Question

If the middle term in the expansion of ${(\frac{1}{x}+x\sin x)}^{10}$ is $\frac{63}{8},$ then the value of $6{\sin }^{2}x+\sin x-2$ is

• A. $0$
• B. $\frac{3}{2}$
• C. $5$
• D. $\frac{61}{8}$

Answer 1

The correct option is A $0$

Middle Term $=T_6={}^{10}{C}_5{\left(\frac{1}{x}\right)}^5(x\sin x{)}^5=252(\sin x{)}^5$
$\therefore 252(\sin x{)}^5=\frac{63}{8}$
$\Rightarrow \sin x=\frac{1}{2}$
$\Rightarrow 2\sin x-1=0$
$\therefore 6{\sin }^{2}x+\sin x-2$
$=(2\sin x-1)(3\sin x+2)=0$