Q92 / 180 · KCET 2012 · Mathematics
If $f: \mathbb{R} \rightarrow \mathbb{R}$ is defined by $f(x)=2 x+3$, then $\mathrm{f}^{-1}(\mathrm{x})$
- Correct answerAis given by $\frac{x-3}{2}$
- Bis given by $\frac{1}{2 x+3}$
- Cdoes not exist because ' $P$ ' is not injective
- Ddoes not exist because ' $\mathrm{P}$ is not surjective
Solution
Let $\mathrm{x}_{1}, \mathrm{x}_{2} \in \mathrm{R}$
$$
\begin{array}{lrl}
\text { Let } & \mathrm{f}\left(\mathrm{x}_{1}\right) & =\mathrm{f}\left(\mathrm{x}_{2}\right) \\
\Rightarrow & 2 \mathrm{x}_{1}+3 & =2 \mathrm{x}_{2}+3 \\
\Rightarrow & \mathrm{x}_{1} & =\mathrm{x}_{2}
\end{array}
$$
Hence, $\mathrm{f}$ is injective.
Let $\mathrm{y} \in \mathrm{codomain} \mathrm{R}$
$\begin{array}{ll}\text { Let } & y=f(x) \\ \Rightarrow & y=2 x+3 \\ \Rightarrow & x=\frac{y-3}{2} \in \text { domain } R\end{array}$
Hence, $f$ is onto (surjective). $\therefore \mathrm{f}^{-1}$ exists.
$$
\therefore \quad \mathrm{f}^{-1}(\mathrm{x})=\frac{\mathrm{x}-3}{2}, \mathrm{x} \in \mathrm{R}
$$
Hence, the correct answer is option: 'A'.