Question

Match List - I with List - II. 

List - I		List - II
A.	Injective function	I.	\(f(x) = x^2\) on \(\mathbb{R}\)
B.	Surjective function	II.	\(f(x) = 2x + 3\) on \(\mathbb{R}\)
C.	Bijective function	III.	\(f(x) = x^3\) on \(\mathbb{R}\)
D.	Non-injective and non-surjective	IV.	Every element of a codomain has a preimage

Choose the correct answer from the options given below:

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

Hint:

On $\mathbb{R}$, $x^2$ is the classic example of a "bad" function: it fails the Horizontal Line Test twice!

Answer 1

The Correct Option is C

Concept:
• Injective (One-to-One): Distinct elements of the domain map to distinct elements of the codomain.
• Surjective (Onto): Every element in the codomain has at least one preimage in the domain.
• Bijective: Both injective and surjective.

Step 1: Analyze f(x) = x\(^2\) on \(\mathbb{R}\) (I)
- Not injective: \(f(1) = 1\) and \(f(-1) = 1\). Two inputs give same output.
- Not surjective: Negative numbers (e.g., -5) in \(\mathbb{R}\) have no real preimage because squares are always non-negative.
So, D-I.

Step 2: Analyze surjective definition (IV)
By definition, a function is surjective (onto) if every element in the codomain has a corresponding preimage in the domain.
So, B-IV.

Step 3: Analyze f(x) = 2x + 3 on \(\mathbb{R}\) (II)
- Injective: \(2x_1 + 3 = 2x_2 + 3 \implies x_1 = x_2\).
- Surjective: For any \(y\), \(x = (y-3)/2\) exists in \(\mathbb{R}\).
Since it is both, it is bijective. In the context of the options, let's look at III first.

Step 4: Analyze f(x) = x\(^3\) on \(\mathbb{R}\) (III)
- Injective: Cube roots are unique.
- Surjective: Range is \((-\infty, \infty)\).
This is also bijective. Looking at Option (C), it pairs A with II and C with III. Since both II and III are technically bijective, they are also injective. This mapping holds.