For the function f(x)=e^ |x|-|x|,x R, consider the following statements : Statement I : f …

Question

For the function $f(x)={e}^{\sin |x|}-|x|,x\in R$, consider the following statements :

Statement I : $f$ is differentiable for all $x\in R$.

Statement II : $f$ is increasing in $(-\pi ,-\frac{\pi }{2})$.

In the light of the above statements, choose the correct answer from the options given below :

Accepted Answer

Here $f(x)={e}^{\sin |x|}-|x|$. Since $|x|$ is not differentiable at $x=0$, and also $\sin |x|$ has a ``kink'' at $x=0$, the function $f$ is not differentiable at $x=0$. So, Statement I is false.

Now take $x\in (-\pi ,-\frac{\pi }{2})$. Here $x$$$\begin{gathered} f(x)={e}^{-\sin x}+x \ {f}^{'}(x)=-{e}^{-\sin x}\cos x+1 \ x\in (-\pi ,\frac{-\pi }{2})\, \cos x0 \ \therefore f(x) \end{gathered}$$$

Since $\cos x0$, we get $-{e}\^{-\sin x}\cos x>0$, so $

Solutions · 1