Question

If in the expansion of ${(\frac{1}{x}+x\tan x)}^5,$ the ratio of $4^{\text{th}}$ term to the $2^{\text{nd}}$ term is $\frac{2}{27}{\pi }^4,$ then the smallest positive value of $x$ is

• A. $\frac{\pi }{8}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is C $\frac{\pi }{3}$

General term is
$T_{r+1}={}^5{C}_r\cdot {\left(\frac{1}{x}\right)}^{5-r}\cdot (x\tan x{)}^r$

$\frac{T_4}{T_2}=\frac{{}^5{C}_3\cdot \frac{1}{x^2}\cdot (x\tan x{)}^3}{{}^5{C}_1\cdot \frac{1}{x^4}\cdot (x\tan x{)}^1}=\frac{2}{27}{\pi }^4$

$\Rightarrow 2x^4\cdot {\tan }^{2}x=\frac{2}{27}{\pi }^4$
$\Rightarrow x^4\cdot {\tan }^{2}x=\frac{{\pi }^4}{27}$
So, the smallest positive value of $x$ is $\frac{\pi }{3}.$