The function $f(x) = [x]^2 - [x^2]$ (where $[x]$ is the greatest integer less than or equal to $x$) is discontinuous at:

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 $\tan \theta =$

If $a$ is the point of discontinuity of the function $f(x) = \begin{cases} \cos 2 x, & \text{for } -\infty < x < 0 \\ e^{3 x}, & \text{for } 0 \leq x < 3 \\ x^2-4 x+3, & \text{for } 3 \leq x \leq 6 \\ \frac{\log (15 x-89)}{x-6}, & \text{for } x>6 \end{cases}$ Then, $\lim _{x \rightarrow a} \frac{x^2-9}{x^3-5 x^2+9 x-9} =$

If a function $f(x)$ defined on $[a, b]$ is discontinuous at $x=\alpha \in(a, b)$,then

The value of $f(0)$ so that the function $f(x) = \frac{2^x - 2^{-x}}{x}$ for $x \neq 0$ is continuous at $x = 0$ is:

Let $f: R \rightarrow R$ be a differentiable function such that $f(3)=3$ and $f^{\prime}(3)=\frac{1}{27}$. If $g(x)=\begin{cases} \int_3^{f(x)} \frac{3t^2}{x-3} dt & \text{if } x \neq 3 \\ K & \text{if } x=3 \end{cases}$ is continuous at $x=3$,then $K=$