Match List-I with List-II, where R is the set of real numbers and N is the set of natural numbers.
| List-I | List-II | ||
| (A) | \(f : R \to R,\ f(x) = 4x - 1\) | (I) | One-one but not onto |
| (B) | \(f : N \to N,\ f(x) = x + 5\) | (II) | Onto but not one-one |
| (C) | \(f : R \to R,\ f(x) = x^{2} - 1\) | (III) | Neither one-one nor onto |
| (D) | \(f : R \to [0, \infty),\ f(x) = x^{2}\) | (IV) | One-one and onto |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
(A) \(f : R \to R,\ f(x) = 4x - 1\)
This matches (IV) One-one and onto.
\(4a - 1 = 4b - 1\) gives \(a = b\)
For any real y, the real number below satisfies \(f(x) = y\):
\( \displaystyle x = \frac{y + 1}{4} \)
(B) \(f : N \to N,\ f(x) = x + 5\)
This matches (I) One-one but not onto.
\(a + 5 = b + 5\) gives \(a = b\)
the smallest image is \(f(1) = 6\)
So 1, 2, 3, 4, 5 have no pre-image in N.
(C) \(f : R \to R,\ f(x) = x^{2} - 1\)
This matches (III) Neither one-one nor onto.
\(f(-1) = f(1) = 0\)
\(x^{2} - 1 \ge -1\), so \(-2\) has no pre-image
(D) \(f : R \to [0, \infty),\ f(x) = x^{2}\)
This matches (II) Onto but not one-one.
\(f(-2) = f(2) = 4\)
every \(y \ge 0\) is \(f(\sqrt{y})\)