If $f(x) = \begin{cases} \frac{\sqrt{1+px} - \sqrt{1-px}}{x}, & -1 \leq x < 0 \\ \frac{2x+1}{x-2}, & 0 \leq x \leq 1 \end{cases}$ is continuous in $[-1, 1]$, then $p = $

Let $f, g: R \rightarrow R$ be functions defined by $f(x) = \begin{cases} x \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}$ and $g(x) = x f(x)$. Consider the following statements: $(i)$ $f(x)$ is continuous at $x = 0$ but not differentiable at $x = 0$. $(ii)$ $g(x)$ is differentiable at $x = 0$, but $g'(x)$ is not continuous at $x = 0$. Then, which one of the following is true?

If $f(x) = \begin{cases} \frac{|x-2|}{x-2}, & x \neq 2 \\ 1, & x = 2 \end{cases}$,then which of the following statements is true?

Let $f(x) = \begin{cases} \frac{2^x + 2^{3-x} - 6}{\sqrt{2^{-x}} - 2^{1-x}} & \text{if } x > 2 \\ \frac{x^2 - 4}{x - \sqrt{3x - 2}} & \text{if } x < 2 \end{cases}$. Determine the nature of the function at $x = 2$.

If the function defined by $f(x) = \frac{\log (1+x)^{1+x}}{x^2} - \frac{1}{x}, x \neq 0$ is continuous at $x=0$,then $6 f(0)$ is equal to

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.