If $f(x)= \begin{cases} \frac{x-[x]}{x-2}, & x>2 \\ b, & x=2 \\ \frac{|x^2-x-2|}{a(2+x-x^2)}, & -1 < x \leq 2 \\ 2a-b, & x \leq -1 \end{cases}$ is continuous on $R$,then $\lim _{x \rightarrow 0} \frac{\sin ^2 ax+x \tan bx}{x^2}=$

Consider $f(x) = \left[ \frac{2(\sin x - \sin^3 x) + |\sin x - \sin^3 x|}{2(\sin x - \sin^3 x) - |\sin x - \sin^3 x|} \right]$ for $x \in (0, \pi), x \neq \frac{\pi}{2}$,and $f(\frac{\pi}{2}) = 3$,where $[ \cdot ]$ denotes the greatest integer function. Then:

If $f(x) = \log(\sec^2 x)^{\cot^2 x}$ for $x \neq 0$ and $f(x) = K+1$ for $x=0$ is continuous at $x=0$,then the value of $K$ is

Let the function $f(x)$ be defined as: $f(x) = \begin{cases} [\tan(\frac{\pi}{4} + x)]^{\frac{1}{x}}, & x \neq 0 \\ k, & x = 0 \end{cases}$. If $f(x)$ is continuous at $x = 0$, then the value of $k$ is...

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?

If $[.]$ denotes the greatest integer function,then $f(x) = [x]^2 - [x^2]$ is discontinuous at