Match List-I with List-II
Where \(\mathrm{R}\) is set of real numbers
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(f: \mathrm{R} \rightarrow \mathrm{R}\) s.t \(f(x) = x^4\) is | (I) | one-one, Into |
| (B) | \(f: \mathrm{R} \rightarrow [0, \infty)\) s.t \(f(x) = x^4\) is | (II) | many-one, into |
| (C) | \(f: [0, \infty) \rightarrow \mathrm{R}\) s.t \(f(x) = x^4\) is | (III) | one-one, onto |
| (D) | \(f: [0, \infty) \rightarrow [0, \infty)\) s.t \(f(x) = x^4\) is | (IV) | many-one, 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)
The rule is the same in all four items; only the domain and codomain change. On R, \(x^4\) is many-one because \(f(-x) = f(x)\) (for example f(−1) = f(1) = 1). On \([0, \infty)\) it is strictly increasing, so one-one. Its range is \([0, \infty)\) in every case, so it is onto exactly when the codomain is \([0, \infty)\).
(A) \(f: \mathrm{R} \rightarrow \mathrm{R}\)
This matches (II) many-one, into: many-one on R, and negative numbers such as −1 have no pre-image.
(B) \(f: \mathrm{R} \rightarrow [0, \infty)\)
This matches (IV) many-one, onto: many-one on R, and every \(y \ge 0\) is the image of \(x = y^{1/4}\).
(C) \(f: [0, \infty) \rightarrow \mathrm{R}\)
This matches (I) one-one, Into: one-one on \([0, \infty)\), but negative numbers have no pre-image.
(D) \(f: [0, \infty) \rightarrow [0, \infty)\)
This matches (III) one-one, onto: one-one on \([0, \infty)\), and the range equals the codomain.
So the matching is (A)-(II), (B)-(IV), (C)-(I), (D)-(III).