Question

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

• A. $\frac{-\pi }{6}$
• B. $\frac{-\pi }{3}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{12}$

Answer 1

Answer: (C)

$\frac{\pi }{3}$

From the above given expression by applying binomial theorem, we get.
$T_4={}^5{C}_{3}x{\tan }^{3}x$
$=10x{\tan }^{3}x$
$T_2={}^5{C}_1{x}^{-3}\tan x$
$=5x^{-3}\tan x$
Therefore $\frac{T_4}{T_2}$
$=\frac{10x{\tan }^{3}x}{5x^{-3}\tan x}$
$=2x^4{\tan }^{2}x$
$=\frac{2{\pi }^4}{27}$
$x^4{\tan }^{2}x=\frac{{\pi }^4}{27}$
Taking root on both sides, we get
$x^2\tan x=\frac{{\pi }^2}{3\sqrt{3}}$
$=\frac{\sqrt{3}{\pi }^2}{9}$
Therefore $x^2=\frac{{\pi }^2}{9}$ ...(i) and $\tan x=\sqrt{3}$...(ii)
Both Eq(i) and Eq (ii) give $x=\frac{\pi }{3}$