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. $2 n \pi+\frac{\pi}{6}$
• 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)^{t h}=6^{t h} $
$\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} $
$\Rightarrow \sin x =\sin \pi / 6$
$\therefore x=n \pi+(-1)^n \frac{\pi}{6}$