JEE Mains · Maths · STD 12 - 1. relation and function
Let \(f: R \rightarrow R\) be a function defined \(f(x)=\frac{2 e^{2 x}}{e^{2 x}+\varepsilon}\). Then \(f\left(\frac{1}{100}\right)+f\left(\frac{2}{100}\right)+f\left(\frac{3}{100}\right)+\ldots .+f\left(\frac{99}{100}\right)\) is equal to
- A \(98\)
- B \(99\)
- C \(100\)
- D \(101\)
Answer & Solution
Correct Answer
(B) \(99\)
Step-by-step Solution
Detailed explanation
\(f(x)+f(1-x)=\frac{2 e^{2 x}}{e^{2 x}+e}+\frac{2 e^{2-2 x}}{e^{2-e x}+e}=\left[\frac{e^{2 x}}{e^{2 x}+e}+\frac{e^{2}}{e^{2}+e^{2 x+1}}\right]\) \(=2\left[\frac{e^{2 x-1}}{e^{2 x-1}+1}+\frac{1}{1+e^{2 x-1}}\right]=2\)…
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