If $f(x) = [x]$,where $[x]$ is the greatest integer not greater than $x$,then $f^{\prime}(1^{+}) = \dots$.

The function $g(x) = \begin{cases} x + b, & x < 0 \\ \cos x, & x \geqslant 0 \end{cases}$ can be made differentiable at $x = 0$.

Let $f(x) = \begin{cases} -1, & -2 \le x < 0 \\ x^2 - 1, & 0 \le x \le 2 \end{cases}$ and $g(x) = |f(x)| + f(|x|)$. Then,in the interval $(-2, 2)$,$g$ is

If $f(x) = \text{sgn}(x^3)$,then

Let $f(x) = \operatorname{Max}\{\cos x, \sin x, 0\}$. If the number of points at which $f(x)$ is not differentiable in $(0, 2024 \pi)$ is $1012 k$,then $k =$

Let $f : (-1, 1) \to \mathbb{R}$ be a function defined by $f(x) = \min\{-|x|, -\sqrt{1 - x^2}\}$. If $K$ is the set of all points at which $f$ is not differentiable,then $K$ has exactly