Given $f(x) = \begin{cases} 1 + x & x < 0 \\ 2 - 3x & x \geq 0 \end{cases}$,find the critical point $x = \dots \dots$.

The function given by $y = ||x| - 1|$ is differentiable for all real numbers except the points

If $f(x) = \begin{cases} 1, & x < 0 \\ 1 + \sin x, & 0 \le x < \frac{\pi}{2} \end{cases}$,then $f'(0) = $

Let $f(x) = x|x|$ and $g(x) = \sin x$.
Statement-$1$: $gof$ is differentiable at $x=0$ and its derivative is continuous at that point.
Statement-$2$: $gof$ is twice differentiable at $x=0$.

Suppose that $f(x)$ is a differentiable function such that $f^{\prime}(x)$ is continuous, $f^{\prime}(0)=1$ and $f^{\prime \prime}(0)$ does not exist. Let $g(x)=x f^{\prime}(x)$. Then,

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