Practicing Success
Which of the following functions is differentiable at x = 0 ? |
$\cos (|x|)+|x|$ $\cos (|x|)-|x|$ $\sin (|x|)+|x|$ $\sin (|x|)-|x|$ |
$\sin (|x|)-|x|$ |
We have, $\cos |x|= \begin{cases} \qquad ~~\cos x~~~ , \qquad x \geq 0 \\ \cos (-x)=\cos x, ~~ x<0\end{cases}$ ⇒ cos |x| = cos x for all x ∈ R Similarly, we have, $\sin |x|= \begin{cases}\sin x & x \geq 0 \\ \sin (-x)=-\sin x, & x<0\end{cases}$ Let $f(x)=\cos (x)+|x|$ and $g(x)=\cos |x|-|x|$. Since $\cos x$ is everywhere differentiable and |x| is not differentiable at x = 0. Therefore, f(x) and g(x) are not differentiable at x = 0. Let $u(x)=\sin (|x|)+|x|$. Then, $u(x)= \begin{cases}\sin x+x, & x \geq 0 \\ -\sin x-x, & x<0\end{cases}$ ∴ (LHD at x = 0) = $\left\{\frac{d}{d x}(-\sin x-x)\right\}_{\text {at } x=0}$ ⇒ (LHD at x = 0) = $\{-\cos x-1\}_{\text {at }} x=0=-\cos 0-1=-2$ and, (RHD at x = 0) = $\left\{\frac{d}{d x}(\sin x+x)\right\}_{\text {at } x=0}=\{\cos x+1\}_{\text {at } x=0}=2$ Clearly, (LHD at x = 0) ≠ (RHD at x = 0) So, u(x) is not differentiable at x = 0. Let $v(x)=\sin |x|-|x|$. Then, $v(x)=\left\{\begin{aligned} -\sin x+x, & x<0 \\ \sin x-x, & x \geq 0 \end{aligned}\right. $ ∴ (LHD at x = 0) = $\left\{\frac{d}{d x}(-\sin x+x)\right\}_{\text {at } x=0}$ ⇒ (LHD at x = 0) = $\{-\cos x+1\}_{\text {at } x=0}=-1+1=0$ ∴ (RHD at x = 0) = $\left\{\frac{d}{d x}(\sin x-x)\right\}_{\text {at } x=0}$ ⇒ (RHD at x = 0) = $\{\cos x-1\}_{\text {at }} x=0=\cos 0-1=0$ ∴ (LHD at x = 0) = (RHD at x = 0) So, v(x) is differentiable at x = 0. |