Let I 1-42-2- I 2-tan -11- e 2-2 e - e 2-1- I 3-tan -1 e 2-2- e 2-1- then which of the following is true-A- I 1- I 3- I 2B- I 3- I 2- I 1C- I 1- I 2- I 3D- I 2- I 1- I 3

The correct option is D I_2 lt; I_1 lt; I_3 Consider f(x) = (tan^ 1 x)^2 + 2/√(x^2+1) then f '(x) gt; 0 ∀ x ε (0, ∞ ) ⇒ f ( 1/e ) lt; f(1) ...

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Derivative of Standard Inverse Trigonometric Functions

Let I 1=π/42+...

Question

Let $I_{1} = {(\frac{π}{4})}^{2} + \sqrt{2}, I_{2} = {(t a n^{- 1} (\frac{1}{e}))}^{2} + \frac{2 e}{\sqrt{e^{2} + 1}}, I_{3} = (t a n^{- 1} e)^{2} + \frac{2}{\sqrt{e^{2} + 1}}$ , then which of the following is true?

$I_{1} < I_{2} < I_{3}$

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$I_{2} < I_{1} < I_{3}$

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$I_{1} < I_{3} < I_{2}$

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$I_{3} < I_{2} < I_{1}$

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Solution

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I_{1} = \int_{0}^{\frac{π}{4}} x^{2008} (t a n x)^{2008} d x, I_{2} = \int_{0}^{\frac{π}{4}} x^{2009} (t a n x)^{2009} d x a n d I_{3} = \int_{0}^{\frac{π}{4}} x^{2010} (t a n x)^{2010} d x

Now what should be the correct relation between $I_{1}$ , $I_{2}$ and $I_{3}$ .

I_{1} = \int_{\frac{π}{3}}^{\frac{π}{6}} \frac{sin x}{x} d x, I_{2} = \int_{\frac{π}{3}}^{\frac{π}{6}} \frac{sin (sin x)}{sin x} d x, I_{3} = \int_{\frac{π}{3}}^{\frac{π}{6}} \frac{sin (tan x)}{tan x} d x

I_{1} = \int_{0}^{\frac{π}{2}} c o s (π s i n^{2} x) d x, I_{2} = \int_{0}^{\frac{π}{2}} c o s (2 π s i n^{2} x) d x a n d I_{3} = \int_{0}^{\frac{π}{2}} c o s (π s i n x) d x,

Let f: R $\to$ R such that f(x + 2y) = f(x) + f(2y) + 4xy, $\forall$ x, y $ϵ$ R and f(0) = 0.

If $I_{1} = \int_{0}^{1} f (d x), I_{2} = \int_{- 1}^{0} f (x) d x, a n d I_{3} = \int_{- 1}^{1} f (x) d x, t h e n$

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