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If $f: R \rightarrow R$ is defined by $f(x)=x-[x]$, where $[x]$ is the greatest integer not exceeding $x$, then the set of discontinuous of $f$ is
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Verified Answer
The correct answer is:
$Z$
We know that,
$$
f(x)=\left\{\begin{array}{cc}
x-(n-1) & \text { for }-1 < x < n \\
0 & \text { for } x=n \\
x-n & \text { for } n < x < n+1
\end{array}\right.
$$
where $n \in z$
Now, we check the continuity at $x=n$
$$
\begin{aligned}
f(n-0) & =\lim _{x \rightarrow n^{-}}[x-(n-1)] \\
& =\lim _{h \rightarrow 0}[n-h-n+1) \\
& =1 \\
f(n+0) & =\lim _{x \rightarrow n^{+}}[x-n] \\
& =\lim _{h \rightarrow 0}(n+h-n)=0
\end{aligned}
$$
and
$$
f(n)=0
$$
$\Rightarrow f(x)$ is discontinuous at $x=n$
$\therefore$ Set of discontinuous of $f$ is $z$
$$
f(x)=\left\{\begin{array}{cc}
x-(n-1) & \text { for }-1 < x < n \\
0 & \text { for } x=n \\
x-n & \text { for } n < x < n+1
\end{array}\right.
$$
where $n \in z$
Now, we check the continuity at $x=n$
$$
\begin{aligned}
f(n-0) & =\lim _{x \rightarrow n^{-}}[x-(n-1)] \\
& =\lim _{h \rightarrow 0}[n-h-n+1) \\
& =1 \\
f(n+0) & =\lim _{x \rightarrow n^{+}}[x-n] \\
& =\lim _{h \rightarrow 0}(n+h-n)=0
\end{aligned}
$$
and
$$
f(n)=0
$$
$\Rightarrow f(x)$ is discontinuous at $x=n$
$\therefore$ Set of discontinuous of $f$ is $z$
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