If f(x)=2^x/2^x+√2, x∈R, then ∑ f(k/82) k ∈ to (k=1, 81) is equal to :

Question

If \(f(x)=\frac{2^{x}}{2^{x}+\sqrt{2}}, x \in R,\) then \(\sum\limits_{k=1}^{81} f\left(\frac{k}{82}\right)\) is equal to :

(1) 41

(2) \(\frac{81}{2}\)

(3) 82

(4) \(81 \sqrt{2}\)

Answer 1

Correct option is (2) \(\frac{81}{2}\)   

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

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

\(=\frac{2^{\mathrm{x}}}{2^{\mathrm{x}}+\sqrt{2}}+\frac{2}{2+\sqrt{2} 2^{\mathrm{x}}}=\frac{2^{\mathrm{x}}+\sqrt{2}}{2^{\mathrm{x}}+\sqrt{2}}=1 \)

\(\text { Now, } \sum\limits_{\mathrm{k}=1}^{81} \mathrm{f}\left(\frac{\mathrm{k}}{82}\right)=\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(\frac{2}{82}\right)+\ldots \ldots+\mathrm{f}\left(\frac{81}{82}\right) \)

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

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

\(=(1+1+\ldots 40 \text { times })+\frac{2^{1 / 2}}{2^{1 / 2}+2^{1 / 2}} \)

\(40+\frac{1}{2}=\frac{81}{2}\)