If f(x) = 2²ˣ/(2²ˣ + 2),x ∈ ℝ, then f( 1/2023 ) + f( 2/2023 ) + ... + f( 2022/2023 ) is equal to

Solution

Maths sir · 4 steps

Terms whose inputs add to 1 pair up, and each such pair of f values adds to exactly 1.

1. Spot the pairing. The inputs k/2023 and (2023-k)/2023 add to 1. So check what f(x) + f(1-x) equals.
2. Simplify f(1 − x). Write 2²ˣ = 4ˣ. Then f(1-x) = (4 · 4⁻ˣ)/(4 · 4⁻ˣ + 2). Multiply the top and bottom by 4ˣ/2 to get f(1-x) = 2/(2 + 4ˣ).

   f(x) + f(1-x) = 4ˣ/(4ˣ + 2) + 2/(4ˣ + 2) = 1

3. Count the pairs. The 2022 terms split into 2022/2 = 1011 pairs: (1/2023, 2022/2023) up to (1011/2023, 1012/2023). No term is left over, because 2023/2 is not a whole number.
4. So. The sum is 1011 × 1 = 1011, which is option (d).

The idea to keep. When a sum runs over f(k/n), test f(x) + f(1-x) first. For aˣ/(aˣ + √a) it is always 1, and the sum is just the number of pairs.

Answer from the NTA final answer key. Solution by PrepWiser.