NIELIT 2019 Feb Scientist C - Section D: 19

Question

If $x=\dfrac{\sqrt{10}+\sqrt{2}}{2}, \: \: y=\dfrac{\sqrt{10}-\sqrt{2}}{2}$ then the value of $\log _{2}(x^{2}+xy+y^{2})$ is:

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

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.