The range of values of the term independent of $x$in the expansion of ${\left(x{\sin }^{-1}\alpha +\frac{{\cos }^{-1}\alpha }{x}\right)}^{10},\alpha \in [-1,1]$, is

The general term in the expansion of ${\left(x{\sin }^{-1}\alpha +\frac{{\cos }^{-1}\alpha }{x}\right)}^{\mathrm{i}0}$ is given by

$T_{r+1}{=}^{10}{C}_r{\left(x{\sin }^{-1}\alpha \right)}^{10-r}{\left(\frac{{\cos }^{-1}\alpha }{x}\right)}^r$

$\Rightarrow T_{r+1}{=}^{10}{C}_r{\left({\sin }^{-1}\alpha \right)}^{10-r}{\left({\cos }^{-1}\alpha \right)}^r{x}^{10-2r}$    …(i)

This will be independent of $x,$ if 

$10-2r=0\Rightarrow r=5$

$\begin{array}{l}{T}_6{=}^{10}{C}_5{\left({\sin }^{-1}\alpha {\cos }^{-1}\alpha \right)}^5\\ \Rightarrow T_6{=}^{10}{C}_5{\left\{{\sin }^{-1}\alpha \left(\frac{\pi }{2}-{\sin }^{-1}\alpha \right)\right\}}^5\\ \Rightarrow T_6{=}^{10}{C}_5{\left\{\frac{\pi }{2}{\sin }^{-1}\alpha -{\left({\sin }^{-1}\alpha \right)}^2\right\}}^5\\ \Rightarrow T_6{=}^{10}{C}_5{\left\{\frac{{\pi }^2}{16}-{\left(\frac{\pi }{4}-{\sin }^{-1}\alpha \right)}^2\right\}}^5\end{array}$

Now,

$\begin{array}{l}-\frac{\pi }{2}\le {\sin }^{-1}\alpha \le \frac{\pi }{2}\\ \Rightarrow -\frac{\pi }{2}\le -{\sin }^{-1}\alpha \le \frac{\pi }{2}\\ =-\frac{\pi }{4}\le \left(\frac{\pi }{4}-{\sin }^{-1}\alpha \right)\le \frac{3\pi }{4}\\ \Rightarrow 0\le {\left(\frac{\pi }{4}-{\sin }^{-1}\alpha \right)}^2\le \frac{9{\pi }^2}{16}\\ \Rightarrow -\frac{9{\pi }^2}{16}\le -{\left(\frac{\pi }{4}-{\sin }^{-1}\alpha \right)}^2\le 0\\ \Rightarrow -\frac{{\pi }^2}{2}\le \frac{{\pi }^2}{16}-{\left(\frac{\pi }{4}-{\sin }^{-1}\alpha \right)}^2\le \frac{{\pi }^2}{16}\\ \Rightarrow {-}^{10}{C}_5{\left(\frac{{\pi }^2}{2}\right)}^5{\le }^{10}{C}_5{\left\{\frac{{\pi }^2}{16}-{\left(\frac{\pi }{4}-{\sin }^{-1}\alpha \right)}^2\right\}}^5\\ {\le }^{10}{C}_5{\left(\frac{{\pi }^2}{16}\right)}^5\end{array}$

$\Rightarrow -\frac{{ }^{10}{C}_5{\pi }^{10}}{2^5}\le T_6\le {10}_{C_5}\frac{{\pi }^{10}}{2^{20}}$