Question

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

• A. [0,1]
• B. [-1,1]
• C. [1,2]
• D. [-1,0]
• E. [0,2]

Hint:

Logic Tip: Always establish the domain of the innermost functions first (like square roots or logarithms) before evaluating the domain of the outer function. This prevents calculating "valid" values for an outer function that would result in complex numbers inside.

Answer 1

The Correct Option is C

Concept:
The function $f(x) = \sin^{-1}(y)$ is defined only when the argument $y$ lies in the interval $[-1, 1]$. Additionally, for a real-valued function, the expression inside a square root must be non-negative.
Step 1: Find the domain for the square root.
[cite_start]The term $\sqrt{2-x}$ requires its argument to be greater than or equal to zero[cite: 73]: $$2 - x \ge 0$$ $$x \le 2$$
Step 2: Find the domain for the inverse sine function.
[cite_start]The argument of the inverse sine function must satisfy[cite: 73]: $$-1 \le \sqrt{2-x} \le 1$$ Since a principal square root is always non-negative, the left side of the inequality ($-1 \le \sqrt{2-x}$) is automatically satisfied. We only need to solve the right side: $$\sqrt{2-x} \le 1$$
Step 3: Solve the inequality.
Squaring both sides of the inequality gives: $$2 - x \le 1$$ $$-x \le 1 - 2$$ $$-x \le -1$$ Multiply by $-1$ and flip the inequality sign: $$x \ge 1$$
Step 4: Combine the conditions.
From Step 1, $x \le 2$. From Step 3, $x \ge 1$. Combining these gives the valid interval: $$1 \le x \le 2$$ [cite_start]Thus, the domain is $[1, 2]$[cite: 79].