JEE Main 2025 Jan 28 Shift 1, Mathematics Q4: Some Important Properties Of Functions

Compute $f(x)+f(1-x)$:

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

Multiplying the second fraction's numerator and denominator by $2^x$ gives $\frac{2}{2+\sqrt{2}{2}^x}$. Combine:

$f(x)+f(1-x)=\frac{2^x}{2^x+\sqrt{2}}+\frac{\sqrt{2}}{\sqrt{2}+2^x}=\frac{2^x+\sqrt{2}}{2^x+\sqrt{2}}=1$.

So $f$ has the symmetry $f(x)+f(1-x)=1$. Pair the sum:

$S=\sum _{k=1}^{81}f\left(\frac{k}{82}\right)$. Pair $k$ with $82-k$ for $k=1,2,\dots ,40$, leaving the middle term $k=41$ unpaired.

Each of the 40 pairs sums to 1. The middle term: $f\left(\frac{41}{82}\right)=f\left(\frac{1}{2}\right)=\frac{\sqrt{2}}{\sqrt{2}+\sqrt{2}}=\frac{1}{2}$.

$\therefore S=40+\frac{1}{2}=\frac{81}{2}$.