Q44 / 160 · AP EAMCET 2005 · Mathematics

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

  • Correct answer
    A
    1
  • B
    2
  • C
    3
  • D
    $\frac{1}{2}$

Solution

Given that or $\begin{aligned} x & =\frac{1}{2}\left(\sqrt{3}+\frac{1}{\sqrt{3}}\right) \\ x^2 & =\frac{1}{4}\left(3+\frac{1}{3}+2\right)=\frac{4}{3} \end{aligned}$ Now, $\begin{aligned} \frac{\sqrt{x^2-1}}{x-\sqrt{x^2-1}} & =\frac{\sqrt{x^2-1}}{x-\sqrt{x^2-1}} \times \frac{x+\sqrt{x^2-1}}{x+\sqrt{x^2-1}} \\ & =\frac{x \sqrt{x^2-1}+\left(x^2-1\right)}{1} \\ & =\frac{1}{2}\left(\sqrt{3}+\frac{1}{\sqrt{3}}\right) \sqrt{\frac{4}{3}-1}+\left(\frac{4}{3}-1\right) \\ & =\frac{1}{2}\left(\frac{4}{\sqrt{3}}\right) \frac{1}{\sqrt{3}}+\frac{1}{3} \Rightarrow \frac{2}{3}+\frac{1}{3}=1\end{aligned}$

Hence, the correct answer is option: 'A'.

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