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← JEE Main 2022 · 27 Jun · Shift 1
MathematicsInteger+4 / -1Question 21 of 90

JEE Main 2022 · 27 Jun · Shift 1, Mathematics question 21

Let f:R→Rf:\mathbb R\to\mathbb R be a function defined by f(x)=2e2xe2x+ef(x)=\dfrac{2e^{2x}}{e^{2x}+e}. Then f(1100)+f(2100)+f(3100)+⋯+f(99100)f\left(\dfrac1{100}\right)+f\left(\dfrac2{100}\right)+f\left(\dfrac3{100}\right)+\dots+f\left(\dfrac{99}{100}\right) is equal to ______.

Answer and worked solution

Official answer: 99

Pairing. f(x)+f(1−x)=2e2xe2x+e+2ee+e2x=2f(x)+f(1-x)=\dfrac{2e^{2x}}{e^{2x}+e}+\dfrac{2e}{e+e^{2x}}=2.

Sum. 49 pairs give 98, and f(12)=1f\left(\frac12\right)=1: total 99.

Answer: 99