JEE Mains · Maths · STD 12 - 1. relation and function
Let \(f(x)=\frac{x-1}{x+1}, x \in R-\{0,-1,1)\). If \(f^{a+1}(x)=f\left(f^{n}(x)\right)\) for all \(n \in N\), then \(f^{\prime}(6)+f(7)\) is equal to
- A \(\frac{7}{6}\)
- B \(-\frac{3}{2}\)
- C \(\frac{7}{12}\)
- D \(-\frac{11}{12}\)
Answer & Solution
Correct Answer
(B) \(-\frac{3}{2}\)
Step-by-step Solution
Detailed explanation
\(f(x)=\frac{x-1}{x+1}\) \(\Rightarrow f^{2}(x)=f(f(x))=\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}=-\frac{1}{x}\) \(f^{3}(x)=f\left(f^{2}(x)\right)=f\left(-\frac{1}{x}\right)=\frac{x+1}{1-x}\) \(\Rightarrow f^{4}(x)=f\left(\frac{x+1}{1-x}\right)=-\frac{1}{x}\)…
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