Question

If, $f\left(x\right)=$ $\left[\begin{array}{cc}x{\tan }^{-1}x+{\sec }^{-1}\frac{1}{x}& ,xϵ(-1,1)-0\\ \frac{\pi }{2}& ifx=0\end{array}\right]$ then $f^{\prime }\left(0\right)$ is

• A. equal to -1
• B. equal to 0
• C. equal to 1
• D. non existent

Answer 1

The correct option is D non existent

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{x+h-x}{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}=\underset{h\to 0}{lim}\frac{h{\tan }^{-1}\left(h\right)+{\sec }^{-1}\left(\frac{1}{h}\right)+\frac{\pi }{2}}{h}=\underset{h\to 0}{lim}({\tan }^{-1}\left(h\right)+\frac{{\sec }^{-1}\left(\frac{1}{h}\right)+\frac{\pi }{2}}{h})=\underset{h\to 0}{lim}{\tan }^{-1}\left(h\right)+\underset{h\to 0}{lim}\frac{-(\frac{\pi }{2}-{\sec }^{-1}\left(\frac{1}{h}\right))}{h}=0+\underset{h\to 0}{lim}\frac{-{\cos }^{-1}\left(\frac{1}{h}\right)}{h}$

Limit does not exist

Hence, $f^{\prime }\left(0\right)$ is non-existent.