Consider the function $f(x) = (x - 2) \log_e x$. Then the equation $x \log_e x = 2 - x$

The function $f(x) = \cot x$ is discontinuous at every point of the set

Statement-$1$: The equation $x \log x = 2 - x$ is satisfied by at least one value of $x$ lying between $1$ and $2$.
Statement-$2$: The function $f(x) = x \log x$ is an increasing function in $[1, 2]$ and $g(x) = 2 - x$ is a decreasing function in $[1, 2]$,and the graphs represented by these functions intersect at a point in $[1, 2]$.

If $f(x) = \begin{cases} \frac{\sqrt{1+mx} - \sqrt{1-mx}}{x}, & -1 \le x < 0 \\ \frac{2x+1}{x-2}, & 0 \le x \le 1 \end{cases}$ is continuous in the interval $[-1, 1]$,then $m$ is equal to:

If the function $f(x) = \frac{\sqrt{1+x}-1}{x}$ is continuous at $x=0$,then $f(0) = $

If the function $f(x) = \frac{2x - \sin^{-1}x}{2x + \tan^{-1}x}, (x \neq 0)$ is continuous at each point of its domain,then the value of $f(0)$ is