Let f : R → R be a function defined by f(x) = 2e²ˣ/(e²ˣ + e). Then f( 1/100 ) + f( 2/100 ) + f( 3/100 ) + ..... + f( 99/100 ) is equal to ______________.

Solution

Maths sir · 4 steps

Pair each term with its mirror across one-half: f(x) + f(1-x) comes out to 2 every time.

1. Find the partner. The inputs run from 1/100 to 99/100, so they pair up as x and 1-x. Compute f(1-x) and multiply its top and bottom by e²ˣ.

   f(1-x) = 2e²⁻²ˣ/(e²⁻²ˣ + e) = 2e²/(e² + e²ˣ⁺¹) = 2e/(e + e²ˣ)

2. Add the pair. Both terms now share the denominator e²ˣ + e. So f(x) + f(1-x) = (2e²ˣ + 2e)/(e²ˣ + e) = 2, whatever x is.
3. Count the pairs. Pair k = 1 with 99, 2 with 98, up to 49 with 51. That makes 49 pairs, worth 49 × 2 = 98. The middle term k = 50 has no partner, and f(1/2) = 2e/(e + e) = 1.
4. So. The sum is 98 + 1 = 99.

The common slip: 98. Pairing everything and forgetting the unpaired middle term f(1/2) = 1 gives 98, not 99.

The idea to keep. When a sum runs over k/n for k = 1 to n-1, test f(x) + f(1-x) first. If it is a constant, pair the terms up and handle the middle term separately.

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