JEE Main 2022 · 30 Jul · Shift 2 · Mathematics Q71

MathematicsSingle correct+4 / -1Question 71 of 90

Let $A_k=\displaystyle\int_{k\pi}^{(k+1)\pi}e^{-x}\sin x\,dx$ , k = 0, 1, 2, ..., 20. Then $\displaystyle\sum_{k=0}^{20}|A_k|$ is equal to

\left(\dfrac{e^\pi+1}{e^\pi-1}\right)\left(1-e^{-20\pi}\right)

\dfrac12\left(\dfrac{e^\pi+1}{e^\pi-1}\right)\left(1-e^{-21\pi}\right)

\left(\dfrac{e^\pi+1}{e^\pi-1}\right)\left(1-e^{-21\pi}\right)

\dfrac12\left(\dfrac{e^\pi+1}{e^\pi-1}\right)\left(1-e^{-20\pi}\right)

Answer and worked solution

Official answer: B

Each piece. $|A_k|=e^{-k\pi}\displaystyle\int_0^\pi e^{-x}\sin x\,dx=e^{-k\pi}\cdot\frac{1+e^{-\pi}}2$ .

Geometric sum $\dfrac{1+e^{-\pi}}2\cdot\dfrac{1-e^{-21\pi}}{1-e^{-\pi}}=\dfrac12\left(\dfrac{e^\pi+1}{e^\pi-1}\right)(1-e^{-21\pi})$ .

(B)

JEE Main 2022 · 30 Jul · Shift 2, Mathematics question 71