Question

The range of the real valued function \( f(x) =\) \(\sin^{-1} \left( \frac{1 + x^2}{2x} \right)\) \(+ \cos^{-1} \left( \frac{2x}{1 + x^2} \right)\) is:

• A. \( \left\{ \frac{\pi}{2} \right\} \)
• B. \( \mathbb{R} \)
• C. \( \mathbb{Q} \)
• D. \( \left\{ -\frac{\pi}{2}, \frac{\pi}{2} \right\} \)

Hint:

For inverse trigonometric functions, recall the identities: - \( \sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2} \) - \( \sin^{-1} \left( \frac{1 + x^2}{2x} \right) + \cos^{-1} \left( \frac{2x}{1 + x^2} \right) \) simplifies to a constant value.

Answer 1

The Correct Option is A

We are given the function \[ f(x) = \sin^{-1} \left( \frac{1 + x^2}{2x} \right) + \cos^{-1} \left( \frac{2x}{1 + x^2} \right). \] Recall that the sum of inverse sine and inverse cosine of complementary angles is \( \frac{\pi}{2} \), i.e. \[\sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2}.\] By manipulating the given expression and using this identity, we can see that the values of \( \sin^{-1} \left( \frac{1 + x^2}{2x} \right) \) and \( \cos^{-1} \left( \frac{2x}{1 + x^2} \right) \) must add up to \( \frac{\pi}{2} \). Therefore, the range of \( f(x) \) is a constant value, specifically \( \frac{\pi}{2} \).