Question

If $x=\frac{1}{2}\left(\sqrt{7}+\frac{1}{\sqrt{7}}\right)$ then $\frac{\sqrt{x^2-1}}{x-\sqrt{x^2-1}}=$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C

$3$

Explanation for the correct answer:

To find the value of $\frac{\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}}{\mathbf{x}\mathbf{-}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}}$:

Given,

$x=\frac{1}{2}\left(\sqrt{7}+\frac{1}{\sqrt{7}}\right)$

squaring on both sides

$$\begin{gathered} \Rightarrow x^2=\frac{1}{4}{\left(\sqrt{7}+\frac{1}{\sqrt{7}}\right)}^2 \\ \Rightarrow x^2=\frac{1}{4}\left(7+\frac{1}{7}+2\right) \\ \Rightarrow 4x^2=4+5+\frac{1}{7} \\ \Rightarrow 4x^2-4=5+\frac{1}{7} \\ \Rightarrow 4\left(x^2-1\right)=\frac{36}{7} \\ \Rightarrow \left(x^2-1\right)=\frac{36}{28}=\frac{9}{7} \end{gathered}$$

Now, on substituting the above value in the equation $\frac{\sqrt{{\mathrm{x}}^2-1}}{\mathrm{x}-\sqrt{{\mathrm{x}}^2-1}}$

we get,

$\begin{array}{rcl}& =& \frac{\sqrt{{\displaystyle \frac{9}{7}}}}{{\displaystyle \frac{1}{2}}\left(\sqrt{7}+{\displaystyle \frac{1}{\sqrt{7}}}\right)-\sqrt{{\displaystyle \frac{9}{7}}}}\\ & =& \frac{\sqrt{\frac{9}{7}}}{{\displaystyle \frac{8}{2\sqrt{7}}}-{\displaystyle \frac{3}{\sqrt{7}}}}\\ & =& \frac{\frac{3}{\sqrt{7}}}{{\displaystyle \frac{8-6}{2\sqrt{7}}}}\\ & =& \frac{6}{2}\\ & =& 3\end{array}$

Hence, the correct option is C.