Q45 / 160 · TS EAMCET 2008 · Mathematics
If $x=\frac{1}{2}\left(\sqrt{7}+\frac{1}{\sqrt{7}}\right)$, then $\frac{\sqrt{x^2-1}}{x-\sqrt{x^2-1}}$ is equal to
- A$1$
- B$2$
- Correct answerC$3$
- D$4$
Solution
Given that,
$\begin{aligned}
x & =\frac{1}{2}\left(\sqrt{7}+\frac{1}{\sqrt{7}}\right) \\
\Rightarrow \quad x^2 & =\frac{1}{4}\left(7+\frac{1}{7}+2\right)=\frac{1}{4}\left(\frac{64}{7}\right) \\
& =\frac{16}{7}
\end{aligned}$
Now,
$\begin{aligned}
\frac{\sqrt{x^2-1}}{x-\sqrt{x^2-1}} & =\frac{\sqrt{x^2-1}}{x-\sqrt{x^2-1}} \times \frac{\left(x+\sqrt{x^2-1}\right)}{\left(x+\sqrt{x^2-1}\right)} \\
& =\frac{x \sqrt{x^2-1}+x^2-1}{1} \\
& =\frac{1}{2}\left(\sqrt{7}+\frac{1}{\sqrt{7}}\right) \sqrt{\frac{16}{7}-1}+\frac{16}{7}-1 \\
& =\frac{1}{2}\left(\sqrt{7}+\frac{1}{\sqrt{7}}\right) \times \frac{3}{\sqrt{7}}+\frac{9}{7} \\
& =\frac{1}{2}\left(3+\frac{3}{7}\right)+\frac{9}{7} \\
& =\frac{1}{2}\left(\frac{24}{7}\right)+\frac{9}{7}=\frac{21}{7} \\
& =3
\end{aligned}$
Hence, the correct answer is option: 'C'.