Question

The domain of the derivative of the function $f(x)=\{\begin{array}{ll}{\tan }^{-1}x& \text{if }|x|\le 1\\ \frac{1}{2}(|x|-1)& \text{if }|x|>1\end{array}$ is

• A. R - {0}
• B. R - {1}
• C. R - {-1}
• D. R - {-1, 1}

Answer 1

Answer: (D)

The given function is $f(x)=\{\begin{array}{ll}{\tan }^{-1}x& \text{if }|x|\le 1\\ \frac{1}{2}(|x|-1)& \text{if }|x|>1\end{array}$
$\Rightarrow f(x)=\{\begin{array}{ll}\frac{1}{2}(-x-1)& \text{if }x<-1\\ {\tan }^{-1}x& \text{if }-1\le x\le 1\\ \frac{1}{2}(x-1)& \text{if }x>1\end{array}$
Clearly L.H.L. at (x = - 1) = $\underset{h\to 0}{\lim }f(-1-h)=0$
R.H.L. at (x = - 1) = $\underset{h\to 0}{\lim }{\tan }^{-1}(-1+h)=3\pi /4$
$\therefore$ L.H.L. $\ne$ R.H.L. at x = - 1
$\therefore$ f (x) is discontinuous at x = - 1
Also we can prove in the same way, that f (x) is discontinuous at x = 1
$\therefore$ f ' (x) can not be found for x = $\pm$1 or domain of f ' (x) = R - {- 1, 1}