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. $2n\pi \pm \frac{\pi }{6}$
• 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 $T_6=7\frac{7}{8}$
$\Rightarrow {}^{10}{C}_5\frac{1}{x^5}\cdot x^5{\sin }^{5}x=\frac{63}{8}$
$\Rightarrow 252\cdot {\sin }^{5}x=\frac{63}{8}$
$\Rightarrow {\sin }^{5}x=\frac{1}{32}$
$\Rightarrow \sin x=\frac{1}{2}$
$\therefore x=n\pi +(-1{)}^n\frac{\pi }{6}$