Question : If $x=\sqrt3+\frac{1}{\sqrt3}$, then the value of $(x-\frac{\sqrt{126}}{\sqrt{42}})(x-\frac{1}{x-\frac{2\sqrt3}{3}})$ is:
Option 1: $\frac{5\sqrt3}{6}$
Option 2: $\frac{2\sqrt3}{3}$
Option 3: $\frac{5}{6}$
Option 4: $\frac{2}{3}$
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Correct Answer: $\frac{5}{6}$
Solution :
$x=\sqrt{3}+\frac{1}{\sqrt{3}}$
⇒ $x=\frac{3+1}{\sqrt{3}}$
⇒ $x=\frac{4}{\sqrt{3}}$
Now,
$(x-\frac{\sqrt{126}}{\sqrt{42}} )(x-\frac{1}{x-\frac{2\sqrt{3}}{3}})$
Putting the value of $x$, we get,
= $(\frac{4}{\sqrt{3}}-\frac{\sqrt{126}}{\sqrt{42}})(\frac{4}{\sqrt{3}}-\frac{1}{\frac{4}{\sqrt{3}}-\frac{2\sqrt{3}}{3}})$
= $(\frac{4\sqrt{42}-\sqrt{126}\sqrt{3}}{\sqrt{42}\sqrt{3}} )(\frac{4}{\sqrt{3}}-\frac{1}{\frac{2}{\sqrt{3}}})$
= $(\frac{4\sqrt{42}-3\sqrt{42}}{\sqrt{42}\sqrt{3}})(\frac{4}{\sqrt{3}}-\frac{\sqrt{3}}{2})$
= $(\frac{\sqrt{42}}{\sqrt{42}\sqrt{3}})(\frac{4}{\sqrt{3}}-\frac{\sqrt{3}}{2})$
= $(\frac{1}{\sqrt{3}})(\frac{8-3}{2\sqrt{3}})$
= $\frac{5}{6}$
Hence, the correct answer is $\frac{5}{6}$.
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