Question

Which of the following function is differentiable at x=0

• A. $\cos \left(|x|\right)+|x|$
• B. $\cos \left(|x|\right)-|x|$
• C. $\sin \left(|x|\right)+|x|$
• D. $\sin \left(|x|\right)-|x|$

Answer 1

The correct option is A $\cos \left(|x|\right)+|x|$

A function is differentiable at 0 only if it's L.H.L &R.H.L is equal.

If $f\left(x\right)=\sin \mid x\mid -\mid x\mid$ then given that

$\sin (-x)=-\sin x$ it can be expressed as,

$\therefore f\left(x\right)=\left\{\begin{array}{c}\sin x-xifx\ge 0\\ -\sin x+xifx<0\end{array}\right\}$

$\Rightarrow f^{{}^{\prime }}\left(x\right)=\left\{\begin{array}{c}\cos x-1ifx\ge 0\\ -\cos x+1ifx<0\end{array}\right\}$

Given that $\cos 0=1$ we have $f_{+}^{{}^{\prime }}\left(0\right)=2$ & $f_{-}^{{}^{\prime }}\left(0\right)=-2$

$\therefore f_{+}^{{}^{\prime }}\left(0\right)\ne f_{-}^{{}^{\prime }}\left(0\right)$

If $f\left(x\right)=\cos \mid x\mid -\mid x\mid$ , then given that $\cos (-x)=\cos \left(x\right)$ can be written as:

$\therefore f\left(x\right)=\left\{\begin{array}{c}\cos x-xifx\ge 0\\ \cos x+xifx<0\end{array}\right\}$

$\Rightarrow f^{{}^{\prime }}\left(x\right)=\left\{\begin{array}{c}-\sin x+1ifx\ge 0\\ -\sin x+1ifx<0\end{array}\right\}$

Given that $\sin 0=0$ we have $f_{+}^{{}^{\prime }}\left(0\right)=-1$ & $f_{-}^{{}^{\prime }}\left(0\right)=1$

$\therefore f_{+}^{{}^{\prime }}\left(0\right)\ne f_{-}^{{}^{\prime }}\left(0\right)$

If $f\left(x\right)=\cos \mid x\mid +\mid x\mid$, then

$\therefore f\left(x\right)=\left\{\begin{array}{c}\cos x+xifx\ge 0\\ \cos x-xifx<0\end{array}\right\}$

$\Rightarrow f^{{}^{\prime }}\left(x\right)=\left\{\begin{array}{c}-\sin x+1ifx\ge 0\\ -\sin x-1ifx<0\end{array}\right\}$

Given that $\sin 0=0$ we have $f_{+}^{{}^{\prime }}\left(0\right)=1$ & $f_{-}^{{}^{\prime }}\left(0\right)=-1$

$\therefore A\Rightarrow \cos (\mid x\mid )+\mid x\mid$