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

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