Medium MCQ +4 / -1 PYQ · JEE Mains 2023

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

  1. A $\frac{9}{4}$ Correct answer
  2. B $\frac{7}{4}$
  3. C $\frac{7}{3}$
  4. D $\frac{9}{2}$

Solution

$\begin{aligned} & \mathrm{f}(\mathrm{x})+\mathrm{f}\left(\frac{1}{1-\mathrm{x}}\right)=1+\mathrm{x} \\\\ & \mathrm{x}=2 \Rightarrow \mathrm{f}(2)+\mathrm{f}(-1)=3 ........(1) \\\\ & \mathrm{x}=-1 \Rightarrow \mathrm{f}(-1)+\mathrm{f}\left(\frac{1}{2}\right)=0 .........(2) \\\\ & \mathrm{x}=\frac{1}{2} \Rightarrow \mathrm{f}\left(\frac{1}{2}\right)+\mathrm{f}(2)=\frac{3}{2} ........(3) \\\\ & (1)+(3)-(2) \Rightarrow 2 \mathrm{f}(2)=\frac{9}{2} \\\\ & \therefore \mathrm{f}(2)=\frac{9}{4}\end{aligned}$

About this question

Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations

This question is part of PrepWiser's free JEE Main question bank. 195 more solved questions on Sets, Relations and Functions are available — start with the harder ones if your accuracy is >70%.

Drill 25 more like these. Every day. Free.

PrepWiser turns these solved questions into a daily practice loop. Chapter-wise drills, full mocks, AI doubt chat. No auto-renew.

Start free →