If $f\left(x\right)$ is a continuous function $\mathrm{\forall }x\in R$ and $f\left(x\right)\in (1,\sqrt{30}),\text{and}g\left(x\right)=\left[\frac{f\left(x\right)}{a}\right],$ where$[.]$ denotes the greatest integer function), is continuous $\mathrm{\forall }x\in R$  then the  least positive integral value of a is

Given g(x) is a continuous function ∀x∈R and the range of f(x) is ($1,\sqrt{30}$​) and g(x)=[$\frac{f\left(x\right)}{a}$​] is continuous ∀x∈R

Hence for g(x) to be continuous, $a>\sqrt{30}$

Because if we take any value of a between ($1,\sqrt{30}$) like 3,4,5 then, when f(x) varies from 1 to  $\sqrt{30}$ and due to greatest integer function g(x) become discontinuous at a=3,4,5.

So, for continuity a=6.

Hence, the least value of a is 6.