QJEEM22Ma1fa3: JEE Main 2022 Mathematics - Class 11 JEEM22Ma1fa3 Solution: Let f : R R be a function defined by . Then is equal to…

Given,

$f(x) = {{2{e^{2x}}} \over {{e^{2x}} + e}}$

$\therefore$ $f(1 - x) = {{2{e^{2(1 - x)}}} \over {{e^{2(1 - x)}} + e}}$

$= {{2\,.\,{{{e^2}} \over {{e^{2x}}}}} \over {{{{e^2}} \over {{e^{2x}}}} + e}}$

$= {{2{e^2}} \over {{e^2} + {e^{2x}}\,.\,e}}$

$= {{2{e^2}} \over {e(e + {e^{2x}})}}$

$= {{2e} \over {e + {e^{2x}}}}$

$\therefore$ $f(x) + f(1 - x) = {{2{e^{2x}}} \over {{e^{2x}} + e}} + {{2e} \over {{e^{2x}} + e}}$

$= {{2({e^{2x}} + e)} \over {{e^{2x}} + e}}$

$= 2$ ...... (1)

Now,

$f\left( {{1 \over {100}}} \right) + f\left( {{{99} \over {100}}} \right)$

$= f\left( {{1 \over {100}}} \right) + f\left( {1 - {1 \over {100}}} \right)$

$= 2$ [as $f(x) + f(1 - x) = 2$ ]

Similarly,

$f\left( {{2 \over {100}}} \right) + f\left( {1 - {2 \over {100}}} \right) = 2$

$\vdots$

$f\left( {{{49} \over {100}}} \right) + f\left( {1 - {{49} \over {100}}} \right) = 2$

$\therefore$ Total sum $= 49 \times 2$

Remaining term $= f\left( {{{50} \over {100}}} \right) = f\left( {{1 \over 2}} \right)$

Put $x = {1 \over 2}$ in equation (1), we get

$f\left( {{1 \over 2}} \right) + f\left( {1 - {1 \over 2}} \right) = 2$

$\Rightarrow 2f\left( {{1 \over 2}} \right) = 2$

$\Rightarrow f\left( {{1 \over 2}} \right) = 1$

$\therefore$ Sum $= 49 \times 2 + 1 = 99$