If $f(x) = \frac{\ln(e^{x^2} + 2\sqrt{x})}{\sqrt{x}}$ is continuous at $x = 0$,then $f(0)$ must be equal to:

Consider the function $f(x) = \begin{cases} \frac{P(x)}{\sin(x-2)}, & x \neq 2 \\ 7, & x = 2 \end{cases}$ where $P(x)$ is a polynomial such that $P''(x)$ is always a constant and $P(3) = 9$. If $f(x)$ is continuous at $x = 2$,then $P(5)$ is equal to:

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

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 $f(x) = \begin{cases} [x] + [-x], & x \neq 2 \\ \lambda, & x = 2 \end{cases}$ is continuous at $x = 2$,then $\lambda = $ (where $[.]$ denotes the greatest integer function).

If a function $f(x) = \begin{cases} \frac{\tan (\alpha + 1)x + \tan 2x}{x}, & \text{if } x > 0 \\ \beta, & \text{at } x = 0 \\ \frac{\sin 3x - \tan 3x}{x^{3}}, & \text{if } x < 0 \end{cases}$ is continuous at $x = 0$,then $|\alpha| + |\beta| =$