If the function $f$ defined on $\left(-\frac{1}{3}, \frac{1}{3}\right)$ by $f(x) = \begin{cases} \frac{1}{x} \log_{e}\left(\frac{1+3x}{1-2x}\right) & x \neq 0 \\ k & x = 0 \end{cases}$ is continuous,then $k$ is equal to

If $f(x) = \begin{cases} x^2, & x \ne 1 \\ 2, & x = 1 \end{cases}$,then which of the following is true?

Discuss the continuity of the function $f$ given by $f(x) = |x|$ at $x = 0$.

If $f(x) = \lim_{n \to \infty} \frac{[x^2] + [(2x)^2] + [(3x)^2] + \cdots + [(nx)^2]}{n^3}$,then $f(x)$ is (Where $[\cdot]$ is the Greatest Integer Function).

Let $\alpha \in R$ be such that the function $f(x) = \begin{cases} \frac{\cos^{-1}(1-\{x\}^2) \sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, & x \neq 0 \\ \alpha, & x=0 \end{cases}$ is continuous at $x=0$,where $\{x\} = x - [x]$ and $[x]$ is the greatest integer less than or equal to $x$. Then:

The function $f(x) = \begin{cases} x, & \text{if } 0 \le x \le 1 \\ 1, & \text{if } 1 < x \le 2 \end{cases}$ is