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JEE Main 2023 · 24 Jan · Shift 2

JEE Main 2023 Mathematics question on Definite Integrals and Areas

If f(x)=22x22x+2,x∈Rf(x) = \dfrac{2^{2x}}{2^{2x}+2}, x \in \mathbb{R}, then f(12023)+f(22023)+…+f(20222023)f\left(\dfrac{1}{2023}\right) + f\left(\dfrac{2}{2023}\right) + \ldots + f\left(\dfrac{2022}{2023}\right) is equal to
  1. A.
    2010
  2. B.
    2011
  3. C.
    1011
  4. D.
    1010
Show answer and solution

Correct option: C

Solution

The key property is f(x) + f(1-x) = 1, since f(x) + f(1-x) = 2^(2x)/(2^(2x)+2) + 2^(2-2x)/(2^(2-2x)+2). Multiplying numerator and denominator of the second term by 2^(2x) gives 2/(2+2^(2x)), so the sum is (2^(2x)+2)/(2^(2x)+2) = 1.

The sum has 2022 terms. Pairing f(k/2023) with f((2023-k)/2023) gives 1011 pairs, each summing to 1, so the total is 1011.

[JEE Main 2023 · 24 Jan · Shift 2]

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