Question : If $x=\frac{1}{x-3},(x>0)$, then the value of $x+\frac{1}{x}$ is:
Option 1: $\sqrt{11}$
Option 2: $\sqrt{17}$
Option 3: $\sqrt{15}$
Option 4: $\sqrt{13}$
Correct Answer: $\sqrt{13}$
Solution :
Given: $x=\frac{1}{x-3},(x>0)$
Or, $x^2-3x-1=0$
Or, $x=\frac{3\pm\sqrt{13}}{2}$
So, $x = \frac{3+\sqrt{13}}{2}$
Now, $x+\frac{1}x = \frac{3+\sqrt{13}}{2}+\frac{2}{3+\sqrt{13}}$
Rationalising,
= $\frac{3+\sqrt{13}}{2}+\frac{2}{3+\sqrt{13}}\times \frac{3-\sqrt{13}}{3-\sqrt{13}}$
= $\frac{3+\sqrt{13}}{2}+2\times \frac{3-\sqrt{13}}{-4}$
= $\sqrt13$
Hence, the correct answer is $\sqrt{13}$.
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