$f(x) = \left[ \frac{x^2 + 1}{x^2[|x|] + 1} \right]$ is discontinuous at (where $[.]$ denotes the greatest integer function):

If $f(x) = \left(\frac{2^{x}-1}{1-3^{x}}\right)$ for $x \neq 0$ is continuous at $x = 0$,then $f(0) = $

Define $f: R \rightarrow R$ by $f(x) = \begin{cases} (x-a) \frac{e^{\frac{1}{x-a}}-1}{e^{\frac{1}{x-a}}+1}, & x \neq a \\ 0, & x=a \end{cases}$. Then which one of the following is true?

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:

If $\{x\} = x - [x]$ where $[x]$ is the greatest integer $\leq x$ and $\lim_{x \rightarrow 0^{-}} \frac{\cos^{-1}(1-\{x\}^2) \sin^{-1}(1-\{x\})}{\{x\}-\{x\}^4} = \theta$,then $\tan \theta =$

Find the relationship between $a$ and $b$ so that the function $f$ defined by $f(x) = \begin{cases} ax + 1, & \text{if } x \le 3 \\ bx + 3, & \text{if } x > 3 \end{cases}$ is continuous at $x = 3$.