Question

If $f : \mathbb{R} \to \mathbb{R}$ is a function defined by $f(x) = x^2$, then which of the following is true?

• A. $f$ is 1-1 but not onto
• B. $f$ is onto but not 1-1
• C. $f$ is neither 1-1 nor onto
• D. $f$ is both 1-1 and onto
• E. $f^{-1} : \mathbb{R} \to \mathbb{R}$ exists

Hint:

Quadratic functions over $\mathbb{R}$ are never one-to-one unless domain is restricted.

Answer 1

The Correct Option is C

Concept:
• One-to-one (injective): $f(a)=f(b) \Rightarrow a=b$
• Onto (surjective): Range = codomain

Step 1: Check one-to-one.
\[ f(x) = x^2 \] Take two different values: \[ f(2) = 4, f(-2) = 4 \] \[ f(2) = f(-2) \text{ but } 2 \neq -2 \] Hence, not one-to-one.

Step 2: Check onto.
Range of $x^2$: \[ [0, \infty) \] Codomain: \[ \mathbb{R} \] Negative numbers are not obtained. Thus, not onto.

Step 3: Inverse existence.
Since function is not one-to-one, inverse does not exist over $\mathbb{R}$.
Final Answer: \[ \boxed{\text{neither 1-1 nor onto}} \]