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1.Relation and Function
hard
If $f(x)=\frac{2^x}{2^x+\sqrt{2}}, x \in R$, then $\sum_{k=1}^{81} f\left(\frac{k}{82}\right)$ is equal to :
A$41$
B$\frac{81}{2}$
C$82$
D$81 \sqrt{2}$
(JEE MAIN-2025)
Solution
$f(x)=\frac{2^x}{2^x+\sqrt{2}}$
$f(x)+f(1-x)=\frac{2^x}{2^x+\sqrt{2}}+\frac{2^{1-x}}{2^{1-x}+\sqrt{2}}$
$=\frac{2^{x}}{2^{x}+\sqrt{2}}+\frac{2}{2+\sqrt{2} 2^{x}}=\frac{2^{x}+\sqrt{2}}{2^{x}+\sqrt{2}}=1$
Now, $\sum_{k=1}^{81} f\left(\frac{k}{82}\right)=f\left(\frac{1}{82}\right)+f\left(\frac{2}{82}\right)+\ldots \ldots .+f\left(\frac{81}{82}\right)$
$=f\left(\frac{1}{82}\right)+f\left(\frac{1}{82}\right)+\ldots \ldots+f\left(1-\frac{2}{82}\right)+f\left(1-\frac{1}{82}\right)$
${\left[f\left(\frac{1}{82}\right)+f\left(1-\frac{1}{82}\right)\right]+\left[f\left(\frac{2}{82}\right)+f\left(1-\frac{2}{82}\right)\right]+\ldots .40 \text { cases }+f\left(\frac{41}{82}\right)}$
$(1+1+\ldots . .+1) 40 \text { times }+\frac{2^{1 / 2}}{2^{1 / 2}+2^{1 / 2}}$
$40+\frac{1}{2}=\frac{81}{2}$
$f(x)+f(1-x)=\frac{2^x}{2^x+\sqrt{2}}+\frac{2^{1-x}}{2^{1-x}+\sqrt{2}}$
$=\frac{2^{x}}{2^{x}+\sqrt{2}}+\frac{2}{2+\sqrt{2} 2^{x}}=\frac{2^{x}+\sqrt{2}}{2^{x}+\sqrt{2}}=1$
Now, $\sum_{k=1}^{81} f\left(\frac{k}{82}\right)=f\left(\frac{1}{82}\right)+f\left(\frac{2}{82}\right)+\ldots \ldots .+f\left(\frac{81}{82}\right)$
$=f\left(\frac{1}{82}\right)+f\left(\frac{1}{82}\right)+\ldots \ldots+f\left(1-\frac{2}{82}\right)+f\left(1-\frac{1}{82}\right)$
${\left[f\left(\frac{1}{82}\right)+f\left(1-\frac{1}{82}\right)\right]+\left[f\left(\frac{2}{82}\right)+f\left(1-\frac{2}{82}\right)\right]+\ldots .40 \text { cases }+f\left(\frac{41}{82}\right)}$
$(1+1+\ldots . .+1) 40 \text { times }+\frac{2^{1 / 2}}{2^{1 / 2}+2^{1 / 2}}$
$40+\frac{1}{2}=\frac{81}{2}$
Standard 12
Mathematics
