The domain of definition of f(x)= sin^-1√(x-1), is

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].