The function $(f)(x)=[x]^2-\left[x^2\right]$ (where, $[x]$ is the greatest integer less than or equal to $x$…

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
  1. all integers.
  2. all integers except 0 .
  3. all integers except 0 and 1 .
  4. all integers except 1 .

Solution

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

Asked in: MHT CET 2022 (10 Aug Shift 1)

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