Question

If $f\left(x\right)=\left\{\begin{array}{c}x\left(\frac{e^{1/x}-e^{-1/x}}{e^{1/x}+e^{-1/x}}\right),\begin{array}{c}x\ne 0\\ x=0\end{array}\\ 0\end{array}\right)$ then at $x=0,f\left(x\right)$ is-

• A. differentiable
• B. not differentiable
• C. $f\left(0^{+}\right)=-1$
• D. $f\left(0^{+}\right)=1$

Answer 1

Answer: (B)

not differentiable

$\int \left(x\right)=\{\begin{array}{c}x\left(\frac{e^{yx}-e^{-yx}}{e^{yx}+e^{-yx}}\right)\\ 0\end{array}\begin{array}{c},x\ne 0\\ ,x=0\end{array}$

$\int \left(x\right)=x\left[\frac{(e^{yx}-e^{-yx})/2}{(e^{yx}+e^{-yx})/2}\right]$

$=x\sin \left(yn\right)$

$\because$ function is not defined at $x=0$.

So, the function is not derivable or differentiable.

Option $\left(B\right)$ is correct.