If p(x) be a polynomial of degree three that has a local maximum value 8 at x = 1 and a local minimum value 4 at x = 2; then p(0) is equal to :

Solution

Maths sir · 5 steps

Start from the derivative: x = 1 and x = 2 are its roots, so integrate, then fix k and C.

1. Build the derivative. p(x) is cubic, so p'(x) is quadratic. It is zero at both turning points x = 1 and x = 2, so p'(x) = k(x-1)(x-2) = k(x² - 3x + 2).
2. Integrate. Integrating gives p(x) with two unknowns, k and the constant C. Note that C = p(0), which is the number we want.

   p(x) = k(x³/3 - 3x²/2 + 2x) + C

3. Use both values. p(1) = 5/6k + C = 8 and p(2) = 2/3k + C = 4. Subtracting gives 1/6k = 4, so k = 24.
4. Check the maximum. p''(x) = k(2x - 3), so p''(1) = -24 < 0. That makes x = 1 a local maximum and x = 2 a local minimum, as required.
5. So. C = 8 - 5/6 × 24 = 8 - 20 = -12, so p(0) = -12, which is option (c).

The idea to keep. When a polynomial's turning points are given, write p'(x) as a product using those roots first. Then integrate and use the given values to find the leftover constants.

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