Question

If $\int\limits^x_1 \frac{dt}{|t\, \sqrt{t^2 - 1}} = \frac{\pi}{6}$ , then $x$ can be equal to

• A. $\frac{ 2}{\sqrt{3}}$
• B. $\sqrt{3}$
• C. $2$
• D. None of these

Answer 1

Answer: (A)

$\int\limits_{1}^{x} \frac{d t}{|t| \sqrt{t^{2}-1}}=\frac{\pi}{6}$
$\Rightarrow \left[\sec ^{-1} t\right]_{1}^{x}=\frac{\pi}{6}$
$\Rightarrow \sec ^{-1} x-\sec ^{-1} 1=\frac{\pi}{6}$
$\Rightarrow \sec ^{-1} x- 0=\frac{\pi}{6}$
$\Rightarrow x=\sec \frac{\pi}{6}$
$\Rightarrow x=\frac{2}{\sqrt{3}}$