Question
The domain of the derivative of the function \[f\left( x \right) = \left\{ \begin{array}{l}
{\tan ^{ - 1}}x\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\,\left| x \right| \le 1\\
\frac{1}{2}\left( {\left| x \right| - 1} \right)\,\,\,{\rm{if}}\,\,\left| x \right| > 1
\end{array} \right.\] is-
A.
$$R - \left\{ 0 \right\}$$
B.
$$R - \left\{ 1 \right\}$$
C.
$$R - \left\{ { - 1} \right\}$$
D.
$$R - \left\{ { - 1,\,1} \right\}$$
Answer :
$$R - \left\{ { - 1,\,1} \right\}$$
Solution :
The given function is
\[\begin{array}{l}
f\left( x \right) = \left\{ \begin{array}{l}
{\tan ^{ - 1}}x\,\,\,\,\,\,\,\,{\rm{if}}\,\,\left| x \right| \le 1\\
\frac{1}{2}\left( {\left| x \right| - 1} \right)\,\,\,{\rm{if}}\,\,\left| x \right| > 1
\end{array} \right.\\
f\left( x \right) = \left\{ \begin{array}{l}
\frac{1}{2}\left( { - x - 1} \right)\,\,\,{\rm{if}}\,x < - 1\\
{\tan ^{ - 1}}x\,\,\,\,\,\,\,\,\,{\rm{if}} - 1 \le x \le 1\\
\frac{1}{2}\left( {x - 1} \right)\,\,\,{\rm{if}}\,x > 1
\end{array} \right.
\end{array}\]
$$\eqalign{
& {\text{Clearly,}}\,\, \cr
& {\text{L}}{\text{.H}}{\text{.L}}{\text{. at}}\left( {x = - 1} \right) = \mathop {\lim }\limits_{h \to 0} f\left( { - 1 - h} \right) = 0 \cr
& {\text{R}}{\text{.H}}{\text{.L}}{\text{.at }}\left( {x = - 1} \right) = \mathop {\lim }\limits_{h \to 0} f\left( { - 1 + h} \right) \cr
& = \mathop {\lim }\limits_{h \to 0} \,{\tan ^{ - 1}}\left( { - 1 + h} \right) = \frac{{3\pi }}{4} \cr
& \therefore {\text{L}}{\text{.H}}{\text{.L}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.L}}{\text{. at }}x = - 1 \cr} $$
$$\therefore f\left( x \right)$$ is discontinuous at $$x =-1$$
Also we can prove in the same way, that $$f\left( x \right)$$ is discontinuous at $$x = 1$$
$$\therefore f'\left( x \right)$$ can not be found for $$x = \pm 1$$ or domain of $$f'\left( x \right) = R - \left\{ { - 1,\,1} \right\}$$