Verified Solution • 2025–26 Session moderate

QJEEM23Ma3d39: JEE Main 2023 Mathematics - Class 11 JEEM23Ma3d39 Solution: If , then is equal to

If f(x)=22x22x+2,x∈Rf(x) = {{{2^{2x}}} \over {{2^{2x}} + 2}},x \in \mathbb{R} , then f(12023)+f(22023) + ... + f(20222023)f\left( {{1 \over {2023}}} \right) + f\left( {{2 \over {2023}}} \right)\, + \,...\, + \,f\left( {{{2022} \over {2023}}} \right) is equal to

Multiple Choice Options

  • A 2011
  • B 2010
  • C 1010
  • D 1011

Step-by-Step Verified Solution

amp;f(x)=4x4x+2amp;f(x)+f(1−x)=4x4x+2+41−x41−x+2amp;=4x4x+2+44+2(4x)amp;=4x4x+2+22+4xamp;=1amp;⇒f(x)+f(1−x)=1amp; Now f(12023)+f(22023)+f(32023)+…….+amp;………..+f(1−32023)+f(1−22023)+f(1−12023)\begin{aligned}& f(x)=\frac{4^x}{4^x+2} \\\\& f(x)+f(1-x)=\frac{4^x}{4^x+2}+\frac{4^{1-x}}{4^{1-x}+2} \\\\& =\frac{4^x}{4^x+2}+\frac{4}{4+2\left(4^x\right)} \\\\& =\frac{4^x}{4^{\mathrm{x}}+2}+\frac{2}{2+4^{\mathrm{x}}} \\\\& =1 \\\\& \Rightarrow \mathrm{f}(\mathrm{x})+\mathrm{f}(1-\mathrm{x})=1 \\\\& \text { Now } \mathrm{f}\left(\frac{1}{2023}\right)+\mathrm{f}\left(\frac{2}{2023}\right)+\mathrm{f}\left(\frac{3}{2023}\right)+\ldots \ldots .+ \\\\& \ldots \ldots \ldots . .+\mathrm{f}\left(1-\frac{3}{2023}\right)+\mathrm{f}\left(1-\frac{2}{2023}\right)+\mathrm{f}\left(1-\frac{1}{2023}\right)\end{aligned}

Now sum of terms equidistant from beginning and end is 1

amp; Sum =1+1+1+………+1 (1011 times )amp;=1011\begin{aligned}& \text { Sum }=1+1+1+\ldots \ldots \ldots+1 \text { (1011 times }) \\\\& =1011\end{aligned}

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