Which one of the following functions is continuous everywhere in its domain but has at least one point where it is not differentiable?

If $f(x) = \begin{cases} x[x], & 0 \le x < 2 \\ (x-1)[x], & 2 \le x \le 4 \end{cases}$,where $[.]$ denotes the greatest integer function,then:

Let $f(x) = \begin{cases} a \sin(x + b) & x \ge 0 \\ 6x^7 - x + 1 & x < 0 \end{cases}$ be differentiable for all real $x$. If $a \in \mathbb{R}$ and $b \in [0, 2\pi]$,then the number of ordered pairs $(a, b)$ is:

Assertion $(A)$: $f(x) = |x|$ is differentiable at $x = a \neq 0$ and continuous but not differentiable at $x = 0$.
Reason $(R)$: If a function is differentiable at a point,then it is continuous at the point. But the converse is not true.

Let the functions $f, g$ and $h$ be defined as follows:
$f(x) = \begin{cases} x \sin \left( \frac{1}{x} \right) & \text{for } -1 \le x \le 1, x \ne 0 \\ 0 & \text{for } x = 0 \end{cases}$
$g(x) = \begin{cases} x^2 \sin \left( \frac{1}{x} \right) & \text{for } -1 \le x \le 1, x \ne 0 \\ 0 & \text{for } x = 0 \end{cases}$
$h(x) = |x|^3$ for $-1 \le x \le 1$.
Which of these functions are differentiable at $x = 0$?

If $f$ is a real-valued differentiable function satisfying $|f(x) - f(y)| \le (x - y)^2$ for all $x, y \in R$ and $f(0) = 0$,then $f(1)$ is equal to: