Question

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

A
π6
B
π3
C
π3
D
π12

Solution

## The correct option is D $$\displaystyle \frac{\pi}{3}$$From the above given expression by applying binomial theorem, we get.$$T_{4}=\:^5C_{3}x\tan^3{x}$$$$=10x\tan^3{x}$$$$T_{2}=\:^5C_{1}x^{-3}\tan{x}$$$$=5x^{-3}\tan{x}$$Therefore $$\displaystyle \frac{T_{4}}{T_{2}}$$$$=\displaystyle \frac{10x\tan^3{x}}{5x^{-3}\tan{x}}$$$$=2x^4\tan^2x$$$$=\displaystyle \frac{2\pi^{4}}{27}$$$$x^4\tan^2x=\displaystyle \frac{\pi^{4}}{27}$$Taking root on both sides, we get $$x^2\tan x=\displaystyle \frac{\pi^2}{3\sqrt3}$$$$=\displaystyle \frac{\sqrt{3}\pi^2}{9}$$Therefore $$x^2=\displaystyle \frac{\pi^2}{9}$$ ...(i) and $$\tan x=\sqrt{3}$$...(ii)Both Eq(i) and Eq (ii) give $$x=\displaystyle \frac{\pi}{3}$$Mathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More