Question

Let \( f: \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \to \mathbb{R} \) be given by \( f(x) = \frac{\pi}{2} + x - \tan^{-1}x \).
Consider the following statements:
P: \( |f(x) - f(y)| < |x - y| \text{ for all } x, y \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \).
Q: \( f \) has a fixed point.
Then:

• A. both P and Q are TRUE
• B. P is TRUE and Q is FALSE
• C. P is FALSE and Q is TRUE
• D. both P and Q are FALSE

Hint:

For a function to be a contraction, its derivative must satisfy \( |f'(x)| < 1 \) in the given domain.

Answer 1

The Correct Option is D

Step 1: Analyze statement P.
The statement \( |f(x) - f(y)| < |x - y| \) implies that the function \( f(x) \) is a contraction mapping. For \( f(x) = \frac{\pi}{2} + x - \tan^{-1}x \), we can check the derivative of \( f(x) \) to verify if the function is a contraction. The derivative is: \[ f'(x) = 1 - \frac{1}{1+x^2}. \] For \( |f'(x)| < 1 \), the function must be a contraction. But for \( f'(x) \), we see that the condition is not satisfied for all \( x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \), so statement P is FALSE.

Step 2: Analyze statement Q.
A fixed point of a function \( f \) is a point \( x \) such that \( f(x) = x \). For the function \( f(x) = \frac{\pi}{2} + x - \tan^{-1}x \), solving \( f(x) = x \) leads to no solution in the given interval. Therefore, statement Q is also FALSE.

Final Answer: (D) both P and Q are FALSE