If f(x) is a continuous function $\forall ~x \in R$ and the range of $f(x)=(2, \sqrt{26})$ and $g(x)$ $=\left[\frac{f(x)}{a}\right]$ is continuous $\forall ~x \in R$ ([.] denotes the greatest integer function), then the least positive integral value of a is:

6

Since $g(x)$ is continuous $\forall ~x \in R, g(x)$ should be constant.

Since $f(x) \in(2, \sqrt{26}), a \geq \sqrt{26}$, (as $\left[\frac{f(x)}{\sqrt{26}}\right]=0 \forall ~x \in R$).

So least integral value of a is 6.

Hence (3) is the correct answer.