Consider the following statements:
(A) For \(-1 \le x \le 0\), \(\cos^{-1} x\) takes all the values in \(\left[\displaystyle\frac{\pi}{2}, \pi\right]\).
(B) The greatest value of \(\tan^{-1} x\) is \(\displaystyle\frac{\pi}{2}\).
(C) The least value of \(\sec^{-1} x\) is 0.
(D) The greatest value of \(\sin^{-1} x\) is \(\displaystyle\frac{\pi}{2}\).
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), (C) and (D) only
A greatest or least value must actually be taken by the function at some x in its domain. An end of the range that is only approached does not count.
\( \displaystyle \cos^{-1}(-1) = \pi, \quad \cos^{-1}(0) = \frac{\pi}{2}, \quad \sec^{-1}(1) = 0, \quad \sin^{-1}(1) = \frac{\pi}{2} \)
So \(\cos^{-1}\) runs from \(\pi\) down to \(\frac{\pi}{2}\) on \([-1, 0]\), \(\sec^{-1}\) reaches 0 and \(\sin^{-1}\) reaches \(\frac{\pi}{2}\), while \(\tan^{-1}\) never reaches \(\frac{\pi}{2}\).
Checking the statements
(A) For \(-1 \le x \le 0\), \(\cos^{-1} x\) takes all the values in \(\left[\frac{\pi}{2}, \pi\right]\). — Correct. \(\cos^{-1}\) is continuous and decreasing, from \(\pi\) at \(x = -1\) to \(\frac{\pi}{2}\) at \(x = 0\).
(B) The greatest value of \(\tan^{-1} x\) is \(\frac{\pi}{2}\). — Incorrect. The range is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\); \(\frac{\pi}{2}\) is approached but never taken, so there is no greatest value.
(C) The least value of \(\sec^{-1} x\) is 0. — Correct. \(\sec^{-1}(1) = 0\), and no value of \(\sec^{-1}\) is negative.
(D) The greatest value of \(\sin^{-1} x\) is \(\frac{\pi}{2}\). — Correct. \(\sin^{-1}(1) = \frac{\pi}{2}\), the right end of \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).