If $$f(x)=[8 x]-3$$, where $$[x]$$ is greatest integer function of $$x$$, then $$f(\pi)=$$ (where $$\pi=3,14$$)

The domain of the function $$f(x)=\sqrt{x-1}+\sqrt{6-x}$$ is

If f(x) = 3[x] + 5{x + 1}, where [x] is greatest integer function of x and {x} is fractional part function of x, then f($$-$$1.32) =

Let $$A=[a, b, c, d], B=[1,2,3]$$. Relation $$R_1, R_2, R_3, R_4$$ are as follows :

$$\begin{aligned} & R_1=[(\mathrm{a}, 1),(\mathrm{b}, 2),(\mathrm{c}, 1),(\mathrm{d}, 2)] \\ & \mathrm{R}_2=[(\mathrm{a}, 1),(\mathrm{b}, 1),(\mathrm{c}, 1),(\mathrm{d}, 1)] \\ & \mathrm{R}_3=[(\mathrm{a}, 2),(\mathrm{b}, 3),(\mathrm{c}, 2),(\mathrm{d}, 2)] \\ & \mathrm{R}_4=[(\mathrm{a}, 1),(\mathrm{b}, 2),(\mathrm{a}, 2),(\mathrm{d}, 3)] \text {, then } \end{aligned}$$