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A function f(x) is given by $$f(x) = {{{5^x}} \over {{5^x} + 5}}$$, then the sum of the series $$f\left( {{1 \over {20}}} \right) + f\left( {{2 \over {20}}} \right) + f\left( {{3 \over {20}}} \right) + ....... + f\left( {{{39} \over {20}}} \right)$$ is equal to :

JEE · Math · previous-year question

  1. A.$${{{39} \over 2}}$$correct
  2. B.$${{{19} \over 2}}$$
  3. C.$${{{49} \over 2}}$$
  4. D.$${{{29} \over 2}}$$

Answer

A. $${{{39} \over 2}}$$

Explanation

$$f(x) = {{{5^x}} \over {{5^x} + 5}}$$ ..... (i) $$f(2 - x) = {{{5^{2 - x}}} \over {{5^{2 - x}} + 5}}$$ $$f(2 - x) = {5 \over {{5^x} + 5}}$$ .... (ii) Adding equation (i) and (ii) $$f(x) + f(2 - x) = 1$$ $$f\left( {{1 \over {20}}} \right) + f\left( {{{39} \over {20}}} \right) = 1$$ $$f\left( {{2 \over {20}}} \right) + f\left( {{{38} \over {20}}} \right) = 1$$ $$\eqalign{ & : \\ & : \\} $$ $$f\left( {{{19} \over {20}}} \right) + f\left( {{{21} \over {20}}} \right) = 1$$ and $$f\left( {{{20} \over {20}}} \right) = f(1) = {1 \over 2}$$ $$ \therefore $$ Sum = $$19 + {1 \over 2} = {{39} \over 2}$$

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