Question

The function $f : A \to B$ given by $f(x) = x, x \in A$, is one to one but not onto. Then:

• A. $B \subset A$
• B. $A = B$
• C. $A' \subset B'$
• D. $A \subset B$
• E. $A \cap B = \emptyset$

Hint:

For $f(x)=x$, the function behaves like an identity mapping.
• Onto $\Rightarrow A = B$
• Not onto $\Rightarrow A$ is a proper subset of $B$

Answer 1

The Correct Option is D

Concept: For the function $f(x)=x$:
• It is always one-to-one (injective) since different inputs give different outputs.
• It is onto (surjective) only if every element of the codomain $B$ is mapped by some element of $A$.

Step 1: Understand the range of the function.
Since $f(x)=x$, every element maps to itself. Thus, the range of $f$ is exactly the set $A$.

Step 2: Apply the "not onto" condition.
For the function to be not onto, there must exist elements in $B$ that are not images of any element in $A$. Since range $=$ $A$, this means: \[ A \neq B \] and some elements of $B$ lie outside $A$.

Step 3: Check the condition for function definition.
For $f: A \to B$ to be valid, every $x \in A$ must satisfy $f(x) \in B$. Since $f(x)=x$, this requires: \[ x \in B \quad \forall x \in A \] Hence, \[ A \subseteq B \]

Step 4: Combine both conditions.
We have: \[ A \subseteq B \quad \text{and} \quad A \neq B \] Therefore, \[ A \subset B \]