Question

The domain of the function \[ f(x) = \frac{\sqrt{4 - x^2}}{\sin^{-1}(2 - x)} \] is:

• A. \( [0, 2] \)
• B. \( (0, 2) \)
• C. \( [1, 2] \)
• D. \( [1, 2] \)

Hint:

When determining the domain of a function involving multiple operations, find the domain of each individual part (like square roots and inverse trigonometric functions) and then take the intersection of these domains.

Answer 1

The Correct Option is C

We are given the function: \[ f(x) = \frac{\sqrt{4 - x^2}}{\sin^{-1}(2 - x)}. \] Step 1: Domain of the square root function. For \( \sqrt{4 - x^2} \) to be defined, the expression inside the square root must be non-negative: \[ 4 - x^2 \geq 0. \] Solving this inequality: \[ x^2 \leq 4, \] \[ -2 \leq x \leq 2. \] Thus, the domain of the square root function is \( x \in [-2, 2] \). Step 2: Domain of the inverse sine function. Next, for \( \sin^{-1}(2 - x) \) to be defined, the argument of the inverse sine function must lie in the range \( [-1, 1] \). So, we must have: \[ -1 \leq 2 - x \leq 1. \] Solving these inequalities: \[ 2 - x \geq -1 \quad \Rightarrow \quad x \leq 3, \] \[ 2 - x \leq 1 \quad \Rightarrow \quad x \geq 1. \] Thus, the domain of \( \sin^{-1}(2 - x) \) is \( x \in [1, 2] \). Step 3: Intersection of the domains. The domain of the function \( f(x) \) is the intersection of the domains of the square root and inverse sine functions. From Step 1, the domain of the square root is \( [-2, 2] \), and from Step 2, the domain of the inverse sine function is \( [1, 2] \). Therefore, the domain of \( f(x) \) is: \[ [1, 2]. \] Step 4: Conclusion. Thus, the domain of the function \( f(x) \) is \( [1, 2] \), and the correct answer is (c).