To solve this problem, we must analyze the behavior of the polynomial \( P(x) = x^4 + ax^3 + bx^2 + cx + d \) given the conditions provided.
The key condition is that \(x = 0\) is the only real root of \(P'(x) = 0\). This indicates that the derivative of \( P(x) \), which is:
\(P'(x) = 4x^3 + 3ax^2 + 2bx + c\)
has only one real root, at \(x = 0\). This implies that the derivative is either always positive or always negative on intervals around \( x = 0 \), except at \( x = 0 \). Since \( P'(x) \) changes signs as it passes through zero (given that there is only one making \( x = 0 \) a local extremum point), it points to \( x = 0 \) being the only critical point where \( P(x) \) changes from increasing to decreasing, or vice versa.
We now need to find the behavior of \( P(x) \) over the interval \([-1, 1]\). Specifically, we evaluate \( P(-1) \) and \( P(1) \), given that \( P(-1) < P(1) \). This means:
\(P(-1) < P(1)\)
To determine if \( P(-1) \) is a minimum or if \( P(1) \) is a maximum, let's consider the possible behaviors:
- Since \( P(x) \) only changes slope at \( x = 0 \) and no other critical points exist in the given range, \( P(-1) \) can be a minimum if the polynomial increases from \( -1 \) to \( 1 \).
- \( P(1) \) cannot be a maximum if \( P(x) \) continues to increase past \( x = 1 \).
With \( P(-1) < P(1) \) and the absence of any critical point other than \( x = 0 \), the only viable scenario for \( P(-1) \) being a minimum and \( P(1) \) not being a maximum is when the polynomial is increasing in the interval \([-1, 0]\) and may continue to increase past \( x = 1 \).
Therefore, based on this analysis, the correct option is:
- $P(-1)$ is the minimum but $P(1)$ is not the maximum of $P$
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