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

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Sets Relations and Functions

Let

f : R - {0, 1} \to R

be a function such that

f (x) + f (\frac{1}{1 - x}) = 1 + x

. Then

f (2)

is equal to :

Answer: (B)

The map

x \mapsto \frac{1}{1 - x}

cycles

2 \to - 1 \to \frac{1}{2} \to 2

. Writing the relation at these three values gives

f (2) + f (- 1) = 3

f (- 1) + f (\frac{1}{2}) = 0

and

f (\frac{1}{2}) + f (2) = \frac{3}{2}

. Adding the first and third,

2 f (2) + f (- 1) + f (\frac{1}{2}) = \frac{9}{2}

, and since the middle relation gives

f (- 1) + f (\frac{1}{2}) = 0

, this leaves

2 f (2) = \frac{9}{2}

. Therefore

f (2) = \frac{9}{4}

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Single correct option