Match List-I with List-II
Let \(f: \mathrm{A} \rightarrow \mathrm{B}\) be a function given by \(f(x) = x^2\)
| List-I | List-II | ||
|---|---|---|---|
| Domain and Co-domain | Kind | ||
| (A) | \(\mathrm{A} = \mathrm{R}\) and \(\mathrm{B} = \mathrm{R}\) | (I) | \(f\) is both one-one and onto |
| (B) | \(\mathrm{A} = \mathrm{R}\) and \(\mathrm{B} = [0, \infty]\) | (II) | \(f\) is one-one but not onto |
| (C) | \(\mathrm{A} = \mathrm{B} = [0, \infty]\) | (III) | \(f\) is not one-one but onto |
| (D) | \(\mathrm{A} = [0, \infty]\) and \(\mathrm{B} = \mathrm{R}\) | (IV) | \(f\) is neither one-one nor onto |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
For \(f(x) = x^2\), whether f is one-one depends on the domain, and whether it is onto depends on the co-domain:
On R: \(f(-1) = f(1) = 1\), so f is not one-one.
On \([0, \infty)\): \(x_1^2 = x_2^2\) with \(x_1, x_2 \ge 0\) gives \(x_1 = x_2\), so f is one-one.
The range is \([0, \infty)\): every \(y \ge 0\) equals \(f(\sqrt{y})\), and no negative number is a value of f.
(A) \(\mathrm{A} = \mathrm{R}\) and \(\mathrm{B} = \mathrm{R}\)
This matches (IV) neither one-one nor onto: the domain is R, and −1 in the co-domain has no pre-image.
(B) \(\mathrm{A} = \mathrm{R}\) and \(\mathrm{B} = [0, \infty]\)
This matches (III) not one-one but onto: the domain is R, and the co-domain equals the range.
(C) \(\mathrm{A} = \mathrm{B} = [0, \infty]\)
This matches (I) both one-one and onto: the domain has no negative numbers, and the co-domain equals the range.
(D) \(\mathrm{A} = [0, \infty]\) and \(\mathrm{B} = \mathrm{R}\)
This matches (II) one-one but not onto: the domain has no negative numbers, but −1 in the co-domain has no pre-image.
So the matching is (A)-(IV), (B)-(III), (C)-(I), (D)-(II).
Note: It seems that there is a mistake in the original NTA paper, which prints the interval as \([0, \infty]\) when it should read \([0, \infty)\), since \(\infty\) is not a real number and cannot be included in an interval. The question can still be solved because \([0, \infty]\) clearly stands for the set of all non-negative real numbers.