Question

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

• A. [0, π/2]
• B. (0, π/6)
• C. [π/6, π/2)
• D. None

Answer 1

Answer: (C)

[π/6, π/2)

Answer 2

Here, $\frac{x^{2} + 1}{x^{2} + 2}$ = $1 - \frac{1}{x^{2} + 2}$

Now, 2 ≤ x² + 2 < ∞ for all x∈R

⇒ $\frac{1}{2} \geq \frac{1}{x^{2} + 2}$ > 0 ⇒ –$\frac{1}{2} \leq \frac{- 1}{x^{2} + 2}$ < 0

⇒ $\frac{1}{2} \leq 1 - \frac{1}{x^{2} + 2}$ < 1 ⇒ $\frac{\pi}{6} \leq \sin^{- 1}\left( 1 - \frac{1}{x^{2} + 2} \right) < \frac{\pi}{2}$