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1.Relation and Function
hard
Let $f : R -\{0,1\} \rightarrow R$ be a function such that $f(x)+f\left(\frac{1}{1-x}\right)=1+x$. Then $f(2)$ is equal to :
A
$\frac{9}{2}$
B
$\frac{9}{4}$
C
$\frac{7}{4}$
D
$\frac{7}{3}$
(JEE MAIN-2023)
Solution
$f ( x )+ f \left(\frac{1}{1- x }\right)=1+ x$
$x =2 \Rightarrow f (2)+ f (-1)=3$
$x =-1 \Rightarrow f (-1)+ f \left(\frac{1}{2}\right)=0$
$x =\frac{1}{2} \Rightarrow f \left(\frac{1}{2}\right)+ f (2)=\frac{3}{2}$
$(1)+(3)-(2) \Rightarrow 2 f (2)=\frac{9}{2}$
$\therefore f (2)=\frac{9}{4}$
Standard 12
Mathematics
