QJEEM25Ma1953: JEE Main 2025 Mathematics - Class 11 JEEM25Ma1953 Solution: If , then is equal to

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

$\begin{array}{r}\text { Now, } \sum_{\mathrm{k}=1}^{81} \mathrm{f}\left(\frac{\mathrm{k}}{82}\right)=\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(\frac{2}{82}\right)+\ldots \ldots+\mathrm{f}\left(\frac{81}{82}\right) \\=\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(\frac{2}{82}\right)+\ldots \ldots .+\mathrm{f}\left(1-\frac{2}{82}\right)+\mathrm{f}\left(1-\frac{1}{82}\right) \\=\left[\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(1-\frac{1}{82}\right)\right]+\left[\mathrm{f}\left(\frac{2}{82}\right)+\mathrm{f}\left(1-\frac{2}{82}\right)\right]+\ldots . .40 \text { cases }+\mathrm{f}\left(\frac{41}{82}\right)\end{array}$

$\begin{aligned}& =(1+1+\ldots 40 \text { times })+\frac{2^{1 / 2}}{2^{1 / 2}+2^{1 / 2}} \\& 40+\frac{1}{2}=\frac{81}{2}\end{aligned}$