Question

If $f:R\to R\mathrm{is}\mathrm{given}\mathrm{by}f\left(x\right)=3x-5,\mathrm{then}{f}^{-1}\left(x\right)$
(a) is given by $\frac{1}{3x-5}$
(b) is given by $\frac{x+5}{3}$
(c) does not exist because f is not one-one
(d) does not exist because f is not onto

Answer 1

Clearly, f is a bijection.
So, f ^-1 exists.
$$\begin{gathered} \text{Let}{f}^{-1}\left(x\right)=y...\left(1\right) \\ \Rightarrow f\left(y\right)=x \\ \Rightarrow 3y-5=x \\ \Rightarrow 3y=x+5 \\ \Rightarrow y=\frac{x+5}{3} \\ \Rightarrow f^{-1}\left(x\right)=\frac{x+5}{3}\text{[from}\left(1\right)\text{]} \end{gathered}$$
So, the answer is (b).