Question

If the middle term of ${(x+\frac{1}{x}{\sin }^{-1}x)}^8$ is $\frac{35{\pi }^4}{8}$, then value of $x$ can be

• A. $\frac{1}{2}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $1$

Answer 1

The correct option is D $1$

Middle term is $T_5$
$T_5={}^8{C}_4\left({\sin }^{-1}\right(x){)}^4$

$=70\left({\sin }^{-1}\right(x){)}^4$

$=\frac{35}{8}{\pi }^4$

Hence, $70\left({\sin }^{-1}\right(x){)}^4=\frac{70}{16}{\pi }^4$

$\Rightarrow \left({\sin }^{-1}\right(x){)}^4=\frac{{\pi }^4}{16}$

$\Rightarrow \left({\sin }^{-1}\right(x)=\pm \frac{\pi }{2}$

$\Rightarrow x=\pm 1$