The domain of the function $sin^{-1}(4x-1)$ is :

[-1, 1] [0, 1] $[0, \frac{1}{2}]$ $[-\frac{1}{2}, \frac{1}{2}]$

$[0, \frac{1}{2}]$

The correct answer is Option (3) → $[0, \frac{1}{2}]$ from definition $-1≤4x-1≤1$ $0≤4x≤2$ $0≤x≤\frac{1}{2}$ $⇒x∈[0,\frac{1}{2}]$