Question

If the middle term in the expansion of $\left(\frac{1}{x}+x \,\sin \,x\right)^{10}$ equals to $7 \frac{7}{8}$ then $x$ is equal to $(n \in I)$

• A. $2 n \pi \pm \frac{\pi}{6}$
• B. $n \pi+\frac{\pi}{6}$
• C. $n \pi+(-1)^{n} \frac{\pi}{6}$
• D. $n \pi+(-1)^{n} \frac{5 \pi}{6}$

Answer 1

Answer: (C)

Middle term in the expansion is $\left(\frac{10}{2}+1\right)^{\text {th }}$ i.e., $6^{\text {th }}$ term.
Thus $T_{6}=7 \frac{7}{8} $
$ \Rightarrow { }^{10} C_{5} \frac{1}{x^{5}} \cdot x^{5} \sin ^{5} x=\frac{63}{8}$
$\Rightarrow 252 \cdot \sin ^{5} x=\frac{63}{8}$
$ \Rightarrow \sin ^{5} x=\frac{1}{32}$
$ \Rightarrow \sin x=\frac{1}{2}$
$\therefore x=n \pi+(-1)^{n} \frac{\pi}{6}$