Question

If $f\left(x\right)=\left[x\sin \pi x\right]in(-1,1)$ then $f\left(x\right)$ is

• A. differentiable at $x=0$
• B. continuous in $(-1,0)$
• C. differentiable in $(-1,1)$
• D. None of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A differentiable at $x=0$
B continuous in $(-1,0)$
C differentiable in $(-1,1)$

By the definition of [x], we have $f\left(x\right)=\left[x\sin \pi x\right]=0$ for $-1\le x\le 1$, because $0\le x\sin \pi x\le 1$ Also, $f\left(x\right)=\left[x\sin \pi x\right]=-1$
for $1<x<1+h$ for some small appropriate h> 0, because
$\sin \pi x$ is negative and $\ge -1$ for $1<x<1+h$.
Thus $f\left(x\right)$ is constant and equal to 0 in the interval $[-1;1]$ and
so it is continuous and differentiable in $(-1,1)$. In particular,
$f\left(x\right)$ is continuous at $x=0$ and in the interval $(-1,0)$. At
$x=1$, $\underset{x\to 1+}{lim}f\left(x\right)=-1$ and
$\underset{x\to 1+}{lim}f\left(x\right)=0$, Hence f is not
continuous at $x=1$ and, in particular, not differentiable at $x=1$.