Q93 / 180 · COMEDK 2018 · Mathematics

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

  • A
    $\cos (|x|)+|x|$
  • B
    $\cos (|x|)-|x|$
  • C
    $\sin (|x|)+|x|$
  • Correct answer
    D
    $\sin (|x|)-|x|$

Solution

(a) We have, $f(x)=\cos |x|+|x|$ $=\left\{\begin{array}{cc}\cos (-x)+(-x), & x < 0 \\ \cos (x)+x, & x \geq 0\end{array}\right.$ $= \begin{cases}\cos x-x, & x < 0 \\ \cos x+x, & x \geq 0\end{cases}$ $\therefore f^{\prime}(x)= \begin{cases}-\sin x-1, & x < 0 \\ -\sin x+1, & x \geq 0\end{cases}$ $\therefore f^{\prime}\left(0^{-}\right)=-1$ and $f^{\prime}\left(0^{+}\right)=1$ $\Rightarrow f^{\prime}\left(0^{-}\right) \neq f^{\prime}\left(0^{+}\right)$ So, it is not differentiable at $x=0$. (b) We have, $$ \begin{aligned} f(x) &=\cos |x|-|x| \\ &=\left\{\begin{array}{rr} \cos (-x)-(-x), & x < 0 \\ \cos (x)-x, & x \geq 0 \end{array}\right.\\ &= \begin{cases}\cos x+x, & x < 0 \\ \cos x-x, & x \geq 0\end{cases} \\ \therefore f^{\prime}(x) &= \begin{cases}-\sin x+1, & x < 0 \\ -\sin x-1, & x \geq 0\end{cases} \\ \therefore f^{\prime}\left(0^{-}\right) &=1 \text { and } f^{\prime}\left(0^{+}\right)=-1 \end{aligned} $$ So, it is not differentiable at $x=0$.

Hence, the correct answer is option: 'D'.

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