The domain of the function \(f(x) = \sec^{-1}(2x)\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\left(-\infty, -\displaystyle\frac{1}{2}\right] \cup \left[\displaystyle\frac{1}{2}, \infty\right)\)
The domain of \(\sec^{-1} t\) is \(\mathrm{R} - (-1, 1)\), that is, \(t \le -1\) or \(t \ge 1\). With \(t = 2x\):
\(2x \le -1\) or \(2x \ge 1\)
Dividing by 2:
\( \displaystyle x \le -\frac{1}{2} \) or \( \displaystyle x \ge \frac{1}{2} \)
Hence the domain is \(\left(-\infty, -\frac{1}{2}\right] \cup \left[\frac{1}{2}, \infty\right)\).
Why the other options are incorrect
Option (1) \(\left[-\frac{1}{2}, \frac{1}{2}\right]\) — Incorrect. This uses the domain rule of \(\sin^{-1}\); for \(x = 0\), \(\sec^{-1}(0)\) is not defined.
Option (3) \((-\infty, -2] \cup [2, \infty)\) — Incorrect. It multiplies by 2 instead of dividing; it misses \(x = 1\), where \(\sec^{-1}(2) = \frac{\pi}{3}\) is defined.
Option (4) \(\mathrm{R} - \{0\}\) — Incorrect. It includes \(x = \frac{1}{4}\), where \(2x = \frac{1}{2}\) lies inside \((-1, 1)\).