Question

The function \[ f:\mathbb R\to\mathbb R,\qquad f(x)=|x| \] (\(\mathbb R\) is the set of real numbers) is:

• A. Injective but not surjective
• B. Surjective but not injective
• C. Both injective and surjective
• D. Neither injective nor surjective

Hint:

For: \[ f(x)=|x| \] \[ f(a)=f(-a) \] so the function is not injective. Also: \[ |x|\geq0 \] therefore it is not surjective over \(\mathbb R\).

Answer 1

The Correct Option is D

Step 1: Check injective property A function is injective if: \[ f(a)=f(b)\Rightarrow a=b \] Given: \[ f(x)=|x| \] Observe: \[ f(1)=|1|=1 \] and: \[ f(-1)=|-1|=1 \] Thus: \[ f(1)=f(-1) \quad \text{but} \quad 1\neq-1 \] Hence the function is: \[ \text{not injective} \]
Step 2: Check surjective property For surjective function: \[ \text{Range}=\text{Codomain} \] Here codomain is: \[ \mathbb R \] But: \[ |x|\geq0 \quad \forall x\in\mathbb R \] So no negative real number can be obtained. Hence range is: \[ [0,\infty) \] which is not equal to: \[ \mathbb R \] Therefore the function is: \[ \text{not surjective} \] Thus the function is: \[ \text{neither injective nor surjective} \] Option analysis:
• Option (A): Incorrect
• Option (B): Incorrect
• Option (C): Incorrect
• Option (D): Correct Hence: \[ \boxed{\text{(D)}} \]