Question

Let $f\left(x\right)=[x{]}^2+\sqrt{\left\{x\right\}}$ where [] & {}respectively denotes the greatest integer and fractional part functions, then which of the following is correct?

• A. $f\left(x\right)$ is continuous at all integral points
• B. $f\left(x\right)$ is not differentiable $\forall$ $xϵI$
• C. $f\left(x\right)$ is discontinuous as $xϵI-\left\{1\right\}$
• D. $f\left(x\right)$ is continuous & differentiable at $x=0$

Answer 1

The correct option is C $f\left(x\right)$ is discontinuous as $xϵI-\left\{1\right\}$

$f\left(x\right)={\left[x\right]}^2+\sqrt{x-\left[x\right]}$

at $x=1$

$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}(0^2+\sqrt{x+0})$

as $(x\to 1^{-};0<x<1;[x]=0)$

$=\sqrt{1}=1$

$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}(1^2+\sqrt{x-1})$

$=1+0=1$

as $x\to 1^{+}\Rightarrow 1<x<2$

$\left[x\right]=1$

LHL $=$ RHL $=f\left(1\right)$

at $x=n,nϵI-\left\{1\right\}$

$\underset{x\to n^{-}}{lim}f\left(x\right)={(n-1)}^2+\sqrt{n-(n-1)}$

$={(n-1)}^2+1$

$=n^2-2n+2$

$\underset{x\to n^{-}}{lim}f\left(x\right)=n^2+\sqrt{n-n}$

$=n^2$

${(n-1)}^2-1\ne n^2$

$\Rightarrow$ LHL $\ne$ RHL

$\therefore$ Discontinuous as $xϵI-\left\{1\right\}$