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
$\therefore\left[\sec ^{-1} t\right]_{\sqrt{2}}^{x}=\frac{\pi}{2}$
$\left[\text { As } \int \frac{d x}{x \sqrt{x^{2}-1}}=\sec ^{-1} x\right]$
$\Rightarrow \sec ^{-1} x-\sec ^{-1} \sqrt{2}=\frac{\pi}{2}$
$\Rightarrow \sec ^{-1} x-\frac{\pi}{4}=\frac{\pi}{2}$
$\Rightarrow \sec ^{-1} x=\frac{\pi}{2}+\frac{\pi}{4}$
$\Rightarrow \sec ^{-1} x=\frac{3 \pi}{4}$
$\Rightarrow x=\sec \frac{3 \pi}{4}$
$\Rightarrow x=-\sqrt{2}$
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and $g(x)=\left(x-\frac{1}{2}\right)^{2}, x \in R .$ Then the area (in sq. units) of the region bounded by the curves, $y=f(x)$ and $y=g(x)$ between the lines, $2 \mathrm{x}=1$ and $2 \mathrm{x}=\sqrt{3},$ is