Let $S = \{ x : \cos^{-1} x = \pi + \sin^{-1} x + \sin^{-1}(2x + 1) \}$. Then $\sum_{x \in S} (2x - 1)^2$ is equal to . . . . . .

The mean of numbers $a, b, 8, 5, 10$ is $6$ and their variance is $6.80$. Then $\operatorname{Tan}^{-1} \frac{1}{a} + \operatorname{Tan}^{-1} \frac{1}{b} =$

If $0 \leq x < \frac{3}{4}$, then the number of values of $x$ satisfying the equation $\operatorname{Tan}^{-1}(2x-1) + \operatorname{Tan}^{-1}(2x) = \operatorname{Tan}^{-1}(4x) - \operatorname{Tan}^{-1}(2x+1)$ is:

Let $g(x) = f(x) + f(1-x)$ and $f''(x) > 0$ for $x \in (0, 1)$. If $g$ is decreasing in the interval $(0, \alpha)$ and increasing in the interval $(\alpha, 1)$,then $\tan^{-1}(2\alpha) + \tan^{-1}\left(\frac{1}{\alpha}\right) + \tan^{-1}\left(\frac{\alpha+1}{\alpha}\right)$ is equal to:

Solve $\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2},$ then $x$ is equal to

The number of positive integral solutions of $\tan ^{-1} x+\cos ^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right)=\sin ^{-1}\left(\frac{3}{\sqrt{10}}\right)$ are