The domain of the function $f(x) = \sqrt{x - 1} + \sqrt{3 - x}$ is ...

The range of the function $f(x) = -\sqrt{-x^2-6x-5}$ is

Find the range of the following function:
$f(x) = 2 - 3x$,where $x \in R$ and $x > 0$.

If $[x]$ represents the greatest integer $\leq x$ and $[\alpha, \beta]$ is the set of all real values of $x$ for which the real function $f(x)=\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{[x]+2}}$ is defined, then $f^2(\alpha+1)+5 f^2(\beta)=$

Let $f$ be a function such that $f(x) = \sum_{r=1}^n [r + \cos(\frac{x}{r})]$,where $[.]$ denotes the greatest integer function and $x \in [0, \pi]$. Then the range of $f(x)$ is:

Let $f(x) = \frac{1}{2} - \tan \left(\frac{\pi x}{2}\right), -1 < x < 1$ and $g(x) = \sqrt{3 + 4x - 4x^2}$, then the domain of $(f + g)$ is