If $x + \frac{1}{x} = \sqrt{13},$ $x + \frac{1}{x} = \sqrt {13} {\text{,}}$ then $\frac{3 x}{(x^{2} - 1)}$ $\frac{{3x}}{{\left( {{x^2} - 1} \right)}}$ equal to?

1 $3 \sqrt{13}$ ${\text{3}}\sqrt {13}$
2 $\frac{\sqrt{13}}{13}$ $\frac{{\sqrt {13} }}{{13}}$
31
43

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given, x + \frac{1}{x} = \sqrt{13} \\ then , \frac{3 x}{(x^{2} - 1)} \\ = \frac{3}{x - \frac{1}{x}} . . . . . . . . . . . . . (i) \\ Now, x + \frac{1}{x} = \sqrt{13} \\ On squaring both side \\ = x^{2} + \frac{1}{x^{2}} \\ = 13 - 2 \\ = 11 \\ = x^{2} + \frac{1}{x^{2}} - 2 \\ = 11 - 2 \\ = 9 \\ \Rightarrow {(x - \frac{1}{x})}^{2} = 9 \\ \Rightarrow {(x - \frac{1}{x})}^{2} = 3^{2} \\ \Rightarrow x - \frac{1}{x} = 3 \\ Put this value in equation (i) \\ \Rightarrow \frac{3}{x - \frac{1}{x}} \\ \Rightarrow \frac{3}{3} \\ \Rightarrow 1 \end{aligned}

$\eqalign{ & {\text{Given, }}x + \frac{1}{x} = \sqrt {13} {\text{ }} \cr & {\text{then ,}}\frac{{3x}}{{\left( {{x^2} - 1} \right)}} \cr & = \frac{3}{{x - \frac{1}{x}}}\,.............(i) \cr & {\text{Now, }}x + \frac{1}{x} = \sqrt {13} \cr & {\text{On squaring both side}} \cr & = {x^2} + \frac{1}{{{x^2}}} \cr & = 13 - 2 \cr & = 11 \cr & = {x^2} + \frac{1}{{{x^2}}} - 2 \cr & = 11 - 2 \cr & = 9 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} = 9 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} = {3^2} \cr & \Rightarrow x - \frac{1}{x} = 3 \cr & {\text{Put this value in equation (i)}} \cr & \Rightarrow \frac{3}{{x - \frac{1}{x}}} \cr & \Rightarrow \frac{3}{3} \cr & \Rightarrow 1 \cr}$