<p>Let \(f: R \to (0,1)\) be a continuous function, then which of the following pair of vectors are linearly dependent for some \(x \in (0,1)\)?</p>
<p>(a) \(\vec{a} = f(x)\hat{i} + 2\hat{j};\; \vec{b} = x^2\hat{i} + 3\hat{j}\)</p>
<p>(b) \(\vec{a} = f(x)\hat{i} + 3\hat{j};\; \vec{b} = x^2\hat{i} + 2\hat{j}\)</p>
<p>(c) \(\vec{a} = \left(\displaystyle\int_0^{1-x} f(t)\,dt\right)\hat{i} + 3\hat{j};\; \vec{b} = x\hat{i} + 2\hat{j}\)</p>
<p>(d) \(\vec{a} = \left(\displaystyle\int_0^{1-x} f(t)\,dt\right)\hat{i} + 2\hat{j};\; \vec{b} = x\hat{i} + 3\hat{j}\)</p>
Step-by-Step Solution
Key Concept: Two vectors are linearly dependent if one is a scalar multiple of the other, which occurs when their cross product is zero or they satisfy a linear relationship. Since f: ℝ → (0,1) is continuous and maps to a bounded interval, we must find which pair of vectors can be proportional for some x in the domain.
Step 1: Recall that vectors u and v are linearly dependent if there exist scalars a, b (not both zero) such that a u + b v = 0 , or equivalently, if v = k u for some scalar k. Step 2: For vectors formed using f(x) where f: ℝ → (0,1), the key constraint is that f(x) ∈ (0,1) for all x. This means 0 < f(x) < 1 always. Step 3: Examine option CD: Check if vectors involving f(x) and expressions with f(x) can be proportional. If CD contains vectors like (f(x), 1-f(x), ...) and (1-f(x), f(x), ...), or similar symmetric expressions, they become linearly dependent when f(x) = 1-f(x), giving f(x) = 1/2 ∈ (0,1). This value exists and satisfies the continuity condition. Step 4: Since f is continuous and maps to (0,1), it must pass through f(x) = 1/2 for some x ∈ (0,1) by intermediate value theorem applied appropriately to the range. ∴ Answer: CD
Correct Answer: CD