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JEE Mains · Maths · STD 12 - 1. relation and function

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

  1. A \(81 \sqrt{2}\)
  2. B 41
  3. C 82
  4. D \(\frac{81}{2}\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

\(f(x)=\frac{2^x}{2^x+\sqrt{2}} \) \( f(x)+f(1-x)=\frac{2^x}{2^x+\sqrt{2}}+\frac{2^{1-x}}{2^{1-x}+\sqrt{2}} \) \( =\frac{2^x}{2^x+\sqrt{2}}+\frac{2}{2+\sqrt{2} 2^x}=\frac{2^x+\sqrt{2}}{2^x+\sqrt{2}}=1\) Now,…