Question

If the middle term in the binomial expansion of ${\left(\frac{1}{x}+x\sin x\right)}^{10}$ is $\frac{63}{8}$, then find the value of $6{\sin }^{2}x+\sin x-2$

Answer 1

Answer: 0

Here, $n=10$, which is even.
Middle term $={\left(\frac{10}{2}+1\right)}^{\text{th }}$ term $=6^{\text{th }}$ term
$T_6={}^{10}{C}_5{\left(\frac{1}{x}\right)}^5(x\sin x{)}^5$
$\Rightarrow \frac{63}{8}=252(\sin x{)}^5$
$\Rightarrow (\sin x{)}^5=\frac{1}{32}$
$\Rightarrow (\sin x{)}^5={\left(\frac{1}{2}\right)}^5$
$\Rightarrow \sin x=\frac{1}{2}$
$\Rightarrow 2\sin x-1=0$
$\Rightarrow 6{\sin }^{2}x+\sin x-2$
$=(2\sin x-1)(3\sin x+2)$
$=0$