Explanation

(b)
$\because\ T_{r + 1} = \ ^{10}C_{r}{(x\sin^{- 1}\alpha)}^{10 - r}\left( \frac{\cos^{- 1}\alpha}{x} \right)^{r}$
$= \ ^{10}C_{r}{(\sin^{- 1}\alpha)}^{10 - r}{(\cos^{- 1}\alpha)}^{r}x^{10 - 2r}$
$\therefore$ For the term independent of $x$,
$10 - 2r = 0\ \Rightarrow \ r = 5$
$T_{5 + 1} = \ ^{10}C_{5}{(\sin^{- 1}\alpha)}^{5}{(\cos^{- 1}\alpha)}^{5}$
$= \ ^{10}C_{5}{(\sin^{- 1}{\alpha\cos^{- 1}\alpha})}^{5}$
Let $f(\alpha) = \sin^{- 1}{\alpha.\cos^{- 1}\alpha}$
$= \sin^{- 1}{\alpha\left( \frac{\pi}{2} - \sin^{- 1}\alpha \right)}$
Put $\sin^{- 1}{\alpha = t}$
$\therefore f(\alpha) = t\left( \frac{\pi}{2} - t \right)$
$= - \left\{ t^{2} - \frac{\pi}{2}t \right\}$
$= - \left\{ \left( t - \frac{\pi}{4} \right)^{2} - \frac{\pi^{2}}{16} \right\}$
$= \frac{\pi^{2}}{16} - \left( t - \frac{\pi}{4} \right)^{2}$
$\therefore\ f(a) = \frac{\pi^{2}}{16} - \left( \sin^{- 1}{\alpha - \frac{\pi}{4}} \right)^{2}$
Maximum value of $f(\alpha)$ is $\frac{\pi^{2}}{16}$, when $\sin^{- 1}{\alpha = \frac{\pi}{4}}$
Also, $- 1 \leq \alpha \leq 1$
$\therefore\ - \frac{\pi}{2} \leq \sin^{- 1}{\alpha \leq \frac{\pi}{2}}$
Minimum value $f(\alpha) = \frac{\pi^{2}}{16} - \left( - \frac{\pi}{2} - \frac{\pi}{4} \right)^{2} = - \frac{\pi^{2}}{2}$
$\therefore Range\ is\ \left\lbrack \ ^{10}C_{5}\left( - \frac{\pi^{2}}{2} \right)^{5},\ \ ^{10}C_{5}\left( \frac{\pi^{2}}{16} \right)^{5} \right\rbrack$
$ie,\ \left\lbrack - \frac{\ ^{10}C_{5}\pi^{10}}{2^{5}},\ \frac{\ ^{10}C_{5}\pi^{10}}{2^{20}} \right\rbrack$