Let the function $f:[0,1] \rightarrow \mathbb{R}$ be defined by $f(x)=\frac{4^{x}}{4^{x}+2}$ Then the value of $f\left(\frac{1}{40}\right)+f\left(\frac{2}{40}\right)+f\left(\frac{3}{40}\right)+\ldots+f\left(\frac{39}{40}\right)-f\left(\frac{1}{2}\right)$ is ____

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Relations And Functions XII

Let the function

f : [0, 1] \to R

be defined by

f (x) = \frac{4 ^{x}}{4 ^{x} + 2}

Then the value of

f (\frac{1}{40}) + f (\frac{2}{40}) + f (\frac{3}{40}) + \dots + f (\frac{39}{40}) - f (\frac{1}{2})

is ____

Answer: 19

First, observe that

f (x) + f (1 - x) = \frac{4 ^{x}}{4 ^{x} + 2} + \frac{4 ^{1 - x}}{4 ^{1 - x} + 2} = \frac{4 ^{x}}{4 ^{x} + 2} + \frac{4/ 4 ^{x}}{4/ 4 ^{x} + 2} = \frac{4 ^{x}}{4 ^{x} + 2} + \frac{4}{4 + 2 \cdot 4 ^{x}} = \frac{4 ^{x}}{4 ^{x} + 2} + \frac{2}{4 ^{x} + 2} = 1

. Pairing terms:

f (1/40) + f (39/40) = 1

f (2/40) + f (38/40) = 1

, ...,

f (19/40) + f (21/40) = 1

. There are 19 such pairs, plus the middle term

f (20/40) = f (1/2)

. So the sum is

19 \cdot 1 + f (1/2) = 19 + f (1/2)

. Thus

f (1/40) + \dots + f (39/40) - f (1/2) = 19 + f (1/2) - f (1/2) = 19

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Numerical answer