Let f(x)=[x^2-x]+|-x+[x]|, where x R and [t] denotes the greatest integer less than or equ…

Question

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

Accepted Answer

We have,

$$\begin{aligned}
f(x) & =\left[x^2-x\right]+|-x+[x]| \\
& =[x(x-1)]+\{x\}
\end{aligned}
$$

$$
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.
\]

At $x=0$,

$$
\[\begin{gathered} \text { LHL }=\lim\limits_{x \rightarrow 0^{-}} f(x)=1 \ \text { RHL } \lim\limits_{x \rightarrow 0^{+}} f(x)=-1 \ f(0)=0 \end{gathered}$$
\]

$\therefore f(x)$ is not continuous at $x=0$

At $x=1$

$$
\begin{aligned}
\mathrm{LHL} & =\lim _{x \rightarrow 1^{-}} f(x)=-1+1=0 \\
\mathrm{RHL} & =\lim _{x \rightarrow 1^{+}} f(x)=1-1=0 \\
f(1) & =0
\end{aligned}
$$

$\therefore f(x)$ is continuous at $x=1$

Hence, $f(x)$ is continuous at $x=1$, but not continuous at $x=0$.

Solutions · 1