Range of $\sin^{-1}(\frac{x^2+1}{x^2+2})$ is:
Answer & explanation
Correct answer: option 3
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.