Let f : [-1, 1] → R be defined as f(x) = ax² + bx + c for all x∈[-1, 1], where a, b, c∈R such that f(-1) = 2, f'(-1) = 1 for x∈(-1, 1) the maximum value of f ''(x) is 1/2. If f(x) ≤ α, x∈[-1, 1], then the least value of α is equal to _________.

Solution

Maths sir · 4 steps

For a quadratic, f'' is a constant, so it gives a at once. Then two conditions fix b and c.

1. Find a. Here f''(x) = 2a is the same for every x, so its maximum value is just 2a. Setting 2a = 1/2 gives a = 1/4.
2. Find b and c. f'(x) = 2ax + b, so f'(-1) = -1/2 + b = 1, which gives b = 3/2. Then f(-1) = 1/4 - 3/2 + c = 2, which gives c = 13/4.

   f(x) = x²/4 + 3x/2 + 13/4

3. Where is the maximum. f'(x) = (x + 3)/2 is positive for every x in [-1, 1], so f is increasing there. The vertex x = -3 lies outside the interval, so the largest value is at the right end, x = 1.
4. So. f(1) = 1/4 + 6/4 + 13/4 = 20/4 = 5. Since f(x) ≤ α must hold on the whole interval, the least value of α is 5.

The common slip: 6. Taking f''(x) = a instead of 2a makes a = 1/2, b = 2, c = 7/2, and gives f(1) = 6, not 5.

The idea to keep. The least α with f(x) ≤ α is the maximum of f on the interval. If the vertex lies outside the interval, check the sign of f' and take the right endpoint.

Answer from the NTA final answer key. Solution by PrepWiser.