Question

The range of the real-valued function \[ f(x) = \cos^{-1}(-x) + \sin^{-1}(-x) + \csc^{-1}(x) \] is

• A. \( \left\{ 0, \dfrac{\pi}{2} \right\} \)
• B. \( \left[ 0, \dfrac{\pi}{2} \right) \cup \left( \dfrac{\pi}{2}, \pi \right] \)
• C. \( \left( 0, \dfrac{\pi}{2} \right) \)
• D. \( \left\{ 0, \pi \right\} \)

Hint:

Use inverse trigonometric identities and known range restrictions to simplify and evaluate composite inverse functions.

Answer 1

The Correct Option is D

We know:
\[ \cos^{-1}(-x) = \pi - \cos^{-1}(x), \quad \sin^{-1}(-x) = -\sin^{-1}(x) \] So the function becomes:
\[ f(x) = \pi - \cos^{-1}(x) - \sin^{-1}(x) + \csc^{-1}(x) \] Using the identity:
\[ \cos^{-1}(x) + \sin^{-1}(x) = \dfrac{\pi}{2} \Rightarrow f(x) = \pi - \dfrac{\pi}{2} + \csc^{-1}(x) = \dfrac{\pi}{2} + \csc^{-1}(x) \] Now \( \csc^{-1}(x) \in \left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right] \setminus \{ 0 \} \), but domain excludes \( -1 < x < 1 \) Hence, range of \( f(x) \) is:
\[ f(x) = \dfrac{\pi}{2} + \csc^{-1}(x) \] For \( x = \pm 1 \Rightarrow \csc^{-1}(x) = \pm \dfrac{\pi}{2} \), then:
\[ f(x) = \dfrac{\pi}{2} + \left( \pm \dfrac{\pi}{2} \right) = 0 \quad \text{or} \quad \pi \] Thus, range is \( \left\{ 0, \pi \right\} \)