If in the expansion of 1-x-x tan x5- the ratio of fourth and second term is 2-27-4 - Then smallest positive value of x is

The correct option is C π/3T_4/T_2=^5C_3 1/x^2(x tan x)^3/^5C_1 ( 1/x )^4(x tan x)= 2x^4.tan^2 x T_4/T_2 = 2/27π^4 ⇒ x^4 tan^2 x = 1/27π^4 x = π/3 ...

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Byju's Answer

Standard XII

Mathematics

Property 5

If in the exp...

Question

If in the expansion of ${(\frac{1}{x} + x t a n x)}^{5}$ , the ratio of fourth and second term is $\frac{2}{27} π^{4}$ . Then smallest positive value of x is

\frac{π}{4}

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\frac{π}{3}

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\frac{π}{6}

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π

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Solution

The correct option is B $\frac{π}{3}$
$\frac{T_{4}}{T_{2}} = \frac{^{5} C_{3} \frac{1}{x^{2}} (x t a n x)^{3}}{^{5} C_{1} {(\frac{1}{x})}^{4} (x t a n x)} = 2 x^{4} . t a n^{2} x$
$\frac{T_{4}}{T_{2}} = \frac{2}{27} π^{4}$
$\Rightarrow x^{4} t a n^{2} x = \frac{1}{27} π^{4}$
$x = \frac{π}{3}$

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The correct option is B π3T4T2=5C31x2(x tan x)35C1(1x)4(x tan x)=2x4.tan2x T4T2=227π4 ⇒x4tan2x=127π4 x=π3

If in the expansion of ${(\frac{1}{x} + x t a n x)}^{5}$ , the ratio of fourth and second term is $\frac{2}{27} π^{4}$ . Then smallest positive value of x is

The correct option is B $\frac{π}{3}$
$\frac{T_{4}}{T_{2}} = \frac{^{5} C_{3} \frac{1}{x^{2}} (x t a n x)^{3}}{^{5} C_{1} {(\frac{1}{x})}^{4} (x t a n x)} = 2 x^{4} . t a n^{2} x$
$\frac{T_{4}}{T_{2}} = \frac{2}{27} π^{4}$
$\Rightarrow x^{4} t a n^{2} x = \frac{1}{27} π^{4}$
$x = \frac{π}{3}$