If in the expansion of 1 x -xtan x 5 the ratio to 4th term to the 2nd term is 2 27 - 4- then the value of x can be

The correct options are A π 3 C π 3 T_4=2π^427T_2 ^5C_3x tan^3x=2π^427 ^5C_11x^3tanx 10x^4tan^2(x)=52π^427 x^4tan^2(x)=π^427 ...

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General Term of Binomial Expansion

If in the exp...

Question

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

\frac{- π}{6}

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\frac{- π}{3}

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\frac{π}{3}

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\frac{π}{12}

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Similar questions

{(\frac{1}{x} + x tan x)}^{5},

4^{th}

2^{nd}

\frac{2}{27} π^{4},

x

(\frac{1}{x} + x tan x)^{5}

4^{t h}

2^{n d}

\frac{2}{27} π^{4}

x

{(\frac{1}{x} + x t a n x)}^{5}

\frac{2}{27} π^{4}

{(\frac{x}{cos θ} + \frac{1}{x sin θ})}^{16}

l_{1}

x

\frac{π}{8} \leq θ \leq \frac{π}{4}

l_{2}

x

\frac{π}{16} \leq θ \leq \frac{π}{8},

l_{2} : l_{1}

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{(1 - {sin}^{3} x)}{3 {cos}^{2} x}, & x < \frac{π}{2} a, & x = \frac{π}{2} \frac{b (1 - sin x)}{(π - 2 x)^{2}}, & x > \frac{π}{2} \end{matrix}

x = \frac{π}{2}

{(\frac{b}{a})}^{5 / 3}

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