The function $f(x) = [x]$,where $[x]$ denotes the greatest integer not greater than $x$,is

Let $f(x) = \begin{cases} x+1, & -1 \leq x \leq 0 \\ -x, & 0 < x \leq 1 \end{cases}$. Which of the following statements is true?

Let $f, g: R \to R$ be two functions defined by $f(x) = \begin{cases} x \sin \left( \frac{1}{x} \right), & x \ne 0 \\ 0, & x = 0 \end{cases}$ and $g(x) = x f(x)$.
Statement $I$: $f$ is a continuous function at $x = 0$.
Statement $II$: $g$ is a differentiable function at $x = 0$.

The function $f(x) = \begin{cases} x + 2, & 1 \le x \le 2 \\ 4, & x = 2 \\ 3x - 2, & x > 2 \end{cases}$ is continuous at

If $f(x) = \frac{x+x^2+x^3+\ldots+x^{n}-n}{x-1}$ for $x \neq 1$ is continuous at $x=1$,then $f(1) =$

The function $f(x) = \frac{\log(1 + ax) - \log(1 - bx)}{x}$ is not defined at $x = 0$. The value which should be assigned to $f$ at $x = 0$ so that it is continuous at $x = 0$ is: