-2}
\end{aligned}\)
\(\begin{aligned}
& y=\frac{2 x-3}{3 x-4} \\
& \Rightarrow 3 x y-4 y=2 x-3 \\
& \Rightarrow x(3 y-2)=4 y-3 \\
& \Rightarrow x=\frac{4 y-3}{3 y-2} \\
& \Rightarrow \mathrm{f}^{-1}(y)=\frac{4 y-3}{3 y-2} \\
& \Rightarrow \mathrm{f}^{-1}(x)=\frac{4 x-3}{3 x-2}
\end{aligned}\)
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