Given x>0 but it has not mentioned x is integer or fraction

Let consider x = 4

now $\sqrt{4}$ and $\frac{\sqrt{4}}{4}$

or 2 > $\frac{1}{2}$ - This staisfy 1st equation and also they are reciprocal.

Now let conside x = $\frac{1}{4}$

Therefore $\sqrt{x}$ and $\frac{\sqrt{x}}{x}$

$\sqrt{\frac{1}{4}}$ and $\frac{\sqrt{\frac{1}{4}}}{\frac{1}{4}}$

= $\frac{1}{2}$ and $\frac{1}{2} *4$

= 1/2 and 2

i.e $\frac{1}{2}$< 2 so this satisfy the second equation i.e$\sqrt{x}$ is less than$\frac{\sqrt{x}}{x}$

Now if we consider again x= 1/4

Then $\sqrt{x}$ and $x$

= $\sqrt{\frac{1}{4}}$ and $\frac{1}{4}$

= $\frac{1}{2}$ > $\frac{1}{4}$ - This satidfy our third equation.

Hence A,B,C are all correct