Q.The domain of the function defined by $f(x)=\sin^{-1}\sqrt{x-1}$ is
(A) $[1,2]$
(B) $[-1,1]$
(C) $[0,1]$
(D) none of these

The key idea is that the argument of $\sin^{-1}$ must lie in $[-1,1]$, and the square root imposes its own non‑negativity condition.

1. For $\sin^{-1}(t)$ to be defined, we need $t \in [-1,1]$. Here $t = \sqrt{x-1}$, so

2. Since a square root is always $\ge 0$, the left inequality $-1 \le \sqrt{x-1}$ is automatically satisfied. The right inequality gives $\sqrt{x-1} \le 1 \quad\Rightarrow\quad 0 \le x-1 \le 1 \quad\Rightarrow\quad 1 \le x \le 2.$ …

The key here is to handle the two layers of restrictions: the square root and the inverse sine. Each imposes its own condition, and the domain is the intersection of both.

1. Condition from the square root

The expression $\sqrt{x-1}$ is defined only when the radicand is non-negative:

So the domain is at least $[1, \infty)$ from this step.

2. Condition from the inverse sine

The function $\sin^{-1}(t)$ is defined only for $t \in [-1, 1]$. Here $t = \sqrt{x-1}$, so we need:

But a square root is always non-negative, so the left inequality ($-1 \leq \sqrt{x-1}$) is automatically satisfied. The real restriction is:

3. Solving the inequality

Square both sides (both sides are non-negative, so squaring preserves the inequality):

4. Combining both conditions

From step 1: $x \geq 1$.

From step 3: $x \leq 2$.

Thus the domain is $[1, 2]$. …

Method: Domain of a composite like $\sin^{-1}\big(\sqrt{g(x)}\big)$

With two nested restrictions — an inner square root and an outer inverse function — the domain is the intersection of both conditions. Handle them from the inside out.

Steps

Step 1: Impose the inner (square-root) condition.

A real square root needs a non-negative radicand:

Solve this first for a preliminary set of $x$.

Step 2: Impose the outer (inverse-sine) condition.

$\sin^{-1}(t)$ needs $-1\le t\le 1$ with $t=\sqrt{g(x)}$. Since a square root is always $\ge 0$, the lower bound $-1\le\sqrt{g(x)}$ is automatic; the only active condition is

$\sqrt{g(x)}\le 1.$ …

Common Mistakes

Mistake 1: Applying the lower bound $-1\le\sqrt{x-1}$ as a real restriction.

Why it's wrong: a square root is always $\ge 0$, so that bound is automatically satisfied and adds nothing; treating it as active leads to a wrong condition like $x\ge 0$. Correct approach: only $\sqrt{x-1}\le 1$ is binding from the $\sin^{-1}$ side.

Mistake 2: Forgetting the square root's own condition $x-1\ge 0$. …