The domain of the function \(f(x) = \mathrm{cosec}^{-1}(x - 2)\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \((-\infty, 1] \cup [3, \infty)\)
\(\mathrm{cosec}^{-1} t\) is defined only when \(t \le -1\) or \(t \ge 1\), because the value of a cosecant is never between \(-1\) and 1. So the argument \(x - 2\) must satisfy one of these.
\(x - 2 \le -1 \;\Rightarrow\; x \le 1\)
\(x - 2 \ge 1 \;\Rightarrow\; x \ge 3\)
Hence the domain is \((-\infty, 1] \cup [3, \infty)\).
Why the other options are incorrect
Option (1) \([1, 3]\) — Incorrect. This is where \(-1 \le x - 2 \le 1\), the domain of \(\sin^{-1}(x - 2)\). For example \(x = 2\) gives \(\mathrm{cosec}^{-1}(0)\), which is not defined.
Option (2) \((-\infty, 1) \cup (3, \infty)\) — Incorrect. It leaves out \(x = 1\) and \(x = 3\), where \(\mathrm{cosec}^{-1}(-1) = -\frac{\pi}{2}\) and \(\mathrm{cosec}^{-1}(1) = \frac{\pi}{2}\) are defined.
Option (3) \((-\infty, -1] \cup [1, \infty)\) — Incorrect. This is the domain of \(\mathrm{cosec}^{-1} x\) itself. It contains \(x = 2\), where the argument is 0.