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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$h(x)=\left\{\begin{array}{lll}\max & \{f(x), g(x)\} & \text { if } x \leq 0, \\ \min & \{f(x), g(x)\} & \text { if } x > 0 .\end{array}\right.$ The number of points at which $h(x)$ is not differentiable is