MCQ
The solution for $x$ ofthe equation $\mathop \smallint \limits_{\sqrt 2 }^x \frac{{dt}}{{t\sqrt {{t^2} - 1} }} = \frac{\pi }{2}$ is
  • A
    $\frac{{\sqrt 3 }}{2}$
  • B
    $\;2\sqrt 2 $
  • C
    $2$
  • none of these

Answer

Correct option: D.
none of these
d
$\int_{\sqrt{2}}^{x} \frac{d t}{t \sqrt{t^{2}-1}}=\frac{\pi}{2}$

$\therefore\left[\sec ^{-1} t\right]_{\sqrt{2}}^{x}=\frac{\pi}{2}$

$\left[\text { As } \int \frac{d x}{x \sqrt{x^{2}-1}}=\sec ^{-1} x\right]$

$\Rightarrow \sec ^{-1} x-\sec ^{-1} \sqrt{2}=\frac{\pi}{2}$

$\Rightarrow \sec ^{-1} x-\frac{\pi}{4}=\frac{\pi}{2}$

$\Rightarrow \sec ^{-1} x=\frac{\pi}{2}+\frac{\pi}{4}$

$\Rightarrow \sec ^{-1} x=\frac{3 \pi}{4}$

$\Rightarrow x=\sec \frac{3 \pi}{4}$

$\Rightarrow x=-\sqrt{2}$

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