Let $f(x)=\left[x^{2}-x\right]+|-x+[x]|$, where $x \in \mathbb{R}$ and $[t]$ denotes the greatest integer less than or equal to $t$. Then, $f$ is :

We have, <br/><br/>$$\begin{aligned} f(x) & =\left[x^2-x\right]+|-x+[x]| \\\\ & =[x(x-1)]+\{x\} \end{aligned} $$ <br/><br/>$$ f(x)=\left\{\begin{array}{ccc} x+1 & ; & -0.5 < x < 0 \\ 0 & ; & x=0 \\ -1+x & ; & 0 < x <1 \\ 0 & ; & x=1 \\ x-1 & ; & 1 < x < 1.5 \end{array}\right. $$ <br/><br/>At $x=0$, <br/><br/>$$ \begin{array}{r} \text { LHL }=\lim\limits_{x \rightarrow 0^{-}} f(x)=1 \\\\ \text { RHL } \lim\limits_{x \rightarrow 0^{+}} f(x)=-1 \\\\ f(0)=0 \end{array} $$ <br/><br/>$\therefore f(x)$ is not continuous at $x=0$ <br/><br/>At $x=1$ <br/><br/>$$ \begin{aligned} \mathrm{LHL} & =\lim _{x \rightarrow 1^{-}} f(x)=-1+1=0 \\\\ \mathrm{RHL} & =\lim _{x \rightarrow 1^{+}} f(x)=1-1=0 \\\\ f(1) & =0 \end{aligned} $$ <br/><br/>$\therefore f(x)$ is continuous at $x=1$ <br/><br/>Hence, $f(x)$ is continuous at $x=1$, but not continuous at $x=0$.