Question
Given P(x) = x4+ ax3 + 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].
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P(–1) is the minimum and P(1) is the maximum of P
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P(–1) is not minimum but P(1) is the maximum of P
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P(–1) is the minimum but P(1) is not the maximum of P
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neither P(–1) is the minimum nor P(1) is the maximum of P
Solution
B.
P(–1) is not minimum but P(1) is the maximum of P
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 ∀ x ∈ [−1, 1].
As P(−1) < P(1) ⇒ P(1) is the max. of P(x) in [−1, 1]
