The domain of definition of $f(x)= sin^{-1}\sqrt{x-1},$ is

[-1,1] [0,1] [1,2] [2,3]

[1,2]

The domain of $sin^{-1}x $ is [-1,1]. So, the domain of $ f(x) = sin^{-1} \sqrt{x-1}$ is the set of values of x satisfying $-1 ≤ \sqrt{x-1}≤ 1$ $⇒ 0 ≤ \sqrt{x-1} ≤1$                 $[∵\sqrt{x-1} ≥0]$ $⇒ 0 ≤ x-1 ≤1 ⇒1 ≤x≤2⇒x ∈[1,2]$ Hence, the domain of $f(x) = sin^{-1}\sqrt{x-1}$ is [1,2].