Question

Match the LIST-I with LIST-II

	LIST-I Function		LIST-II Range (Principal value)
A.	\(\tan^{-1}x\)	I.	\([0,\pi]-\left\{\frac{\pi}{2}\right\}\)
B.	\(\sec^{-1}x\)	II.	\(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)
C.	\(\operatorname{cosec}^{-1}x\)	III.	\(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]-\{0\}\)
D.	\(\sin^{-1}x\)	IV.	\(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\)

Choose the correct answer from the options given below:

• A. A-IV, B-I, C-III, D-II
• B. A-II, B-III, C-I, D-IV
• C. A-IV, B-III, C-I, D-II
• D. A-II, B-IV, C-III, D-I

Hint:

Recall the principal value branches. Remove the angle where the trigonometric ratio is undefined.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Each inverse trigonometric function has a fixed principal value branch, chosen so that the function is one-one and onto. These ranges are standard facts from the NCERT chapter on inverse trigonometric functions.

Step 2: Check A: tan inverse x.
\(\tan^{-1}x\) takes all real \(x\) and gives angles strictly between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\). The end points are not reached. So the range is the open interval in IV.

Step 3: Check B: sec inverse x.
\(\sec^{-1}x\) is defined for \(|x| \geq 1\). Its range is \([0,\pi]\) but \(\frac{\pi}{2}\) is left out, because \(\sec\frac{\pi}{2}\) is undefined. This is I.

Step 4: Check C: cosec inverse x.
\(\operatorname{cosec}^{-1}x\) is defined for \(|x| \geq 1\). Its range is \([-\frac{\pi}{2},\frac{\pi}{2}]\) without 0, because \(\operatorname{cosec}0\) is undefined. This is III.

Step 5: Check D: sin inverse x.
\(\sin^{-1}x\) is defined for \(-1 \leq x \leq 1\). Its range is the closed interval \([-\frac{\pi}{2},\frac{\pi}{2}]\). This is II.

Step 6: Compare with the options.
We have A-IV, B-I, C-III, D-II. This is exactly option 1. The other options pair A with II, or B with III, or D with I, and these are all wrong.

Final Answer:
The matching is A-IV, B-I, C-III, D-II. \[ \boxed{\text{Option 1}} \]