JEE Advance - Mathematics (2020 - Paper 2 Offline - No. 16)
Let the function f : [0, 1] $$ \to $$ R be defined by
$$f(x) = {{{4^x}} \over {{4^x} + 2}}$$
Then the value of $$f\left( {{1 \over {40}}} \right) + f\left( {{2 \over {40}}} \right) + f\left( {{3 \over {40}}} \right) + ... + f\left( {{{39} \over {40}}} \right) - f\left( {{1 \over 2}} \right)$$ is ..........
$$f(x) = {{{4^x}} \over {{4^x} + 2}}$$
Then the value of $$f\left( {{1 \over {40}}} \right) + f\left( {{2 \over {40}}} \right) + f\left( {{3 \over {40}}} \right) + ... + f\left( {{{39} \over {40}}} \right) - f\left( {{1 \over 2}} \right)$$ is ..........
Answer
19
Explanation
The given function f : [0, 1] $$ \to $$ R be define by
$$f(x) = {{{4^x}} \over {{4^x} + 2}}$$
$$ \because $$ $$f(1 - x) = {{{4^{1 - x}}} \over {{4^{1 - x}} + 2}} = {2 \over {2 + {4^x}}}$$
$$ \therefore $$ $$f(x) + f(1 - x) = {{{4^x}} \over {{4^x} + 2}} + {2 \over {2 + {4^x}}}$$
$$ = {{{4^x} + 2} \over {{4^x} + 2}}$$
So, f(x) + f(1 $$-$$ x) = 1 .....(i)
$$ \therefore $$ $$f\left( {{1 \over {40}}} \right) + f\left( {{2 \over {40}}} \right) + f\left( {{3 \over {40}}} \right) + ... + f\left( {{{39} \over {40}}} \right) - f\left( {{1 \over 2}} \right)$$
$$ = \left[ {f\left( {{1 \over {40}}} \right) + f\left( {{{39} \over {40}}} \right)} \right] + \left[ {f\left( {{2 \over {40}}} \right) + f\left( {{{38} \over {40}}} \right)} \right] + ... + \left[ {f\left( {{{18} \over {40}}} \right) + f\left( {{{22} \over {40}}} \right)} \right] + \left[ {f\left( {{{19} \over {40}}} \right) + f\left( {{{21} \over {40}}} \right)} \right] + \left[ {f\left( {{{20} \over {40}}} \right) - f\left( {{1 \over 2}} \right)} \right]$$
$$ = \{ 1 + 1 + ... + 1 + 1\} + f\left( {{1 \over 2}} \right) - f\left( {{1 \over 2}} \right)$$
$$ = \{ 1 + 1 + ... + 1 + 1\}$$(19 times) {from Eq. (i)}
= 19.
$$f(x) = {{{4^x}} \over {{4^x} + 2}}$$
$$ \because $$ $$f(1 - x) = {{{4^{1 - x}}} \over {{4^{1 - x}} + 2}} = {2 \over {2 + {4^x}}}$$
$$ \therefore $$ $$f(x) + f(1 - x) = {{{4^x}} \over {{4^x} + 2}} + {2 \over {2 + {4^x}}}$$
$$ = {{{4^x} + 2} \over {{4^x} + 2}}$$
So, f(x) + f(1 $$-$$ x) = 1 .....(i)
$$ \therefore $$ $$f\left( {{1 \over {40}}} \right) + f\left( {{2 \over {40}}} \right) + f\left( {{3 \over {40}}} \right) + ... + f\left( {{{39} \over {40}}} \right) - f\left( {{1 \over 2}} \right)$$
$$ = \left[ {f\left( {{1 \over {40}}} \right) + f\left( {{{39} \over {40}}} \right)} \right] + \left[ {f\left( {{2 \over {40}}} \right) + f\left( {{{38} \over {40}}} \right)} \right] + ... + \left[ {f\left( {{{18} \over {40}}} \right) + f\left( {{{22} \over {40}}} \right)} \right] + \left[ {f\left( {{{19} \over {40}}} \right) + f\left( {{{21} \over {40}}} \right)} \right] + \left[ {f\left( {{{20} \over {40}}} \right) - f\left( {{1 \over 2}} \right)} \right]$$
$$ = \{ 1 + 1 + ... + 1 + 1\} + f\left( {{1 \over 2}} \right) - f\left( {{1 \over 2}} \right)$$
$$ = \{ 1 + 1 + ... + 1 + 1\}$$(19 times) {from Eq. (i)}
= 19.
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