PrepRung

JEE Advanced 2022 Paper 2 · Mathematics · Indefinite and Definite Integrals · hard

For a non-negative integer n let In=∫0^π/2sin ^nx dx.

One or more options correct

For a non-negative integer nn let In=∫0π/2sin⁡nx dxI_n=\displaystyle\int_0^{\pi/2}\sin^{n}x\,dx. Which of the following statements is/are correct?
  1. A. I5=815I_5=\dfrac8{15}
  2. B. I6=3π16I_6=\dfrac{3\pi}{16}
  3. C. The series ∑n=1∞In\displaystyle\sum_{n=1}^{\infty}I_n converges
  4. D. lim⁡n→∞n In=π2\displaystyle\lim_{n\to\infty}\sqrt n\,I_n=\frac\pi2
Show answer and solution

Answer: A

Integration by parts gives In=n−1nIn−2I_n=\frac{n-1}{n}I_{n-2}, with I0=π2I_0=\frac\pi2 and I1=1I_1=1. Then I3=23I_3=\frac23 and I5=45⋅23=815I_5=\frac45\cdot\frac23=\frac8{15}; A is true. Also I2=π4I_2=\frac\pi4, I4=3π16I_4=\frac{3\pi}{16} and I6=56⋅3π16=5π32I_6=\frac56\cdot\frac{3\pi}{16}=\frac{5\pi}{32}, so B is false (it quotes I4I_4). From nInIn−1=π2nI_nI_{n-1}=\frac\pi2 and In∼In−1I_n\sim I_{n-1} we get nIn2→π2nI_n^2\to\frac\pi2, so In∼π/(2n)I_n\sim\sqrt{\pi/(2n)}; the series therefore diverges like ∑n−1/2\sum n^{-1/2}, making C false, and n In→π/2≠π2\sqrt n\,I_n\to\sqrt{\pi/2}\neq\frac\pi2, making D false.
Report a mistake

Original question written in the JEE Advanced 2022 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake