Question

$f\left(x\right)=(\left[x\right]-[-x])si{n}^{-1}|x-1|$.
Which of the following statements is/are correct ?
(Note : $[.]$ denotes the greatest integer function)

• A. $f\left(x\right)$ is continuous at $x=1.$
• B. $f\left(x\right)$ is differentiable at $x=1.$
• C. $f\left(x\right)$ is not differentiable at $x=1.$
• D. $f\left(x\right)$ is discontinuous at $x=1.$

Answer 1

Answer: (A), (C)

The correct options are
A $f\left(x\right)$ is not differentiable at $x=1.$
C $f\left(x\right)$ is continuous at $x=1.$

We know, $\left[x\right]-[-x]=2\left[x\right]+1$ for non-integral values of $x$
$\therefore f\left(x\right)=\{\begin{array}{cc}{\sin }^{-1}(1-x)& 0\le x<1\\ 0& x=1\\ 3{\sin }^{-1}(x-1)& 1<x\le 2\end{array}$
Clearly $f\left(x\right)$ is continuous at $x=1$
$f^{\prime }\left(x\right)=\frac{-1}{\sqrt{1-(1-x{)}^2}};x<1$
$\frac{3}{\sqrt{1-(x-1{)}^2}};x>1$
$\Rightarrow \begin{array}{c}{f}^{\prime }\left(1^{-}\right)=1\\ f^{\prime }\left(1^{+}\right)=3\end{array}\}$ Hence $f\left(x\right)$ is not differentiable at $x=1$