Which of the following functions is differentiable at x = 0 - Vidyadip

MCQ

Which of the following functions is differentiable at $x = 0$

A
$\cos (|x|)\, + |x|$
B
$\cos (|x|)\, - |x|$
C
$\sin (|x|)\, + |x|$
✓
$\sin (|x|)\, - |x|$

✓

Answer

Correct option: D.

$\sin (|x|)\, - |x|$

d
(d) $\cos |x|\, = \cos x$ which is differentiable everywhere and $|x|$ is not differentiable at $x = 0$. So, both $\cos (|x|) \pm |x|$ are not differentiable at $x = 0$.
For $x < 0,\,\sin (|x|)\, + |x|\, = \sin ( - x) - x = - \sin x - x$
For $x > 0,\,\sin (|x|)\, + |x|\, = \sin x + x$.
$\therefore$ Its $LHD$ $x = 0$ is ${[ - \cos x - 1]_{x = 0}} = - 2$
and its $RHD$ at $x = 0$ is ${[\cos x + 1]_{x = 0}} = 2$.
$\therefore$ $\sin (|x|)\, + |x|$ is not differentiable at $x = 0$.
For $x < 0,\,\sin (|x|)\,\, - |x|\, = \sin ( - x) - ( - x) = - \sin x + x$
For $x > 0,\,\sin (|x|)\, - |x|\, = \sin x - x$
$\therefore$ its $LHD$ at $x = 0$ is ${[ - \cos x + 1]_{x = 0}} = 0$ and its $RHD$ at $x = 0$ is ${[\cos x - 1]_{x = 0}} = 0$
$\therefore$ $\left( {\sin (|x|)\,\, - |x|} \right)$ is differentiable at $x = 0$.

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