Question

If \(A\) and \(B\) are the entire domain and range, respectively, of the real-valued function \[ f(x)=\cos^{-1}\!\left(\frac{2-x^2}{2+x^2}\right), \] then \(A\cap B=\)

• A. \( [0, \sqrt{2}) \)
• B. \( [0, \frac{\pi}{2}) \)
• C. \( [0, \pi) \)
• D. \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \)

Hint:

Use the substitution \( x = \sqrt{2}\tan\theta \). Then \( \frac{2-x^2}{2+x^2} = \frac{2-2\tan^2\theta}{2+2\tan^2\theta} = \cos 2\theta \). This simplifies the function to \( f(x) = \cos^{-1}(\cos 2\theta) \), making it easier to visualize the range.

Answer 1

The Correct Option is C

Concept: The domain of \( \cos^{-1}(u) \) is \( [-1, 1] \) and its principal range is \( [0, \pi] \).
• Domain (\( A \)): Find all \( x \) such that \( -1 \leq \frac{2-x^2}{2+x^2} \leq 1 \).
• Range (\( B \)): Find the set of all possible values of \( f(x) \).
• Intersection: Find the common elements between set \( A \) and set \( B \).

Step 1: Finding the Domain \( A \).
For \( f(x) \) to be defined, we require: \[ -1 \leq \frac{2-x^2}{2+x^2} \leq 1 \] Since \( 2+x^2 > 0 \) for all \( x \in \mathbb{R} \), we can multiply through: 1) \( 2-x^2 \leq 2+x^2 \Rightarrow 0 \leq 2x^2 \), which is true for all \( x \in \mathbb{R} \). 2) \( -2-x^2 \leq 2-x^2 \Rightarrow -2 \leq 2 \), which is also always true. Thus, the domain \( A = \mathbb{R} \).

Step 2: Finding the Range \( B \).
Let \( y = \frac{2-x^2}{2+x^2} \). We can rewrite this as \( y = \frac{4}{x^2+2} - 1 \). Since \( x^2 \in [0, \infty) \), then \( x^2+2 \in [2, \infty) \). Thus, \( \frac{4}{x^2+2} \in (0, 2] \). Subtracting 1: \( y \in (-1, 1] \). The range of \( \cos^{-1}(y) \) for \( y \in (-1, 1] \) is \( [0, \pi) \). Note that \( \pi \) is excluded because \( y \) never actually reaches \( -1 \). Thus, \( B = [0, \pi) \).

Step 3: Finding \( A \cap B \).
\( A \cap B = \mathbb{R} \cap [0, \pi) = [0, \pi) \).