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

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(B) Given $P(x) = x^4 + ax^3 + bx^2 + cx + d$.$P'(x) = 4x^3 + 3ax^2 + 2bx + c$.Since $x=0$ is the only real root of $P'(x)=0$,we have $P'(0) = 0$,whic

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

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(B) Given $P(x) = x^4 + ax^3 + bx^2 + cx + d$.$P'(x) = 4x^3 + 3ax^2 + 2bx + c$.Since $x=0$ is the only real root of $P'(x)=0$,we have $P'(0) = 0$,which implies $c=0$.Thus,$P'(x) = x(4x^2 + 3ax + 2b)$.Since $x=0$ is the only real root,the quadratic factor $4x^2 + 3ax + 2b$ must have no real roots,meaning its discriminant $D $D = (3a)^2 - 4(4)(2b) = 9a^2 - 32b Since the leading coefficient $4 > 0$,the quadratic $4x^2 + 3ax + 2b > 0$ for all $x \in \mathbb{R}$.Therefore,the sign of $P'(x)$ is determined by the sign of $x$.For $x \in [-1, 0)$,$P'(x) For $x \in (0, 1]$,$P'(x) > 0$,so $P(x)$ is strictly increasing.Since $P(x)$ decreases on $[-1, 0]$ and increases on $[0, 1]$,the global minimum on $[-1, 1]$ occurs at $x=0$.Thus,$P(-1)$ is not the minimum.Since $P(x)$ is increasing on $(0, 1]$ and $P(-1) Therefore,$P(-1)$ is not the minimum,but $P(1)$ is the maximum.

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