Question

Let $\left[x\right]$ denote the greatest integer less than or equal to $x$. If $f\left(x\right)=\left[x\sin \pi x\right]$ , then $f\left(x\right)$ is :

• A. continuous at $x=0$
• B. continuous in $(-1,0)$
• C. differentiable at $x=1$
• D. differentiable in $(-1,1)$

Answer 1

Answer: (A), (B), (D)

continuous at $x=0$
continuous in $(-1,0)$
differentiable in $(-1,1)$

We have $-1<x<1$$\Rightarrow 0\le x\sin \pi x\le \frac{1}{2}$

Also $x\sin \pi x$ becomes negative and numerically less
than 1 when x is slightly greater than 1 and so by definition of $\left[x\right]$

$f\left(x\right)=\left[x\sin \pi x\right]=1$ when $1<x<1+h$
Thus $f\left(x\right)$ is continuous and equal to 0 in the closed interval $[-1,1]$
and so $f\left(x\right)$ is continuous in the open interval $(-1,1)$

At $x=1$, $f\left(x\right)$ is discontinuous
since $\underset{h\to 0}{lim}(1-h)=0$ and $\underset{h\to 0}{lim}(1+h)=-1$
$\therefore f\left(x\right)$ is not differentiable at $x=1$