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
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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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