Mathematics · 2022
JEE Main · 27 June 2022, Shift 1 · Q21
Let f: R → R be a function defined by f(x)=(2 e^2 x)/(e^2 x+ e). Then f(1/100)+f(2/100)+f(3/100)+…+f(99/100) is equal to \ \ \ \.
Let $\displaystyle f: \mathbf{R} \rightarrow \mathbf{R}$ be a function defined by $\displaystyle f(x)=\frac{2 \mathrm{e}^{2 x}}{\mathrm{e}^{2 x}+\mathrm{e}}$. Then $\displaystyle f\left(\frac{1}{100}\right)+f\left(\frac{2}{100}\right)+f\left(\frac{3}{100}\right)+\ldots+f\left(\frac{99}{100}\right)$ is equal to \_\_\_\_.
Official answer
99
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.