Let P(x) be a real polynomial of degree 3 which vanishes at x = -3. Let P(x) have local minima at x = 1, local maxima at x = -1 and ∫₋₁¹ P(x)dx = 18, then the sum of all the coefficients of the polynomial P(x) is equal to _________.

Solution

Maths sir · 4 steps

Build P from its derivative: the two turning points fix P', then integrate and use the two conditions.

1. Start from P'. P'(x) has degree 2 and vanishes at x = 1 and x = -1, so P'(x) = a(x-1)(x+1) = a(x² - 1). A minimum at x = 1 needs P''(1) = 2a > 0, so a > 0.

   P(x) = a(x³/3 - x) + c

2. Use the integral. The part a(x³/3 - x) is odd, so its integral from -1 to 1 is zero. Only the constant remains: ∫₋₁¹ c dx = 2c = 18, so c = 9.
3. Use the root. P(-3) = a(-9 + 3) + 9 = -6a + 9 = 0, so a = 3/2, which is positive as needed. This gives P(x) = 1/2x³ - 3/2x + 9.
4. So. The sum of all the coefficients is P(1) = 1/2 - 3/2 + 9 = 8.

The common slip: 16. Taking ∫₋₁¹ c dx as c instead of 2c gives c = 18 and a = 3, which leads to 16, not 8.

The idea to keep. When the turning points are given, write P' first and integrate it. Over a symmetric interval the odd terms integrate to zero. The sum of the coefficients of any polynomial is P(1).

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