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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

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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.

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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.

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

Question: A function f : R₊ → R (where R₊ is the set of all non-negative real numbers) is defined by f(x) = 4x + 3. Then f is:
(A) one–one but not onto (B) onto but not one–one (C) both one–one and onto (D) neither one–one nor onto

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Solution:
One–one: Let f(x₁) = f(x₂). Then 4x₁ + 3 = 4x₂ + 3 ⇒ x₁ = x₂. Hence, f is one–one.

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Onto: For f to be onto, for every y ∈ R there must exist x ∈ R₊ such that y = 4x + 3 ⇒ x = (y − 3)/4. Since x ≥ 0, we must have y ≥ 3. Thus, values y < 3 are not obtained. Hence, f is not onto R.

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Conclusion: The function is one–one but not onto.
Correct option: (A)

**Correct Option if MCQ:** B **Reasoning:** * Calculate the dot product: \(\vec{a} \cdot \vec{b} = (3)(1) + (2)(-1) + (-1)(1) = 3 - 2 - 1 = 0\). * Since the dot product is zero, the vectors are perpendicular. * \(\vec{a} \perp \vec{b}\)

\((\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|^{2}})\vec{b}\)

\(\frac{2\pi}{3}\)

The position vector of the midpoint \(D\) is the average of the position vectors of \(B\) and \(C\) relative to \(A\):\[\vec{AD} = \frac{\vec{AB} + \vec{AC}}{2}\] \[\vec{AD} = \frac{3}{2}\hat{i} + \frac{5}{2}\hat{k}\] The length of the median is the magnitude of the vector \(\vec{AD}\), denoted as \(|\vec{AD}|\):\[|\vec{AD}| = \sqrt{\left(\frac{3}{2}\right)^2 + \left(0\right)^2 + \left(\frac{5}{2}\right)^2}\] \[|\vec{AD}| = \frac{\sqrt{34}}{2}\]