Q104 / 160 · AP EAMCET 2021 (24 Aug Shift 2) · Mathematics
If [.] denotes the greatest integer function, then $f(x)=[x]^2-\left[x^2\right]$ is discontinuous at
- Aall integers
- Ball integers except 0 and 1
- Correct answerCall integers except 1
- Dall integers except 0
Solution
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 .
Hence, the correct answer is option: 'C'.