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

Solution (By Examveda Team)

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