Question

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

• A. [-1, -\frac{1}{\sqrt{2}}] \cup [\frac{1}{\sqrt{2}}, 1]
• B. [-1, 1]
• C. (-\infty, -\frac{1}{\sqrt{2}}] \cup [\frac{1}{\sqrt{2}}, +\infty]
• D. [\frac{1}{\sqrt{2}}, 1]

Hint:

Be extremely careful when squaring inequalities! Squaring can introduce extraneous solutions. Always check the basic sign constraints beforehand (like noticing \(x \ge \text{positive radical}\) forces \(x \ge 0\)) to instantly rule out misleading options like choice (1).

Answer 1

The Correct Option is D

Concept: The domain of a real-valued function consists of all real values of \(x\) for which the expression is mathematically well-defined. For square root functions of the form \(\sqrt{g(x)}\) to produce real outputs, the radicand expression inside the radical sign must be non-negative: \[ g(x) \ge 0 \] In this problem, we have nested radicals, which requires establishing and solving a system of simultaneous inequalities to find their shared interval intersection. Step 1: Analyze the condition for the internal square root.
The inner radical contains the expression \(\sqrt{1 - x^2}\). For this term to be real: \[ 1 - x^2 \ge 0 \quad \Rightarrow \quad x^2 \le 1 \] Taking the square root across the inequality establishes our primary bounding interval constraint: \[ -1 \le x \le 1 \quad \implies \quad x \in [-1, 1] \quad \cdots (1) \]

Step 2: Analyze the condition for the primary external square root.
The main outermost radical requires its complete underlying expression to be non-negative: \[ x - \sqrt{1 - x^2} \ge 0 \quad \Rightarrow \quad x \ge \sqrt{1 - x^2} \quad \cdots (2) \] Let us interpret this inequality logically before performing algebraic squaring:
• The right side, \(\sqrt{1 - x^2}\), represents a principal square root, which is always non-negative by definition (\(\ge 0\)).
• For the variable \(x\) to be greater than or equal to a non-negative number, \(x\) itself must be strictly non-negative: \[ x \ge 0 \quad \cdots (3) \] This observation immediately eliminates any negative intervals from our potential solution set.

Step 3: Square both sides to solve the algebraic inequality.
Since both sides of inequality (2) are non-negative for \(x \ge 0\), squaring both sides preserves the inequality sign: \[ x^2 \ge \left(\sqrt{1 - x^2}\right)^2 \] \[ x^2 \ge 1 - x^2 \] Add \(x^2\) to both sides: \[ 2x^2 \ge 1 \quad \Rightarrow \quad x^2 \ge \frac{1}{2} \] Solving this inequality yields: \[ x \le -\frac{1}{\sqrt{2}} \quad \text{or} \quad x \ge \frac{1}{\sqrt{2}} \quad \cdots (4) \]

Step 4: Find the intersection of all constraint intervals.
We now find the common overlap of all four derived interval constraints:
• \(x \in [-1, 1]\) (From the inner radical domain)
• \(x \ge 0\) (From the positive sign requirement)
• \(x \in \left(-\infty, -\frac{1}{\sqrt{2}}\right] \cup \left[\frac{1}{\sqrt{2}}, +\infty\right]\) (From the algebraic solution) Intersecting these intervals step-by-step:
• Combining \(x \in [-1, 1]\) with \(x \ge 0\) reduces the domain space to \([0, 1]\).
• Intersecting \([0, 1]\) with \(\left(-\infty, -\frac{1}{\sqrt{2}}\right] \cup \left[\frac{1}{\sqrt{2}}, +\infty\right]\) completely discards the negative sub-interval, leaving only: \[ x \in \left[\frac{1}{\sqrt{2}}, 1\right] \]