If f : R→R is given by f(x)=3x−5,


If $f: R \rightarrow R$ is given by $f(x)=3 x-5$, then $f^{-1}(x)$

(a) is given by $\frac{1}{3 x-5}$

(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


Clearly, $f$ is a bijection.

So, $f^{-1}$ exists.

Let $f^{-1}(x)=y$                ...(1)

$\Rightarrow f(y)=x$

$\Rightarrow 3 y-5=x$

$\Rightarrow 3 y=x+5$

$\Rightarrow y=\frac{x+5}{3}$

$\Rightarrow f^{-1}(x)=\frac{x+5}{3} \quad[$ from $(1)]$

So, the answer is (b).

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