Question:
The domain of \( \sin^{-1}(1 - 2x) \) is :
The domain of \( \sin^{-1}(1 - 2x) \) is :
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Always remember to flip the inequality signs whenever you divide or multiply by a negative quantity. A quick check of boundary values (like plugging in \( x=0 \) and \( x=1 \)) can confirm correctness instantly.
Updated On: Jul 4, 2026
- \( [-1, 1] \)
- \( [-1, 3] \)
- \( [-2, 2] \)
- \( [0, 1] \)
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The Correct Option is D
Solution and Explanation
Concept:
The domain of the standard inverse sine function \( \sin^{-1}(\theta) \) is restricted to the interval \( [-1, 1] \). Therefore, for the function \( \sin^{-1}(1 - 2x) \) to be well-defined in the real number system, its argument must lie within these closed boundaries:
\[
-1 \le 1 - 2x \le 1
\]
Step 1: Set up and solve the compound inequality.
Subtract 1 from all parts of the inequality chain: \[ -1 - 1 \le -2x \le 1 - 1 \] \[ -2 \le -2x \le 0 \]
Step 2: Divide by the negative coefficient.
Divide the entire inequality by \( -2 \). Remember that dividing or multiplying an inequality by a negative number reverses the direction of the inequality signs: \[ \frac{-2}{-2} \ge \frac{-2x}{-2} \ge \frac{0}{-2} \] \[ 1 \ge x \ge 0 \] Rewriting this in the standard low-to-high interval notation: \[ 0 \le x \le 1 \quad \Rightarrow \quad x \in [0, 1] \] This range matches option (D).
Step 1: Set up and solve the compound inequality.
Subtract 1 from all parts of the inequality chain: \[ -1 - 1 \le -2x \le 1 - 1 \] \[ -2 \le -2x \le 0 \]
Step 2: Divide by the negative coefficient.
Divide the entire inequality by \( -2 \). Remember that dividing or multiplying an inequality by a negative number reverses the direction of the inequality signs: \[ \frac{-2}{-2} \ge \frac{-2x}{-2} \ge \frac{0}{-2} \] \[ 1 \ge x \ge 0 \] Rewriting this in the standard low-to-high interval notation: \[ 0 \le x \le 1 \quad \Rightarrow \quad x \in [0, 1] \] This range matches option (D).
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