Let f : [-π/3, 2π/3] → [0, 4] be a function defined as $f(x) = \sqrt{3}sin x- cosx+2$. Then f^-1(x) is given by

$\sin^{-1}(\frac{x-2}{2})+\frac{\pi}{6}$

$f(x) = \sqrt{3}sin x- cosx+2 = 2 sin(x-\frac{\pi}{6}) + 2$.

Since f(x) is one-one and onto, f is invertible.

Now $fof ^{-1}(x) = x ⇒ 2 sin (f ^{-1}(x)-\frac{\pi}{6})+ 2 = x$

$⇒ sin(f ^{-1}(x)-\frac{\pi}{6}) =\frac{x}{2} – 1 ⇒ f^{ -1}(x) = sin^{-1}(\frac{x}{2} – 1)+\frac{\pi}{6}$,

because $|\frac{x}{2} – 1|≤1$ for all x ∈ [0, 4].

Hence (B) is the correct answer.