JEE Main 2025 — 28 January, Morning Shift — Question 4

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

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

$=\frac{2^{x}}{2^{x}+\sqrt{2}}+\frac{2}{2+\sqrt{2}.2^{x}}=\frac{2^{x}+\sqrt{2}}{2^{x}+\sqrt{2}}=1$

Now, $\sum_{k=1}^{81} \mathrm{f}\left(\frac{\mathrm{k}}{82}\right)=\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(\frac{2}{82}\right)+\ldots \ldots .+\mathrm{f}\left(\frac{81}{82}\right)$

$=f\left(\frac{1}{82}\right)+f\left(\frac{1}{82}\right)+\ldots \ldots .+f\left(1-\frac{2}{82}\right)+f\left(1-\frac{1}{82}\right)$

$\left[\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(1-\frac{1}{82}\right)\right]+\left[\mathrm{f}\left(\frac{2}{82}\right)+\mathrm{f}\left(1-\frac{2}{82}\right)\right]+\ldots . .40$ cases $+\mathrm{f}\left(\frac{41}{82}\right)$

$(1+1+\ldots .+1) 40$ times $+\frac{2^{1 / 2}}{2^{1 / 2}+2^{1 / 2}}$

$40+\frac{1}{2}=\frac{81}{2}$