Question:
The domain of the real function $f(x) = \sin^{-1}\left(\log_2\left(\frac{x^2}{2}\right)\right)$ is
The domain of the real function $f(x) = \sin^{-1}\left(\log_2\left(\frac{x^2}{2}\right)\right)$ is
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For nested functions, work from the outside in.
Set up the standard boundaries for the outer function and solve step-by-step using inverse operations.
Set up the standard boundaries for the outer function and solve step-by-step using inverse operations.
Updated On: Jun 16, 2026
- $[-2, -1] \cup [1, 2]$
- $[-2, -1]$
- $[1, 2]$
- $[-2, 2]$
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The problem requires finding the domain of the real-valued function $f(x)$.
We must find the set of all real numbers $x$ for which both the logarithmic function and the inverse sine function are well-defined.
Step 2: Key Formula or Approach:
For a composite function $g(x) = \sin^{-1}(u)$, the domain of the inverse sine function requires:
\[ -1 \le u \le 1 \]
For the logarithmic function $\log_2(v)$, the argument must satisfy:
\[ v \gt 0 \]
Step 3: Detailed Explanation:
• First, let us set up the inequality for the inverse sine function:
\[ -1 \le \log_2\left(\frac{x^2}{2}\right) \le 1 \]
• We can eliminate the logarithm by applying the base 2 exponentiation to all parts of the inequality:
\[ 2^{-1} \le \frac{x^2}{2} \le 2^1 \implies \frac{1}{2} \le \frac{x^2}{2} \le 2 \]
• Multiplying the entire inequality by 2 gives:
\[ 1 \le x^2 \le 4 \]
• Now we split this inequality into two parts:
1) $x^2 \ge 1 \implies |x| \ge 1 \implies x \in (-\infty, -1] \cup [1, \infty)$
2) $x^2 \le 4 \implies |x| \le 2 \implies x \in [-2, 2]$
• Taking the intersection of these two solution sets:
\[ x \in ([-2, 2]) \cap ((-\infty, -1] \cup [1, \infty)) \implies x \in [-2, -1] \cup [1, 2] \]
• Since $\frac{x^2}{2} \ge \frac{1}{2} \gt 0$ for all $x$ in this domain, the logarithmic argument is automatically positive.
Step 4: Final Answer:
The domain of the function is $[-2, -1] \cup [1, 2]$.
The problem requires finding the domain of the real-valued function $f(x)$.
We must find the set of all real numbers $x$ for which both the logarithmic function and the inverse sine function are well-defined.
Step 2: Key Formula or Approach:
For a composite function $g(x) = \sin^{-1}(u)$, the domain of the inverse sine function requires:
\[ -1 \le u \le 1 \]
For the logarithmic function $\log_2(v)$, the argument must satisfy:
\[ v \gt 0 \]
Step 3: Detailed Explanation:
• First, let us set up the inequality for the inverse sine function:
\[ -1 \le \log_2\left(\frac{x^2}{2}\right) \le 1 \]
• We can eliminate the logarithm by applying the base 2 exponentiation to all parts of the inequality:
\[ 2^{-1} \le \frac{x^2}{2} \le 2^1 \implies \frac{1}{2} \le \frac{x^2}{2} \le 2 \]
• Multiplying the entire inequality by 2 gives:
\[ 1 \le x^2 \le 4 \]
• Now we split this inequality into two parts:
1) $x^2 \ge 1 \implies |x| \ge 1 \implies x \in (-\infty, -1] \cup [1, \infty)$
2) $x^2 \le 4 \implies |x| \le 2 \implies x \in [-2, 2]$
• Taking the intersection of these two solution sets:
\[ x \in ([-2, 2]) \cap ((-\infty, -1] \cup [1, \infty)) \implies x \in [-2, -1] \cup [1, 2] \]
• Since $\frac{x^2}{2} \ge \frac{1}{2} \gt 0$ for all $x$ in this domain, the logarithmic argument is automatically positive.
Step 4: Final Answer:
The domain of the function is $[-2, -1] \cup [1, 2]$.
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