Match List-I with List-II, where R is the set of real numbers.
| List-I | List-II | ||
| (A) | \( \displaystyle f : \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \to [-1, 1],\ f(x) = \sin x \) | (I) | Onto but not one-one |
| (B) | \(f : [0, \pi] \to R,\ f(x) = \cos x\) | (II) | One-one and onto |
| (C) | \(f : [0, \pi] \to [0, 1],\ f(x) = \sin x\) | (III) | Neither one-one nor onto |
| (D) | \( \displaystyle f : \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \to R,\ f(x) = \cos x \) | (IV) | One-one but not onto |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
(A) \(\sin x\) on \([-\pi/2, \pi/2]\) to \([-1, 1]\)
This matches (II) One-one and onto.
\(\sin x\) increases from \(\sin(-\pi/2) = -1\) to \(\sin(\pi/2) = 1\)
So each value in \([-1, 1]\) is taken exactly once.
(B) \(\cos x\) on \([0, \pi]\) to R
This matches (IV) One-one but not onto.
\(\cos x\) decreases from \(\cos 0 = 1\) to \(\cos \pi = -1\)
So it is one-one, but its range \([-1, 1]\) misses real numbers such as 2.
(C) \(\sin x\) on \([0, \pi]\) to \([0, 1]\)
This matches (I) Onto but not one-one.
\(\sin(\pi/6) = \sin(5\pi/6) = 1/2\)
The values run from 0 up to 1 and back to 0, so every number in \([0, 1]\) is taken.
(D) \(\cos x\) on \([-\pi/2, \pi/2]\) to R
This matches (III) Neither one-one nor onto.
\(\cos(-\pi/3) = \cos(\pi/3) = 1/2\)
The range is \([0, 1]\), which misses real numbers such as \(-1\).