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Let f: \mathbf{R} \rightarrow \mathbf{R} be a function defined by f(x)=\frac{2 \mathrm{e}^{2 x}}{\mathrm{e}^{2 x}+\mathrm{e}}. 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________________.

Option: 1

99


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

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

\mathrm{\Rightarrow f(x)+f(1-x)=2}

\mathrm{s=f\left(\frac{1}{100}\right)+f\left(\frac{2}{100}\right)+\cdots f\left(\frac{50}{100}\right)+\cdots-f\left(\frac{93}{100}\right)+f\left(\frac{99}{100}\right)} \\

\mathrm{s=f\left(\frac{99}{100}\right)+f\left(\frac{98}{100}\right)+\cdots f\left(\frac{50}{100}\right)+\cdots f\left(\frac{2}{100}\right)+f\left(\frac{1}{100}\right) }

Adding

\mathrm{2 s =99 \times(f(r)+f(1-r))} \\

\mathrm{\Rightarrow s =\frac{99}{2} \times 2=99 . }

Hence answer is \mathrm{99 . }

Posted by

Shailly goel

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