Question

Which of the following function is differentiable at $x=0$.

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

Answer 1

The correct option is D $\sin \left(\left|x\right|\right)-\left|x\right|$

A function is differentiable at 0 only if it's left and derivatives at 0 are equal, i.e.,

${lim}_{x\to 0^{-}}{f}^{\prime }\left(x\right)={lim}_{x\to 0^{+}}{f}^{\prime }\left(x\right)$

If $f\left(x\right)=\sin \left|x\right|-\left|x\right|$

Now, as we know that

$\sin (-x)=-\sin x$

Therefore, it can be expressed as,

$f\left(x\right)=\{\begin{array}{ccc}\sin x-x& & \text{if}x\ge 0\\ x-\sin x& & \text{if}x<0\end{array}$

$f^{\prime }\left(x\right)=\{\begin{array}{ccc}\cos x-1& & \text{if}x\ge 0\\ 1-\cos x& & \text{if}x<0\end{array}$

Now,

${lim}_{x\to 0^{-}}{f}^{\prime }\left(x\right)=\cos 0-1=0$

${lim}_{x\to 0^{+}}{f}^{\prime }\left(x\right)=1-\cos 0=0$

$\therefore {lim}_{x\to 0^{-}}{f}^{\prime }\left(x\right)={lim}_{x\to 0^{+}}{f}^{\prime }\left(x\right)$

Hence the function $\sin \left|x\right|-\left|x\right|$ will be differentiable at $x=0$.