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

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(B) 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 the minimum and $P(1)$ is the maximum value of $P$.

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(B) 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 $P'(0) = 0$,we have $c = 0$. Thus,$P'(x) = 4x^3 + 3ax^2 + 2bx = x(4x^2 + 3ax + 2b)$. Since $x=0$ is the only real root,the quadratic $4x^2 + 3ax + 2b = 0$ must have no real roots or a repeated root at $x=0$. If $4x^2 + 3ax + 2b = 0$ has no real roots,then the discriminant $D = (3a)^2 - 4(4)(2b) = 9a^2 - 32b Since $P'(x)$ changes sign at $x=0$ (as it is a cubic with only one real root),$x=0$ is a point of local extremum. For $x  0$,$P'(x) > 0$,so $x=0$ is a local minimum. Since $P(x)$ is a polynomial of degree $4$ with a positive leading coefficient,it is continuous on $[-1, 1]$. The absolute minimum must occur at $x=0$ (since it is the only critical point). Given $P(-1) Therefore,$P(-1)$ is the minimum value and $P(1)$ is the maximum value of $P(x)$ on $[-1, 1]$.

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