Quadratic Equations Question - None

Question

The second degree polynomial $f(x)$, satisfying $f(0) = 0$, $f(1) = 1$, $f'(x) \geq 0; x \in (0, 1)$ is

Accepted Answer

Step 1: Set up the general form

Let f(x) = Ax² + Bx + C (second-degree polynomial)

Step 2: Apply boundary condition f(0) = 0

f(0) = C = 0, so f(x) = Ax² + Bx

Step 3: Apply boundary condition f(1) = 1

f(1) = A + B = 1, so B = 1 - A

Thus f(x) = Ax² + (1-A)x

Step 4: Apply monotonicity condition f'(x) ≥ 0 on (0,1)

f'(x) = 2Ax + (1-A)

For f'(x) ≥ 0 on (0,1), we need: 2Ax + (1-A) ≥ 0 for all x ∈ (0,1)

Step 5: Find constraints on A

The minimum of f'(x) on (0,1) occurs at x = 0 (if A > 0) or is constantly negative (if A < 0):

• At x = 0: f'(0) = 1 - A ≥ 0 ⟹ A ≤ 1

• At x = 1: f'(1) = 2A + 1 - A = A + 1. For this to be non-negative: A ≥ -1

• Since we need a true second-degree polynomial: A ≠ 0

• For f'(x) ≥ 0 throughout (0,1) with A > 0: minimum value 1 - A ≥ 0 ⟹ A ≤ 1

• Combining: 0 < A ≤ 1, or setting a = A, we get a ∈ (0,1]

Step 6: Verify the answer form

With substitution a = A, we get f(x) = ax + (1-a)x² where a ∈ (0,1)

Note: a cannot equal 1 (degenerate to linear), and a must be positive (need second-degree term)

∴ Answer: B

Question: