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The function $(f)(x)=[x]^2-\left[x^2\right]$
(where, $[x]$ is the greatest integer less than or equal to $x$ ), is discontinuous at
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(where, $[x]$ is the greatest integer less than or equal to $x$ ), is discontinuous at
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The correct answer is:
all integers except 1 .
for $x=0$
$\left.\begin{array}{l}\text { L.H.L }=\left[O^{-}\right]^2-\left[\left(O^{-}\right)^2\right]=(-1)^2-(0)=1 \\ \text { R.H.L }=\left[O^{+}\right]^2-\left[O^{+}\right]^2=0^2-(0)=0\end{array}\right\}$ discount at $x=0$
for $x=1$
$\left.\begin{array}{l}\text { L.H.L }=\left[1^{-}\right]^2-\left[\left(1^{-}\right)^2\right]=0^2-0=0 \\ \text { R.H.L }=\left[1^{+}\right]^2-\left[\left(1^{+}\right)^2\right]=1^2-1=0 \\ \text { F.V. }=[1]^2-\left[1^2\right]=1-1=0\end{array}\right\}$ continuous at $x=1$
the function $f(x)$ is continuous at $x=1$ and discontinuous at all other integral point
$\left.\begin{array}{l}\text { L.H.L }=\left[O^{-}\right]^2-\left[\left(O^{-}\right)^2\right]=(-1)^2-(0)=1 \\ \text { R.H.L }=\left[O^{+}\right]^2-\left[O^{+}\right]^2=0^2-(0)=0\end{array}\right\}$ discount at $x=0$
for $x=1$
$\left.\begin{array}{l}\text { L.H.L }=\left[1^{-}\right]^2-\left[\left(1^{-}\right)^2\right]=0^2-0=0 \\ \text { R.H.L }=\left[1^{+}\right]^2-\left[\left(1^{+}\right)^2\right]=1^2-1=0 \\ \text { F.V. }=[1]^2-\left[1^2\right]=1-1=0\end{array}\right\}$ continuous at $x=1$
the function $f(x)$ is continuous at $x=1$ and discontinuous at all other integral point
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