Question:medium

Find the domain of the function \( f(x)=\sin^{-1(3x-1) \).}

Show Hint

For inverse trigonometric functions, always restrict the inner expression according to the valid range of the original trigonometric function.
Updated On: Jun 3, 2026
  • \( [-1,1] \)
  • \( [0,\frac{2}{3}] \)
  • \( [-\frac13,\frac13] \)
  • \( [\frac13,1] \)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
The domain of a mathematical function represents the complete set of all possible real input values ($x$) for which the function is defined and outputs a valid real number. For the standard inverse sine function, $y = \sin^{-1}(\theta)$, the function is only defined when the internal input argument $\theta$ falls within the closed interval from $-1$ to $1$. This restriction exists because the regular sine function only yields output values between $-1$ and $1$. Therefore, to find the domain of a composite inverse sine function, you must set up and solve a compound inequality keeping the inner expression bounded within those exact limits.
Step 2: Key Formula or Approach:
For the function $f(x) = \sin^{-1}(g(x))$, the domain restriction inequality rule is: $$ -1 \le g(x) \le 1 $$ In this specific problem, our internal argument function is: $$ g(x) = 3x - 1 $$
Step 3: Detailed Explanation:
Let's apply the domain inequality boundaries to our specific function argument: $$ -1 \le 3x - 1 \le 1 $$ To isolate the variable $x$ in the middle, perform identical algebraic operations across all three sections of the compound inequality simultaneously: 1. Step 1: Add 1 to all sides of the inequality to remove the constant term: $$ -1 + 1 \le 3x - 1 + 1 \le 1 + 1 $$ $$ 0 \le 3x \le 2 $$ 2. Step 2: Divide all sides of the inequality by 3 to isolate $x$: $$ \frac{0}{3} \le \frac{3x}{3} \le \frac{2}{3} $$ $$ 0 \le x \le \frac{2}{3} $$ The variable $x$ must be greater than or equal to $0$ and less than or equal to $\frac{2}{3}$. Writing this solution segment in standard interval notation gives $[0, 2/3]$. This matches option (B).
Step 4: Final Answer:
The domain of the function is [0, 2/3].
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