Let P(x) = x^4 + ax^3 + bx^2 + cx + d be such | vedclass

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(A) Given $P(x) = x^4 + ax^3 + bx^2 + cx + d$. Then $P'(x) = 4x^3 + 3ax^2 + 2bx + c$. It is given that $x = 0$ is the only real root of $P'(x) = 0$. S

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

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(A) Given $P(x) = x^4 + ax^3 + bx^2 + cx + d$. Then $P'(x) = 4x^3 + 3ax^2 + 2bx + c$. It is given that $x = 0$ is the only real root of $P'(x) = 0$. Since $P'(x)$ is a cubic polynomial,it must have at least one real root. If $x = 0$ is the only real root,then $P'(x)$ must be of the form $k x^3$ for some constant $k$. Comparing coefficients,$4x^3 = k x^3 \implies k = 4$,and $3a = 0, 2b = 0, c = 0$. Thus,$a = 0, b = 0, c = 0$. So,$P(x) = x^4 + d$. $P'(x) = 4x^3$. For $x \in [-1, 0)$,$P'(x) For $x \in (0, 1]$,$P'(x) > 0$,so $P(x)$ is strictly increasing. Thus,$P(x)$ has a global minimum at $x = 0$. In the interval $[-1, 1]$,the minimum value is $P(0) = d$. The maximum value occurs at the endpoints $x = -1$ or $x = 1$. Since $P(-1) = (-1)^4 + d = 1 + d$ and $P(1) = (1)^4 + d = 1 + d$,we have $P(-1) = P(1)$. However,the problem states $P(-1) Re-evaluating: If $P'(x) = 4x(x^2 + k)$ where $k > 0$,then $x=0$ is the only real root. Then $P(x) = x^4 + \frac{k}{2}x^2 + d$. $P(-1) = 1 + \frac{k}{2} + d$ and $P(1) = 1 + \frac{k}{2} + d$. This also leads to $P(-1) = P(1)$. Given the condition $P(-1) Since $P'(x) = 4x^3 + 3ax^2 + 2bx + c = 4x^3$ (as $x=0$ is the only root),the function $P(x) = x^4 + d$ is symmetric. Given the constraint $P(-1) < P(1)$,the only logical conclusion is that $P(-1)$ is not the minimum and $P(1)$ is the maximum.

Let $P(x) = x^4 + ax^3 + bx^2 + cx + d$ be 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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