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Saturday, 31 October

JEE Main 2022 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 30 July 2022, Shift 2, Maths Q71

Question 71 of 90 in this shift, Maths question 11 of 30, Section A.

Let Ak=∫kπ(k+1)πe−xsin⁡x dxA_k = \int_{k\pi}^{(k+1)\pi} e^{-x} \sin x\, dx , k=0,1,2,...,20k = 0, 1, 2, ..., 20. Then ∑k=020∣Ak∣\sum_{k=0}^{20} |A_k| is equal to
  1. (1)(eπ+1eπ−1)(1−e−20π)\left(\frac{e^{\pi}+1}{e^{\pi}-1}\right)\left(1 - e^{-20\pi}\right)
  2. (2)12(eπ+1eπ−1)(1−e−21π)\frac{1}{2}\left(\frac{e^{\pi}+1}{e^{\pi}-1}\right)\left(1 - e^{-21\pi}\right)Official answer
  3. (3)(eπ+1eπ−1)(1−e−21π)\left(\frac{e^{\pi}+1}{e^{\pi}-1}\right)\left(1 - e^{-21\pi}\right)
  4. (4)12(eπ+1eπ−1)(1−e−20π)\frac{1}{2}\left(\frac{e^{\pi}+1}{e^{\pi}-1}\right)\left(1 - e^{-20\pi}\right)

Official answer

Option 2

NTA final key.

Chapter
Integrals