Question

If the middle term in binomial expression of $(\frac{1}{x}+x\sin x{)}^{10}$ is equal to $\frac{63}{8}$, find the value of x.

Answer 1

In the binomial expansion of $(\frac{1}{x}+x\sin x{)}^{10},(\frac{10}{2}+1{)}^{th}$ i.e., 6th term is the middle term.

It is given that :

$T_6=\frac{63}{8}$

${10}_{C_5}\left(\frac{1}{x}{)}^{10-5}\right(x\sin x{)}^5=\frac{63}{8}$

$\frac{10!}{5!5!}(\sin x{)}^5=\frac{63}{8}$

$(\sin x{)}^5=(\frac{1}{2}{)}^5$

$\sin x=\frac{1}{2}=\sin \frac{\pi }{6}$

$x=n\pi +(-1{)}^n\frac{\pi }{6},n\in Z$