JEE Mains · Maths · STD 12 - 1. relation and function
Let \(f ^1( x )=\frac{3 x +2}{2 x +3}, x \in R -\left\{\frac{-3}{2}\right\}\) For \(n \geq 2\), define \(f ^{ n }( x )= f ^1 0 f ^{ n -1}( x )\). If \(f ^5( x )=\frac{ ax + b }{ bx + a }, \operatorname{gcd}( a , b )=1\), then \(a + b\) is equal to \(............\).
- A \(3124\)
- B \(3123\)
- C \(3126\)
- D \(3125\)
Answer & Solution
Correct Answer
(D) \(3125\)
Step-by-step Solution
Detailed explanation
\(f^1(x)=\frac{3 x+2}{2 x+3}\) \(\Rightarrow f^2(x)=\frac{13 x+12}{12 x+13}\) \(\Rightarrow f^3(x)=\frac{63 x+62}{62 x+63}\) \(\therefore f^5(x)=\frac{1563 x+1562}{1562 x+1563}\) \(a+b=3125\)
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