$f(x)=\frac{3 x+2}{5 x-3}, x \in R-\left\{\frac{3}{5}\right\}$, then

$f(x)=\frac{3 x+2}{5 x-3}, x \in R-\left\{\frac{3}{5}\right\}$, then
  1. $\mathrm{f}^{-1}(x)=\mathrm{f}(x)$
  2. $\mathrm{f}^{-1}(x)$ does not exist.
  3. $\mathrm{f}[\mathrm{f}(x)]=-x$
  4. $\mathrm{f}^{-1}(x)=-\mathrm{f}(x)$

Solution

$\begin{array}{l} \text { Let } y=f(x)=\frac{3 x+2}{5 x-3} \\ \therefore y(5 x-3)=3 x+2 \Rightarrow 5 x y-3 y=3 x+2 \Rightarrow(5 y-3) x=3 y+2 \\ x=\frac{3 y+2}{5 y-3} \\ \Rightarrow f^{-1}(y)=\frac{3 y+2}{5 y-3} \\ \Rightarrow f^{-1}(x)=\frac{3 x+2}{5 x-3} \end{array}$ Thus $f^{-1}(x)=f(x)$

Asked in: MHT CET 2020 (20 Oct Shift 1)

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