Mathematics - Continuity - Differentiability Question with Solution | TestHub

Given,

$f\left(x\right)=\left[x^2-x\right]+\left|-x+\left[x\right]\right|$

$\Rightarrow f\left(x\right)=\left[x^2-x\right]+\left|x-\left[x\right]\right|, \left\{ \text{as} \left|-A\right|=\left|A\right|\right\}$

$\Rightarrow f\left(x\right)=\left[x^2-x\right]+\left|\left\{x\right\}\right|, \left\{\text{as} \left[x\right]+\left\{x\right\}=x\right\}$ 

$\Rightarrow f\left(x\right)=\left[x^2-x\right]+\left\{x\right\} (\because \left\{x\right\}\ge 0)$

Now at $x=0$, $f\left(0\right)=0$

And $f\left(0^{+}\right)=-1, \left\{\text{as} x^2-x<0 \text{for} x\to 0^{+}\right\}$

So, function is discontinuous at $x=0$

Now at $x=1$, $f\left(1\right)=0$, 

$f\left(1^{+}\right)=0+0=0$ and $f\left(1^{-}\right)=-1+1=0$

Hence, the function is continuous at $x=1$.