Question

If the range of the value of the term independent of $x$ in the expansion of ${\{x{\sin }^{-1}\alpha +\frac{{\cos }^{-1}\alpha }{x}\}}^{10},\alpha \in [-1,1]$, is :

• A. $[1,2]$
• B. $(1,2)$
• C. $[-\frac{{}^{10}C{}_5{\pi }^2}{2^5},\frac{{}^{10}C{}_5{\pi }^2}{2^{20}}]$
• D. $[-\frac{{}^{10}C{}_5{\pi }^{10}}{2^5},\frac{{}^{10}C{}_5{\pi }^{10}}{2^{20}}]$

Answer 1

The correct option is D $[-\frac{{}^{10}C{}_5{\pi }^{10}}{2^5},\frac{{}^{10}C{}_5{\pi }^{10}}{2^{20}}]$

The term independent of x, will be the 6th term
$=T_{5+1}$
$=I\left(\alpha \right)$
$={}^{10}{C}_5\left(si{n}^{-1}\right(\alpha ).co{s}^{-1}(\alpha ){)}^5$.
Now
$I\left(\alpha \right)$ is minimum when $\alpha =-1$
Hence $I(\alpha {)}_{minimum}={}^{10}{C}_5.(-\frac{\pi }{2}.\pi {)}^5$
$=-{}^{10}{C}_5\frac{{\pi }^{10}}{2^5}$ ..(i)
And $I\left(\alpha \right)$ is maximum when $\alpha =\frac{1}{\sqrt{2}}$
$I(\alpha {)}_{maximum}={}^{10}{C}_5.(\frac{\pi }{4}.\frac{\pi }{4}{)}^5$
$={}^{10}{C}_5\frac{{\pi }^{10}}{2^{20}}$
Hence
$I\left(\alpha \right)ϵ[-{}^{10}{C}_5\frac{{\pi }^{10}}{2^5},{}^{10}{C}_5\frac{{\pi }^{10}}{2^{20}}]$.