Question

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

• A. ${60}^0$
• B. ${30}^0$
• C. ${45}^0$
• D. ${90}^0$

Answer 1

Answer: (B)

${30}^0$

Solution:

Middle term of the expansion ${(\frac{1}{x}+x\sin x)}^{10}$ is $T_6.$

$T_6={}^{10}{C}_5\times {\left(\frac{1}{x}\right)}^5\times (x\sin x{)}^5=7\frac{7}{8}$

or, ${}^{10}{C}_5\times {\sin }^{5}x=\frac{63}{8}$

or, $252\times {\sin }^{5}x=\frac{63}{8}$

or, ${\sin }^{5}x=\frac{1}{32}$

or, $\sin x=\frac{1}{2}$

or, $x={30}^{\circ }$

Hence, B is the correct option.