Question

A real valued function $ f: A \to B $ defined by $ f(x) = \frac{4 - x^2}{4 + x^2} \ \forall x \in A $ is a bijection. If $ -4 \in A $, then $ A \cap B = $

• A. \((-1, 1]\)
• B. \([0, 1]\)
• C. \([0, \infty)\)
• D. \((-1, 0]\)

Hint:

For a function to be bijective, check both injectivity and surjectivity. Use function values at given points to test range intersections.

Answer 1

The Correct Option is D

Given \( f(x) = \frac{4 - x^2}{4 + x^2} \). Since the function is bijective, its domain and codomain must match in range and uniqueness. The expression \( \frac{4 - x^2}{4 + x^2} \) always gives values less than or equal to 1, and as \( x \to \infty \) or \( x \to -\infty \), the value approaches \(-1\). Evaluate at \( x = -4 \): \[ f(-4) = \frac{4 - 16}{4 + 16} = \frac{-12}{20} = -\frac{3}{5} \] So this value is included, and since it’s bijective, the function maps onto this interval. The range is \( (-1, 0] \). Since \( -4 \in A \), the image of this \( x \) lies in \( B \). So, \[ A \cap B = (-1, 0] \]