Let $f: R \rightarrow R$ be a function given by $f(x) = \begin{cases} \frac{1-\cos 2x}{x^2} & , x < 0 \\ \alpha & , x = 0 \\ \frac{\beta \sqrt{1-\cos x}}{x} & , x > 0 \end{cases}$. If $f$ is continuous at $x = 0$,then $\alpha^2 + \beta^2$ is equal to:

$f(x) = \begin{cases} [x^2] - [-x^2], & x \neq 3 \\ k, & x = 3 \end{cases}$ is continuous at $x = 3$,then $k = $ where $[\cdot]$ is the greatest integer function.

Let $f(x) = \frac{1-\tan x}{4x-\pi}$,where $x \neq \frac{\pi}{4}$ and $x \in [0, \frac{\pi}{2}]$. If $f(x)$ is continuous in $[0, \frac{\pi}{2}]$,then $f(\frac{\pi}{4})$ is:

Let $f: \left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \rightarrow \mathbb{R}$ be defined as
$f(x) = \begin{cases} (1+|\sin x|)^{\frac{3a}{|\sin x|}}, & -\frac{\pi}{4} < x < 0 \\ b, & x = 0 \\ e^{\frac{\cot 4x}{\cot 2x}}, & 0 < x < \frac{\pi}{4} \end{cases}$
If $f$ is continuous at $x = 0$,then the value of $6a + b^2$ is equal to:

The function $f(x) = \frac{x^2 - 1}{x^3 - 1}$ is not defined at $x = 1$. If the function is continuous at $x = 1$,then the value of $f(1)$ will be:

For what value of $\lambda$ is the function defined by $f(x) = \begin{cases} \lambda(x^2 - 2x), & \text{if } x \le 0 \\ 4x + 1, & \text{if } x > 0 \end{cases}$ continuous at $x=0$? What about continuity at $x=1$?