Question

Let $f: \mathbb{R} \to (0,1)$ be a continuous function, then which of the following pair of vectors are linearly dependent for some $x \in (0,1)$?

• A. $\vec{a} = f(x)\hat{i} + 2\hat{j}; \quad \vec{b} = x^2\hat{i} + 3\hat{j}$
• B. $\vec{a} = f(x)\hat{i} + 3\hat{j}; \quad \vec{b} = x^2\hat{i} + 2\hat{j}$
• C. $\vec{a} = \Bigl(\int_{0}^{1-x} f(t)\,dt\Bigr)\hat{i} + 3\hat{j}; \quad \vec{b} = x\hat{i} + 2\hat{j}$
• D. $\vec{a} = \Bigl(\int_{0}^{1-x} f(t)\,dt\Bigr)\hat{i} + 2\hat{j}; \quad \vec{b} = x\hat{i} + 3\hat{j}$

Answer 1

Answer: (B), (C), (D)

B, C and D

Answer 2

To test linear dependence of $\vec a=(a_1,a_2)$ and $\vec b=(b_1,b_2)$, set the determinant to zero:

$$D = \begin{vmatrix} a_1 & b_1\\ a_2 & b_2 \end{vmatrix} = a_1b_2 - a_2b_1 = 0.$$

1. Option A: $D = f(x)\cdot3 - 2\cdot x^2 = 3f(x) - 2x^2$. Since $f(x)\in(0,1)$, $3f(x)\in(0,3)$ and $2x^2\in(0,2)$, so $D>0$. Never zero.
2. Option B: $D = f(x)\cdot2 - 3\cdot x^2 = 2f(x) - 3x^2$. At $x\to0$, $D\approx2f(0)>0$. At $x\to1$, $D=2f(1)-3<2-3=-1<0$. By the Intermediate Value Theorem, $D=0$ for some $x\in(0,1)$.
3. Option C: $D = \bigl(\int_{0}^{1-x}f(t)\,dt\bigr)\cdot2 - 3\cdot x = 2\!\int_{0}^{1-x}f(t)\,dt - 3x$. At $x=0$, $D = 2\!\int_{0}^{1}f>0$. At $x=1$, $D = -3<0$. Hence zero in $(0,1)$.
4. Option D: $D = \bigl(\int_{0}^{1-x}f(t)\,dt\bigr)\cdot3 - 2\cdot x = 3\!\int_{0}^{1-x}f(t)\,dt - 2x$. At $x=0$, positive; at $x=1$, $-2<0$. Again zero by continuity.
   Therefore, options B, C and D are correct.