If $\tan ^{-1}(x+1)+\tan ^{-1} x+\tan ^{-1}(x-1)=\tan ^{-1} 3$, then for $x<0$ the value of $500 x^4+270 x^2+997=$

Let $\mathrm{f}: \mathbb{R}-\{2\} \rightarrow \mathbb{R}-\{1\}$ defined by $\mathrm{f}(x)=\frac{x-3}{x-2}$ and $\mathrm{g}: \mathbb{R} \rightarrow \mathbb{R}$ defined by $\mathrm{g}(x)=3 x-2$, then sum of all values of $x$ for which $\mathrm{f}^{-1}(x)+\mathrm{g}^{-1}(x)=\frac{19}{6}$ is

There are 11 points in a plane of which 5 points are collinear. Then the total number of distinct quadrilaterals with vertices at these points is

$\int \frac{x^4 \cos \left(\tan ^{-1} x^5\right)}{1+x^{10}} d x$ equals