If the middle term of

Question:

If the middle term of $\left(\frac{1}{x}+x \sin x\right)^{10}$ is equal to $7 \frac{7}{8}$, then the 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) $n \pi+(-1)^{n} \frac{\pi}{3}$

Solution:

Given middle term of $\left(\frac{1}{x}+x \sin x\right)^{10}$ is $7 \frac{7}{8}$

Since n = 10

i.e. middle term is $\left(\frac{n}{2}+1\right)^{\text {th }}$ term is 6th term.

$\therefore T_{6}=T_{5+1}={ }^{10} C_{5}\left(\frac{1}{x}\right)^{10-5}(x \sin x)^{5}$

i. e. $\frac{63}{8}={ }^{10} C_{5} \frac{1}{x^{5}} \times x^{5}(\sin x)^{5}$

i. e. $\frac{63}{8}=\frac{10 !}{5 ! 5 !} \sin ^{5} x$

i. e. $\frac{63}{8}=252 \sin ^{5} x$

i.e. $\sin ^{5} x=\frac{1}{3^{2}}=\left(\frac{1}{2}\right)^{5}$

i. e. $\sin x=\frac{1}{2}$

i. e. $x=n \pi+(-1)^{n} \frac{\pi}{6}$

Hence, the correct answer is option C.