Given that: $x=\frac{\sqrt10+\sqrt2}{2},y=\frac{\sqrt10-\sqrt2}{2}$
first find:$x^2+xy+y^2$
$\implies x^2+xy+y^2+xy-xy$
$\implies x^2+2xy+y^2-xy$
$\implies (x+y)^2-xy….(1)$
Let find $x+y$ first:
$\implies \frac{\sqrt10+\sqrt2}{2}+\frac{\sqrt10-\sqrt2}{2}$
$\implies \frac{2*\sqrt10}{2}$
$\implies x+y= \sqrt10….(2)$
Let find $x*y$:
$\implies \frac{\sqrt10+\sqrt2}{2}*\frac{\sqrt10-\sqrt2}{2}$
$\implies \frac{(10-2)}{4}$
$\implies x*y= 2...(3)$
Substitute eq (2) &(3) into (1), we get:
$\implies (\sqrt{10})^2-2$
$\implies 8$
Now find $log_2(8)$,which is equal to $3$
Option (D) is correct.