Search any question & find its solution
Question:
Answered & Verified by Expert
If [.] denotes the greatest integer function, then $f(x)=[x]^2-\left[x^2\right]$ is discontinuous at
Options:
Solution:
1832 Upvotes
Verified Answer
The correct answer is:
all integers except 1
We have,
$f(x)=[x]^2-\left[x^2\right]$
$\begin{aligned} \lim _{x \rightarrow 0^{-}} f(x) & =\lim _{h \rightarrow 0}[0-h]^2-\left[(0-h)^2\right] \\ & =(-1)^2-0=1\end{aligned}$
$\begin{aligned} \lim _{x \rightarrow 0^{+}} f(x) & =\lim _{h \rightarrow 0}(0+h)^2-\left[(0+h)^2\right] \\ & =0-0=0\end{aligned}$
$\because f(x)$ is discontinuous of $x=0$
Now,
$\begin{aligned} \lim _{x \rightarrow 1^{-}} f(x) & =\lim _{h \rightarrow 0}[1-h]^2-\left[(1-h)^2\right] \\ & =0-0=0\end{aligned}$
$\begin{aligned} \lim _{x \rightarrow 1^{+}} f(x) & =\lim _{h \rightarrow 0}[1+h]^2-\left[(1+h)^2\right] \\ & =1-1=0\end{aligned}$
$\because f(x)$ is continuous at $x=1$.
$\therefore f(x)$ is discontinuous at all integer except 1 .
$f(x)=[x]^2-\left[x^2\right]$
$\begin{aligned} \lim _{x \rightarrow 0^{-}} f(x) & =\lim _{h \rightarrow 0}[0-h]^2-\left[(0-h)^2\right] \\ & =(-1)^2-0=1\end{aligned}$
$\begin{aligned} \lim _{x \rightarrow 0^{+}} f(x) & =\lim _{h \rightarrow 0}(0+h)^2-\left[(0+h)^2\right] \\ & =0-0=0\end{aligned}$
$\because f(x)$ is discontinuous of $x=0$
Now,
$\begin{aligned} \lim _{x \rightarrow 1^{-}} f(x) & =\lim _{h \rightarrow 0}[1-h]^2-\left[(1-h)^2\right] \\ & =0-0=0\end{aligned}$
$\begin{aligned} \lim _{x \rightarrow 1^{+}} f(x) & =\lim _{h \rightarrow 0}[1+h]^2-\left[(1+h)^2\right] \\ & =1-1=0\end{aligned}$
$\because f(x)$ is continuous at $x=1$.
$\therefore f(x)$ is discontinuous at all integer except 1 .
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.