Let $f:R \to R$ be a function defined by $f(x) = \text{Min}\{x + 1, |x| + 1\}$. Then which of the following is true?

If $f(x)=|x|+|\sin x|$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,then its left hand derivative at $x=0$ is

Consider the following statements.
$(a)$ If a function is differentiable at a point $p$ then it is not continuous at $p$.
$(b)$ If a function is not continuous at $x = a$, then it is not differentiable at $x = a$.
$(c)$ If $f(x) = |x|$ then $f(x)$ is not differentiable but continuous on $R$.
$(d)$ If $f(x) = x - [x]$, then $f'(1) = 1$.
Which of the above statements are (is) correct?

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

Suppose $f: R \rightarrow R$ is given by $f(x) = \begin{cases} 1, & \text{if } x=1 \\ e^{(x^{10}-1)} + (x-1)^2 \sin \frac{1}{x-1}, & \text{if } x \neq 1 \end{cases}$. Then:

The function $f(x) = \begin{cases} 2x + 1, & x \in \mathbb{Q} \\ x^2 - 2x + 5, & x \notin \mathbb{Q} \end{cases}$ is