If $f$ is a continuous real-valued function defined on a closed interval $[a, b]$,then the range of the function is . . . . . .

Let $f:[-1,2] \rightarrow R$ be defined by $f(x)=[x^2-3]$ where $[\cdot]$ denotes the greatest integer function. The number of points of discontinuity for the function $f$ in the interval $(-1,2)$ is:

Let $f(x) = \begin{cases} \frac{\sin \pi x}{5x}, & x \ne 0 \\ k, & x = 0 \end{cases}$. If $f(x)$ is continuous at $x = 0$,then $k =$

Let $f(x) = \begin{cases} 1 + 6x - 3x^2, & x \leq 1 \\ x + \log_2(b^2 + 7), & x > 1 \end{cases}$. Then the set of all possible values of $b$ such that $f(1)$ is the maximum value of $f(x)$ is

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 $\lim _{x \rightarrow 0} \frac{\sin ^2 ax+x \tan bx}{x^2}=$

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, $\lim _{x \rightarrow a} \frac{x^2-9}{x^3-5 x^2+9 x-9} =$