If $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$,then $x(y+z)+y(z+x)+z(x+y)$ equals to

$x, y, z$ are in $G$.$P$. and $\tan^{-1} x, \tan^{-1} y, \tan^{-1} z$ are in $A$.$P$.,then

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

Let $f:[0, 4\pi] \rightarrow [0, \pi]$ be defined by $f(x) = \cos^{-1}(\cos x)$. The number of points $x \in [0, 4\pi]$ satisfying the equation $f(x) = \frac{10-x}{10}$ is

If $\alpha=3 \sin ^{-1} \frac{6}{11}$ and $\beta=3 \cos ^{-1}\left(\frac{4}{9}\right)$,where the inverse trigonometric functions take only the principal values,then the incorrect option is

If $\alpha$ and $\beta$ (where $\alpha > \beta$) are two zeroes of the equation $3\cos^{-1}\left(x^2 - 5x - \frac{11}{2}\right) = \pi$,then $(\alpha^2 + \beta^3)$ is -