Question : If $x+ \frac{1}{x} =\sqrt{13}$, then $\frac{3x}{{x}^{2} -1}$ equals to:
Option 1: $\sqrt[3]{13}$
Option 2: $\frac{\sqrt{13}}{3}$
Option 3: 1
Option 4: 3
Correct Answer: 1
Solution :
Given:
$⇒x+ \frac{1}{x} =\sqrt{13}$
By squaring both sides,
$⇒({x}+ \frac{1}{x})^{2} =(\sqrt{13})^{2}$
$⇒{x}^{2}+ \frac{1}{x^{2}}+2×x×\frac{1}{x} =13$
$⇒{x}^{2}+ \frac{1}{x^{2}}=13-2 =11$ ---(1)
Also, $({x}- \frac{1}{x})^{2} ={x}^{2}+ \frac{1}{x^{2}}-2×x×\frac{1}{x}$
By using the value of equation (1), we get,
$⇒({x}- \frac{1}{x})^{2} = 11 - 2=9$
$⇒({x}- \frac{1}{x}) = \sqrt{9}=3$
Multiplying by $x$ in the whole expression,
$⇒{x}^{2} -1 = 3x$
$⇒\frac{3x}{{x}^{2} -1} =1$
Hence, the correct answer is 1.
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