Question

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

• A. $2n\pi +\frac{\pi }{6}$
• B. $n\pi +\frac{\pi }{6}$
• C. $n\pi +(-1{)}^n\frac{\pi }{6}$
• D. $n\pi +(-1{)}^n\frac{\pi }{3}$

Answer 1

Answer: (C)

${\left(\frac{1}{x}+x\sin x\right)}^{10}$
Here, $n=10$ (even)
$\Rightarrow$ Middle term $={\left(\frac{10}{2}+1\right)}^{th}=6^{th}$
$\Rightarrow T_6={}^{10}{C}_5{\left(\frac{1}{x}\right)}^{10-5}(x\sin x{)}^5$
$\Rightarrow 252(\sin x{)}^5=\frac{63}{8}$
$\Rightarrow \sin x{)}^5=\frac{1}{32}$
$\Rightarrow \sin x=\frac{1}{2}$
$\Rightarrow \sin x=\sin \pi /6$
$\therefore x=n\pi +(-1{)}^n\frac{\pi }{6}$