<p>Given function \(f: \mathbb{N} \to \mathbb{Z}\) such that<br>
\[f(x) = \begin{cases} \dfrac{n-1}{2}, & \text{when } n \text{ is odd} \\ -\dfrac{n}{2}, & \text{when } n \text{ is even} \end{cases}\]<br>
Which of the following is correct?</p>
<p>(A) \(f\) is one-one and onto</p>
<p>(B) \(f\) is many-one and onto</p>
<p>(C) \(f\) is one-one and into</p>
<p>(D) \(f\) is many-one and into</p>
Step-by-Step Solution
Key Concept: Systematically evaluate f for multiple natural numbers to determine the range and verify surjectivity onto ℤ. Recognize that odd n produces non-negative integers while even n produces negative integers, together covering all of ℤ.
<p><strong>Step 1:</strong> Evaluate f for odd natural numbers.</p><p>For n = 1: f(1) = (1-1)/2 = 0</p><p>For n = 3: f(3) = (3-1)/2 = 1</p><p>For n = 5: f(5) = (5-1)/2 = 2</p><p>For n = 7: f(7) = (7-1)/2 = 3</p><p>Pattern: Odd inputs {1, 3, 5, 7, ...} map to {0, 1, 2, 3, ...} = ℕ ∪ {0}</p></p><p><strong>Step 2:</strong> Evaluate f for even natural numbers.</p><p>For n = 2: f(2) = -2/2 = -1</p><p>For n = 4: f(4) = -4/2 = -2</p><p>For n = 6: f(6) = -6/2 = -3</p><p>For n = 8: f(8) = -8/2 = -4</p><p>Pattern: Even inputs {2, 4, 6, 8, ...} map to {-1, -2, -3, -4, ...} = ℤ⁻</p></p><p><strong>Step 3:</strong> Determine the range.</p><p>Range of f = {0, 1, 2, 3, ...} ∪ {-1, -2, -3, -4, ...} = ℤ</p><p>Therefore f is <strong>surjective</strong> (onto).</p><p><strong>Step 4:</strong> Check injectivity.</p><p>For distinct m, n ∈ ℕ: if m and n have the same parity, different values yield different outputs. Since odd and even inputs map to disjoint sets (non-negative and negative respectively), f is <strong>injective</strong> (one-to-one).</p><p>∴ f is <strong>bijective</strong>. Answer: B</p>
Correct Answer: B