Question:
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
Solution:
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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