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WBJEE · Maths · Functions

For every real number \(x \neq-1\), let \(f(x)=\frac{x}{x+1}\). Write \(f_1(x)=f(x)\) \& for \(n \geq 2, f_n(x)=f\left(f_{n-1}(x)\right)\). Then \(f_1(-2) . f_2(-2)\) \(\qquad\) must be

  1. A \(\frac{2^n}{1.3 .5 \ldots \ldots .(2 n-1)}\)
  2. B 1
  3. C \(\frac{1}{2}\binom{2 n}{n}\)
  4. D \(\binom{2 n}{n}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{2^n}{1.3 .5 \ldots \ldots .(2 n-1)}\)

Step-by-step Solution

Detailed explanation

\begin{aligned} & \text { Hint: } f(x)=\frac{x}{x+1}, \quad f_2(x)=\frac{x}{\frac{x+1}{\frac{x}{x+1}+1}}=\frac{x}{2 x+1} \\ & f_3(x)=\frac{x}{3 x+1}, \Rightarrow f_1(-2) \cdot f_2(-2) \ldots \ldots \ldots . f_n(-2) \\ & =\frac{-2}{-1} \times \frac{-2}{-3} \times \frac{-2}{-5}…