If Iₙ=_{π/4}^{π/2} ⁿ x \, dx, then [25 Feb 2021 Shift 2]

Correct Answer is : 1I2+I4,1I3+I5,1I4+I6{1}{I₂+I₄}, {1}{I₃+I₅}, {1}{I₄+I₆}I2+I41,I3+I51,I4+I61 are in AP

{1}{I₂+I₄}, {1}{I₃+I₅}, {1}{I₄+I₆} are in AP

I₂+I₄, I₃+I₅, I₄+I₆ are in AP

{1}{I₂+I₄}, {1}{I₃+I₅}, {1}{I₄+I₆} are in GP

I₂+I₄, (I₃+I₅)², I₄+I₆ are in GP

Correct Answer is : 1I2+I4,1I3+I5,1I4+I6{1}{I₂+I₄}, {1}{I₃+I₅}, {1}{I₄+I₆}I2+I41,I3+I51,I4+I61 are in AP

Test Index

JEE Main 25 Feb 2021 Shift 2 Solved Paper

Section: Mathematics

Question : 64 of 90

Marks: +1, -0

I_{n}=\int\limits_{\pi/4}^{\pi/2} \cot^{n} x \, dx

Solution:

I_{n}=\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{n} x \, dx=\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{n-2} x (\cot^{2} x) \, dx

I_{n}=\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{n-2} x \csc^{2} x \, dx-\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{n-2} x \, dx

I_{n}+I_{n-2}=\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{n-2} x \cdot \csc^{2} x \, dx

Now, let

\cot x = t

, then

\csc^{2} x \, dx = -dt

, limit will be

I_{n}+I_{n-2} = \int\limits_{1}^{0} -t^{n-2} \, dt

= \frac{-(t)^{n-1}}{n-1}\Big]_{1}^{0} = -\left\{ \frac{0}{n-1} - \frac{(1)^{n-1}}{n-1} \right\}

I_{n}+I_{n-2} = \frac{1}{n-1}

Now, put

n = 4

\Rightarrow I_{2}+I_{4} = \frac{1}{3}

, then

\frac{1}{I_{2}+I_{4}} = 3

. . . (i)

Put

n = 5

\Rightarrow I_{5}+I_{3} = \frac{1}{4}

, then

\frac{1}{I_{3}+I_{5}} = 4

. . . (ii)

Put

n = 6

\Rightarrow I_{6}+I_{4} = \frac{1}{5}

, then

\frac{1}{I_{4}+I_{6}} = 5

. . . (iii)

Here, from Eqs. (i), (ii) and (iii), we conclude

\frac{1}{I_{2}+I_{4}}, \frac{1}{I_{3}+I_{5}}

and

\frac{1}{I_{4}+I_{6}}

are in AP with common difference

1.

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