Question

If \( f(x) = \cos^{-1 } \left( \frac{\sqrt{2x^2 + 1}}{x^2 + 1} \right) \), then the range of \( f(x) \) is:

• A. \( [0, \pi] \)
• B. \( \left[ 0, \frac{\pi}{4} \right] \)
• C. \( \left[ 0, \frac{\pi}{3} \right] \)
• D. \( \left[ 0, \frac{\pi}{2} \right] \)

Hint:

For inverse trigonometric functions like \( \cos^{-1} \), the range is typically \( [0, \pi] \), but the values inside the function must stay within the domain of -1 to 1.

Answer 1

The Correct Option is D

Given \( f(x) = \cos^{-1} \left( \frac{\sqrt{2x^2 + 1}}{x^2 + 1} \right) \), and knowing the range of the inverse cosine function is \( [0, \pi] \), the expression within the inverse cosine yields values between -1 and 1. Consequently, the range of \( f(x) \) is determined to be \( \left[ 0, \frac{\pi}{2} \right] \).