Q575: If $f:N\rightarrow W$ is defined as \(f(n)=\begin{cases}\f...

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1 Marks 2025 AISSCE(Board Exam) MCQ SINGLE

If $f:N\rightarrow W$ is defined as $f(n)=\begin{cases}\frac{n}{2},&if~n~is~even\\ 0,&if~n~is~odd\end{cases}$, then f is:

(A) injective only

(B) surjective only

(D) neither surjective nor injective

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Correct Answer: B

Explanation

Solution

Question: If f : N → W is defined by
f(n) = { n/2, if n is even; 0, if n is odd }, then f is:

A. injective only
B. surjective only
C. a bijection
D. neither surjective nor injective

Solution:

Injective (one–one):
By definition, f is injective if f(x₁)=f(x₂) ⇒ x₁=x₂.
Here, f(1)=0, f(3)=0, f(5)=0.
Different natural numbers have the same image, hence f is not injective.

Surjective (onto):
A function f : N → W is surjective if for every y ∈ W, there exists n ∈ N such that f(n)=y.
For y=0, choose any odd n.
For y≥1, choose n=2y ⇒ f(2y)=y.
Thus every element of W has a preimage, so f is surjective.

Conclusion: The function is surjective but not injective.
Correct option: B

AI Tutor Explanation

Detailed Solution

Step 1: Check for Injectivity

A function is injective (one-to-one) if different elements in the domain map to different elements in the codomain. Consider $f(1) = 0$ and $f(3) = 0$. Since $f(1) = f(3)$ but $1 \neq 3$, the function is not injective.

Step 2: Check for Surjectivity

A function is surjective (onto) if every element in the codomain has a corresponding element in the domain. The codomain is $W$ (whole numbers), which includes all non-negative integers. Let $w \in W$. We need to find an $n \in N$ such that $f(n) = w$. If $w = 0$, then $f(1) = 0$, so $0$ is in the range. If $w > 0$, then let $n = 2w$. Since $w$ is a whole number, $n = 2w$ is a natural number and is even. Then $f(n) = f(2w) = \frac{2w}{2} = w$. Thus, every element in $W$ has a pre-image in $N$. Therefore, the function is surjective.

Step 3: Conclusion

Since the function is not injective but is surjective, the correct option is (B).

Final Answer: surjective only

AI Suggestion: Option B

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Pedagogical Audit

Bloom's Analysis: This is an UNDERSTAND question because the student needs to comprehend the definitions of injective and surjective functions and apply them to the given function.

Knowledge Dimension: CONCEPTUAL

Justification: The question requires understanding the concepts of functions, injectivity, and surjectivity, rather than recalling specific facts or performing routine procedures.

Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's understanding of the definitions of injective and surjective functions, which are core concepts in the Relations and Functions chapter.

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