If the function $f : R \to R$ 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

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Solution

f(x) = x - [x]
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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If the function f : R→R be such that fx=x-x, where [x] denotes the greatest integer less than or equal to x, then f-1 x is (a) 1x-x (b) [x] − x (c) not defined (d) none of these

f(x) = x - [x] We know that the range of f is [0, 1). Co-domain of f = R As range of f ≠ Co-domain of f, f is not onto. ⇒ f is not a bijective function. So, f -1 does not exist. Thus, the answer is (c).

If the function $f : R \to R$ 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

f(x) = x - [x]
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).