The graph of \(y = \cot^{-1} x\) (principal value branch) lies entirely between the lines
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(y = 0\) and \(y = \pi\)
\(\cot^{-1} x\) is defined for every real x, and its range is \((0, \pi)\).
The function is decreasing: its values come close to \(\pi\) for large negative x and close to 0 for large positive x, but never reach them, because \(\cot 0\) and \(\cot \pi\) are not defined.
\( \displaystyle \cot^{-1}(-1) = \frac{3\pi}{4}, \quad \cot^{-1}(0) = \frac{\pi}{2}, \quad \cot^{-1}(1) = \frac{\pi}{4} \)
So the whole graph lies between the lines \(y = 0\) and \(y = \pi\).
Why the other options are incorrect
Option (1) \(y = -\frac{\pi}{2}\) and \(y = \frac{\pi}{2}\) — Incorrect. These lines bound the graph of \(\tan^{-1} x\); the point \(\left(-1, \frac{3\pi}{4}\right)\) of the \(\cot^{-1}\) graph is above \(y = \frac{\pi}{2}\).
Option (2) \(y = 0\) and \(y = \frac{\pi}{2}\) — Incorrect. \(\cot^{-1}(-1) = \frac{3\pi}{4}\) is above \(\frac{\pi}{2}\).
Option (4) \(x = -1\) and \(x = 1\) — Incorrect. \(\cot^{-1} x\) is defined for all real x, so the graph runs along the whole x-axis, for example through \(\left(\sqrt{3}, \frac{\pi}{6}\right)\).