Question:medium

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

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Whenever inverse trigonometric functions and square roots appear together, solve both restrictions separately and then take their intersection.
Updated On: May 20, 2026
  • $(-\infty,1]\cup[2,\infty)$
  • $[0,1]$
  • $[-1,1]$
  • $[1,2]$
Show Solution

The Correct Option is D

Solution and Explanation

Understanding the Concept: The domain of a function is the set of all real values of $x$ for which the function is defined. In the given function, \[ f(x)=\sin^{-1}(\sqrt{x-1}), \] two separate conditions must be satisfied:
The expression inside the square root must be non-negative.
The input of $\sin^{-1}(x)$ must lie between $-1$ and $1$ inclusive.
Therefore, we solve both conditions carefully.
Step 1: Applying the square root condition.
For the square root to exist, \[ x-1\ge0 \] Adding $1$ on both sides: \[ x\ge1 \] So the function is defined only when: \[ x\in[1,\infty) \]
Step 2: Applying the inverse sine condition.
For $\sin^{-1}(u)$ to exist, \[ -1\le u\le1 \] Here, \[ u=\sqrt{x-1} \] Therefore: \[ -1\le\sqrt{x-1}\le1 \] Since square roots are always non-negative, the lower inequality is automatically satisfied. So we only need: \[ \sqrt{x-1}\le1 \]
Step 3: Solving the inequality.
Squaring both sides: \[ x-1\le1 \] Adding $1$: \[ x\le2 \]
Step 4: Finding the common interval.
From Step 1: \[ x\ge1 \] From Step 3: \[ x\le2 \] Combining both: \[ 1\le x\le2 \] Hence, the domain is: \[ \boxed{[1,2]} \]
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