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If \( f(x)=\frac{e^{2 x-1}}{1+e^{2 x-1}} \), then the value of \( f\left(\frac{1}{2009}\right)+f\left(\frac{2}{2009}\right)+f\left(\frac{3}{2009}\right)+\ldots . .+f\left(\frac{2008}{2009}\right) \) is (1) \( 1002.5 \) (2) \( 1001.5 \) (3) 1003 (4) 1004 |
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Answer» f(x) = \(\frac{e^{2x-1}}{1+e^{2x-1}}\) f(1-x) = \(\frac{e^{2(1-x)-1}}{1+e^{2(1-x)-1}}\) = \(\frac{e^{1-2x}}{1+e^{1-2x}}\) = \(\frac{e^{-(2x-1)}}{1+e^{-(2x+1)}}\) = \(\frac1{e^{2x-1}+1}=\frac1{1+e^{2x-1}}\) \(\therefore\) f(x) + f(1 - x) = \(\frac{e^{2x-1}+1}{1+e^{2x-1}}=1\) \(\therefore\) f(1/2009) + f(2008/2009) = 1 \(\therefore\) f(2/2009) + f(2007/2009) = 1 f(1004/2009) + f(1005/2009) = 1 \(\therefore\) f(1/2009) + f(2008/2009) + f(2/2009) + f(2007/2009) +....f(1004/2009) +f(1005/2009) = 1 + 1 + ...1 (10004 times) = 1004 ⇒ f(1/2009) + f(2/2009) + .... + f(1004/2009) + f(1005/2009) + ....+ f(2007/2009) + f(2008/2009) = 1004 |
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