If \(x \le -1\), then the set of all values of \(\sec^{-1} x\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\left(\displaystyle\frac{\pi}{2}, \pi\right]\)
\(\sec^{-1} x = \theta\) means \(\sec\theta = x\) with \(\theta\) in \([0, \pi] - \left\{\frac{\pi}{2}\right\}\). For \(x \le -1\) the secant is negative, so \(\cos\theta = \frac{1}{x}\) is negative:
\( \displaystyle -1 \le \cos\theta \lt 0 \)
In \([0, \pi]\) this happens exactly for \(\frac{\pi}{2} \lt \theta \le \pi\).
The closed end \(\pi\) is taken: \(\sec^{-1}(-1) = \pi\). The open end \(\frac{\pi}{2}\) is only approached as x becomes a large negative number, because \(\sec\frac{\pi}{2}\) is not defined.
Hence the set of values is \(\left(\frac{\pi}{2}, \pi\right]\).
Why the other options are incorrect
Option (2) \(\left[\frac{\pi}{2}, \pi\right]\) — Incorrect. It includes \(\frac{\pi}{2}\), but \(\sec\frac{\pi}{2}\) is not defined, so \(\sec^{-1}\) never takes this value.
Option (3) \(\left(\frac{\pi}{2}, \pi\right)\) — Incorrect. It leaves out \(\pi\), which is taken at \(x = -1\).
Option (4) \(\left[0, \frac{\pi}{2}\right)\) — Incorrect. These are the values for \(x \ge 1\), for example \(\sec^{-1} 2 = \frac{\pi}{3}\).