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← JEE Main 2022 · 25 Jun · Shift 2
MathematicsSingle correct+4 / -1Question 12 of 90

JEE Main 2022 · 25 Jun · Shift 2, Mathematics question 12

If bn=∫0π2cos⁡2nxsin⁡xdxb_n=\displaystyle\int_0^{\frac\pi2}\frac{\cos^2nx}{\sin x}dx, n∈Nn\in\mathbb N, then

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Answer and worked solution

Official answer: D

Difference. cos⁡2nx−cos⁡2(n−1)x=−sin⁡(2n−1)x sin⁡x\cos^2nx-\cos^2(n-1)x=-\sin(2n-1)x\,\sin x, so bn−bn−1=−∫0π/2sin⁡(2n−1)x dx=−12n−1b_n-b_{n-1}=-\displaystyle\int_0^{\pi/2}\sin(2n-1)x\,dx=-\frac1{2n-1}.

Values. −15,−17,−19-\frac15,-\frac17,-\frac19; reciprocals −5,−7,−9-5,-7,-9: an A.P. with common difference −2-2.

(D)

Topics:Properties of definite integralsTrigonometric ratios and identitiesArithmetic progression
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