If $[x]$ denotes the greatest integer not exceeding the number $x$,then $f(x)$ defined by $f(x) = \begin{cases} [x], & \text{if } x < 2 \\ [x]-1, & \text{if } x \geq 2 \end{cases}$ is continuous in the interval.

Let $f(x) = \begin{cases} x+a \sqrt{2} \sin x, & 0 \leq x < \frac{\pi}{4} \\ 2x \cot x+b, & \frac{\pi}{4} \leq x < \frac{\pi}{2} \\ a \cos 2x-b \sin x, & \frac{\pi}{2} \leq x \leq \pi \end{cases}$. If $f(x)$ is continuous for $0 \leq x \leq \pi$,then:

If the function $f(x) = \frac{\log(1 + ax) - \log(1 - bx)}{x}$,$x \neq 0$ is continuous at $x = 0$,then $f(0) = $ . . . . . .

Show that the function defined by $g(x)=x-[x]$ is discontinuous at all integral points. Here $[x]$ denotes the greatest integer less than or equal to $x$.

If $\lim _{x \rightarrow a^{+}} f(x)=p, \lim _{x \rightarrow a^{-}} f(x)=m$ and $f(a)=k$,then which one of the following is true?

Consider $f(x) = [x]|x^3 - 2x^2 - x + 2|$ in $[-\frac{3}{2}, \frac{9}{2}]$. The number of points where $f(x)$ is discontinuous is (where $[.]$ denotes the greatest integer function).