Question : If $a=\frac{1}{a-\sqrt{6}}$ and $(a>0)$, then the value of $\left(a+\frac{1}{a}\right)$ is:
Option 1: $\sqrt{6}$
Option 2: $\sqrt{10}$
Option 3: $\sqrt{15}$
Option 4: $\sqrt{7}$
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Correct Answer: $\sqrt{10}$
Solution :
$a=\frac{1}{a-\sqrt{6}}(x>0)$
$⇒a^2-a\sqrt 6-1=0$
$⇒a^2-1=a\sqrt 6$
Multiplying both sides by $\frac{1}{a}$, we get,
$⇒a-\frac{1}{a}=\sqrt 6$
Squaring both sides, we get,
$⇒(a-\frac{1}{a})^2=(\sqrt 6)^2$
$⇒a^2+\frac{1}{a^2}-2=6$
Adding 4 to both sides, we get,
$⇒a^2+\frac{1}{a^2}+2=6+4$
$⇒(a+\frac{1}{a})^2=10$
$\therefore a+\frac{1}{a}=\sqrt{10}$
Hence, the correct answer is $\sqrt{10}$.
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