If the function f : R→Rf : R→R be such that f(x)=x−[x]fx=x-x,


If the function $f: R \rightarrow R$ be such that $f(x)=x-[x]$, where $[x]$ denotes the greatest integer less than or equal to $x$, then $f^{-1}(x)$ is

(a) $\frac{1}{x-[x]}$

(b) $[x]-x$

(c) not defined

(d) none of these



We know that the range of $f$ is $[0,1)$.

Co-domain of $f=R$

As range of $f \neq$ Co-domain of $f, f$ is not onto.

$\Rightarrow f$ is not a bijective function.

So, $f^{-1}$ does not exist.

Thus, the answer is (c).

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