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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:

  1. $0$
  2. $1$
  3. $2$
  4. $3$
in Quantitative Aptitude 9.1k points 21 371 778
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1 Answer

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$\textrm{Given that:}$ $x=\frac{\sqrt10+\sqrt2}{2},y=\frac{\sqrt10-\sqrt2}{2}$

$\textrm{first find $x^2+xy+y^2$}$

$\implies$  $x^2+xy+y^2+xy-xy$

$\implies$ $x^2+2xy+y^2-xy$

$\implies$ $(x+y)^2-xy….(1)$

$\textrm{Let find x+y first:}$

$\implies$ $\frac{\sqrt10+\sqrt2}{2}+\frac{\sqrt10-\sqrt2}{2}$

$\implies$ $\frac{2*\sqrt10}{2}$

$\implies$ $x+y=$ $\sqrt10...(2)$

$\textrm{Let find x*y:}$

$\implies$ $\frac{\sqrt10+\sqrt2}{2}*\frac{\sqrt10-\sqrt2}{2}$

$\implies$ $\frac{(10-2)}{4}$

$\implies$ $x*y=$$2...(3)$

$\textrm{Substitute eq (2) &(3) into (1), we get:}$

$\implies$ $(\sqrt{10})^2-2$

$\implies$ $8$

$\textrm{Now find l$og_2(8)$,which is equal to 3}$

$\textrm{Option D is correct.}$
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