Range of $\sin^{-1}(\frac{x^2+1}{x^2+2})$ is:

$[0, π/2]$ $[0, π/6]$ $[π/6, π/2]$ none of these

$[π/6, π/2]$

Here, $\frac{x^2+1}{x^2+2}=1-\frac{1}{x^2+2}$ Now, 2 ≤ x2 + 2 < ∞  for all x ∈ R $⇒\frac{1}{2}≥\frac{1}{x^2+2}>0$ $⇒-\frac{1}{2}≤\frac{-1}{x^2+2}<0$ $⇒\frac{1}{2}≤1-\frac{1}{x^2+2}<1$ $⇒\frac{\pi}{6}≤\sin^{-1}(1-\frac{1}{x^2+2})<\frac{\pi}{2}$. Hence (C) is the correct answer.