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JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Sequences and Series

Let

f : R \to R

be a function defined

f (x) = \frac{2 e ^{2 x}}{e ^{2 x} + e}

.

Then

f (\frac{1}{100}) + f (\frac{2}{100}) + f (\frac{3}{100}) + \dots + f (\frac{99}{100})

is equal to ____.

Answer: 99

Solution

First observe that:

f (x) + f (1 - x) = \frac{2 e ^{2 x}}{e ^{2 x} + e} + \frac{2 e ^{2 - 2 x}}{e ^{2 - 2 x} + e} = 2

This can be verified by simplifying the sum.

Now pair the terms:

k = 1 \sum 99 f (\frac{k}{100}) = k = 1 \sum 49 [f (\frac{k}{100}) + f (\frac{100 - k}{100})] + f (\frac{50}{100})

Each pair sums to 2, and there are 49 pairs:

49 \times 2 + f (\frac{1}{2}) = 98 + 1 = 99

where

f (\frac{1}{2}) = \frac{2 e}{e + e} = 1

.

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Integer answer

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