JEE MAIN - Mathematics (2021 - 25th February Evening Shift - No. 2)
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 :
$${{{39} \over 2}}$$
$${{{19} \over 2}}$$
$${{{49} \over 2}}$$
$${{{29} \over 2}}$$
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)
$$f(x) + f(2 - x) = 1$$
$$f\left( {{1 \over {20}}} \right) + f\left( {{{39} \over {20}}} \right) = 1$$
$$f\left( {{2 \over {20}}} \right) + f\left( {{{38} \over {20}}} \right) = 1$$
$$\eqalign{ & : \cr & : \cr} $$
$$f\left( {{{19} \over {20}}} \right) + f\left( {{{21} \over {20}}} \right) = 1$$
and $$f\left( {{{20} \over {20}}} \right) = f(1) = {1 \over 2}$$
$$ \therefore $$ Sum = $$19 + {1 \over 2} = {{39} \over 2}$$
$$f(2 - x) = {{{5^{2 - x}}} \over {{5^{2 - x}} + 5}}$$
$$f(2 - x) = {5 \over {{5^x} + 5}}$$ .... (ii)
Adding equation (i) and (ii)
$$f(x) + f(2 - x) = 1$$
$$f\left( {{1 \over {20}}} \right) + f\left( {{{39} \over {20}}} \right) = 1$$
$$f\left( {{2 \over {20}}} \right) + f\left( {{{38} \over {20}}} \right) = 1$$
$$\eqalign{ & : \cr & : \cr} $$
$$f\left( {{{19} \over {20}}} \right) + f\left( {{{21} \over {20}}} \right) = 1$$
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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