QJEEAdv20Ma485c: JEE Advanced 2020 Mathematics - Class 11 JEEAdv20Ma485c Solution: Let the function f : [0, 1] R be defined by Then the value of…

Question

Verified Solution • 2025–26 Session moderate

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 ..........

Accepted Answer

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.

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