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) $2n\pi +\frac{\mathrm{\pi }}{6}$

(b) $n\pi +\frac{\mathrm{\pi }}{6}$

(c) $n\pi +{\left(-1\right)}^n\frac{\mathrm{\pi }}{6}$

(d) $n\pi +{\left(-1\right)}^n\frac{\mathrm{\pi }}{3}$

Answer 1

Given middle term of ${\left(\frac{1}{x}+x\sin x\right)}^{10}\mathrm{is}7\frac{7}{8}$
Since n = 10
i.e. middle term is ${\left(\frac{n}{2}+1\right)}^{\mathrm{th}}$ term is 6th term.
$$\begin{gathered} \therefore T_6=T_{5+1}{=}^{10}{C}_5{\left(\frac{1}{x}\right)}^{10-5}{(x\sin x)}^5 \\ \mathrm{i}.\mathrm{e}.\frac{63}{8}{=}^{10}{C}_5\frac{1}{x^5}\times x^5{\left(\sin x\right)}^5 \\ \mathrm{i}.\mathrm{e}.\frac{63}{8}=\frac{10!}{5!5!}{\sin }^{5}x \\ \mathrm{i}.\mathrm{e}.\frac{63}{8}=252{\sin }^{5}x \\ \mathrm{i}.\mathrm{e}.{\sin }^{5}x=\frac{1}{3^2}={\left(\frac{1}{2}\right)}^5 \\ \mathrm{i}.\mathrm{e}.\sin x=\frac{1}{2} \\ \mathrm{i}.\mathrm{e}.x=n\mathrm{\pi }+{(-1)}^n\frac{\mathrm{\pi }}{6} \end{gathered}$$

Hence, the correct answer is option C.