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Question:

Given P(x) = x⁴ + ax³ + bx² + cx + d such that x = 0 is the only real root of P'(x) = 0. If P(-1) < P(1), then in the interval [-1, 1], P(-1) is the minimum and P(1) is the maximum of P?

P(-1) is the minimum and P(1) is the maximum of P

P(-1) is not minimum but P(1) is the maximum of P

neither P(-1) is the minimum nor P(1) is the maximum of P

P(-1) is the minimum but P(1) is not the maximum of P

Solution:

P(x) = x⁴ + ax³ + bx² + cx + d ⇒ P'(x) = 4x³ + 3ax² + 2bx + c
As P'(x) = 0 has only root x = 0 ⇒ c = 0 ⇒ P(x) = x(4x² + 3ax + 2b) ⇒ 4x³ + 3ax + 2b = 0 has non real root. and 4x² + 3ax + 2b > 0 ∀ ∈ [-1, 1].
At x = 0, f'(x) changes sign from negative to positive. Hence, x = 0 is the point of local minima.
There is no local maxima. Hence, we check at the end points.
As P(-1) < P(1) ⇒ P(1) is the max. of P(x) in [-1, 1]
Hence, option 'B' is correct.