Question:

Step 1: Two vectors $\vec{a} = p\hat{i} + q\hat{j}$ and $\vec{b} = r\hat{i} + s\hat{j}$ are linearly dependent iff $ps - qr = 0$. Step 2: Check option 1. Linear dependence requires $3f(x) - 2x^2 = 0$, i.e., $f(x) = \dfrac{2x^2}{3}$. For $x \in (0,1)$, $\dfrac{2x^2}{3} \in (0, \frac{2}{3}) \subset (0,1)$. This could hold, but we need to verify using IVT. Step 3: Check option 2. Linear dependence requires $2f(x) - 3x^2 = 0$, i.e., $f(x) = \dfrac{3x^2}{2}$. For $x \in (0,1)$, $\dfrac{3x^2}{2}$ can exceed 1 (e.g., at $x$ near 1), so $f(x) = \dfrac{3x^2}{2}$ may not always have a solution in $(0,1)$ since $f: R \to (0,1)$. Step 4: Check option 3. Let $h(x) = 2\displaystyle\int_0^{1-x} f(t)\,dt - 3x$. Linear dependence requires $h(x) = 0$. - At $x = 0$: $h(0) = 2\int_0^1 f(t)\,dt > 0$ (since $f > 0$). - At $x = 1$: $h(1) = 2\int_0^0 f(t)\,dt - 3 = -3 < 0$. By IVT, $\exists\, x \in (0,1)$ such that $h(x) = 0$. Option 3 is correct. Step 5: Check option 4. Let $h(x) = 3\displaystyle\int_0^{1-x} f(t)\,dt - 2x$. Linear dependence requires $h(x) = 0$. - At $x = 0$: $h(0) = 3\int_0^1 f(t)\,dt > 0$. - At $x = 1$: $h(1) = 3\int_0^0 f(t)\,dt - 2 = -2 < 0$. By IVT, $\exists\, x \in (0,1)$ such that $h(x) = 0$. Option 4 is correct.