JEE Main 2025 (Online) 28th January Morning ShiftQuestion 1
If $f(x)=\frac{2^{x}}{2^{x}+\sqrt{2}}$, $x\in\mathbb{R}$, then $\sum_{k=1}^{81}f\left(\frac{k}{82}\right)$ is equal to
Functions
Solution
$f(x)+f(1-x)=\frac{2^x}{2^x+\sqrt2}+\frac{\sqrt2}{\sqrt2+2^x}=1$ and $f(\tfrac12)=\tfrac12$. The 81 terms give 40 pairs summing to 1 plus the middle term $\tfrac12$: $40+\tfrac12=\frac{81}{2}$.