Verified Solution • 2025–26 Session moderate

QJEEM23Mad638: JEE Main 2023 Mathematics - Class 11 JEEM23Mad638 Solution: Let be a function such that . Then is equal to

Let f:R−0,1→Rf:\mathbb{R}-{0,1}\to \mathbb{R} be a function such that f(x)+f(11−x)=1+xf(x)+f\left(\frac{1}{1-x}\right)=1+x . Then f(2)f(2) is equal to

Multiple Choice Options

  • A 94\frac{9}{4}
  • B 74\frac{7}{4}
  • C 73\frac{7}{3}
  • D 92\frac{9}{2}

Step-by-Step Verified Solution

amp;f(x)+f(11−x)=1+xamp;x=2⇒f(2)+f(−1)=3........(1)amp;x=−1⇒f(−1)+f(12)=0.........(2)amp;x=12⇒f(12)+f(2)=32........(3)amp;(1)+(3)−(2)⇒2f(2)=92amp;∴f(2)=94\begin{aligned} & \mathrm{f}(\mathrm{x})+\mathrm{f}\left(\frac{1}{1-\mathrm{x}}\right)=1+\mathrm{x} \\\\ & \mathrm{x}=2 \Rightarrow \mathrm{f}(2)+\mathrm{f}(-1)=3 ........(1) \\\\ & \mathrm{x}=-1 \Rightarrow \mathrm{f}(-1)+\mathrm{f}\left(\frac{1}{2}\right)=0 .........(2) \\\\ & \mathrm{x}=\frac{1}{2} \Rightarrow \mathrm{f}\left(\frac{1}{2}\right)+\mathrm{f}(2)=\frac{3}{2} ........(3) \\\\ & (1)+(3)-(2) \Rightarrow 2 \mathrm{f}(2)=\frac{9}{2} \\\\ & \therefore \mathrm{f}(2)=\frac{9}{4}\end{aligned}

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