JEE Mains · Maths · STD 12 - 1. relation and function
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 \(\frac{7}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{9}{4}\)
Step-by-step Solution
Detailed explanation
\(f ( x )+ f \left(\frac{1}{1- x }\right)=1+ x\) \(x =2 \Rightarrow f (2)+ f (-1)=3\) \(x =-1 \Rightarrow f (-1)+ f \left(\frac{1}{2}\right)=0\) \(x =\frac{1}{2} \Rightarrow f \left(\frac{1}{2}\right)+ f (2)=\frac{3}{2}\) \((1)+(3)-(2) \Rightarrow 2 f (2)=\frac{9}{2}\)…
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