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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
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B
π3
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C
π3
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D
π12
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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

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