Consider the following statements about the principal value branches:
(A) The graph of \(y = \sec^{-1} x\) comes closer and closer to the line \(y = \displaystyle\frac{\pi}{2}\) as \(|x|\) increases, but never meets it.
(B) \(\mathrm{cosec}^{-1} x\) is defined at \(x = \displaystyle\frac{1}{2}\).
(C) For \(x \ge 1\), \(\sec^{-1} x\) increases as x increases.
(D) The graph of \(y = \sec^{-1} x\) is symmetric about the y-axis.
Which of the above statements are correct?
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) and (C) only
\(\sec^{-1} x = \theta\) means \(\cos\theta = \frac{1}{x}\) with \(\theta\) in \([0, \pi] - \left\{\frac{\pi}{2}\right\}\). Both \(\sec^{-1}\) and \(\mathrm{cosec}^{-1}\) are defined only for \(|x| \ge 1\). Some values:
\( \displaystyle \sec^{-1} 1 = 0, \quad \sec^{-1} 2 = \frac{\pi}{3} \)
\( \displaystyle \sec^{-1}(-2) = \frac{2\pi}{3}, \quad \sec^{-1}(-1) = \pi \)
So for \(x \ge 1\) the values rise from 0 towards \(\frac{\pi}{2}\), for \(x \le -1\) they lie above \(\frac{\pi}{2}\), the value \(\frac{\pi}{2}\) itself is never taken, and nothing is defined for \(|x| \lt 1\).
Checking the statements
(A) The graph of \(y = \sec^{-1} x\) comes closer and closer to \(y = \frac{\pi}{2}\) as \(|x|\) increases, but never meets it. — Correct. As \(|x|\) grows, \(\cos\theta = \frac{1}{x}\) tends to 0, so \(\theta\) tends to \(\frac{\pi}{2}\), which is excluded.
(B) \(\mathrm{cosec}^{-1} x\) is defined at \(x = \frac{1}{2}\). — Incorrect. A cosecant is never between \(-1\) and 1, so \(\frac{1}{2}\) is outside the domain.
(C) For \(x \ge 1\), \(\sec^{-1} x\) increases as x increases. — Correct. As x increases, \(\cos\theta = \frac{1}{x}\) decreases, so \(\theta\) increases; for example \(\sec^{-1} 1 = 0 \lt \sec^{-1} 2 = \frac{\pi}{3}\).
(D) The graph of \(y = \sec^{-1} x\) is symmetric about the y-axis. — Incorrect. \(\sec^{-1} 2 = \frac{\pi}{3}\) but \(\sec^{-1}(-2) = \frac{2\pi}{3}\).