The function $f(x) = [x]^2 - [x^2]$ (where $[x]$ is the greatest integer less than or equal to $x$) is discontinuous at:

Let $f(x) = \begin{cases} 0, & \text{if } -1 \leq x < 0 \\ 1, & \text{if } x = 0 \\ 2, & \text{if } 0 < x \leq 1 \end{cases}$ and let $F(x) = \int_{-1}^{x} f(t) \, dt, -1 \leq x \leq 1$. Then:

Let $f(x) = \begin{cases} 0, & x=0 \\ 2-x, & 0 < x < 1 \\ 2, & x=1 \\ \frac{1}{2}-x, & 1 < x < 2 \\ \frac{-3}{2}, & x \geq 2 \end{cases}$ then which of the following is true?

Let $f(x) = \begin{cases} \frac{\tan^2 \{x\}}{x^2 - [x]^2} & \text{for } x > 0 \\ 1 & \text{for } x = 0 \\ \sqrt{\{x\} \cot \{x\}} & \text{for } x < 0 \end{cases}$ where $[x]$ is the greatest integer function and $\{x\}$ is the fractional part function of $x$,then:

If $f(x) = \frac{3^{x+3} - 3^{-x} - 2}{\tan x \cdot \log(1+x)}$ for $x \neq 0$, is continuous at $x = 0$, then the value of $f(0)$ is equal to ...

If function $f(x) = \begin{cases} x, & \text{if } x \text{ is rational} \\ 1 - x, & \text{if } x \text{ is irrational} \end{cases}$,then $f(x)$ is continuous at how many points?