Key Concepts and Formulas

• First Derivative Test: The sign of the first derivative $P'(x)$ indicates whether the function $P(x)$ is increasing or decreasing. If $P'(x) > 0$, $P(x)$ is increasing; if $P'(x) < 0$, $P(x)$ is decreasing. Local extrema occur where $P'(x) = 0$ or is undefined.
• Quadratic Discriminant: For a quadratic equation $ax^2 + bx + c = 0$, the discriminant is $D = b^2 - 4ac$. If $D < 0$, the quadratic has no real roots.
• Polynomial Roots: A polynomial of degree $n$ has at most $n$ real roots.

Step-by-Step Solution

Step 1: Find the derivative and use the given information.

We are given the polynomial $P(x) = x^4 + ax^3 + bx^2 + cx + d$. We need to find its derivative, $P'(x)$, and use the information that $x=0$ is the only real root of $P'(x) = 0$. $P'(x) = 4x^3 + 3ax^2 + 2bx + c$ Since $x = 0$ is a root of $P'(x) = 0$, we substitute $x=0$ into the derivative: $P'(0) = 4(0)^3 + 3a(0)^2 + 2b(0) + c = 0$ This implies $c = 0$. Now we have: $P'(x) = 4x^3 + 3ax^2 + 2bx = x(4x^2 + 3ax + 2b)$ Since $x=0$ is the only real root of $P'(x) = 0$, the quadratic $4x^2 + 3ax + 2b$ must have no real roots.

Step 2: Analyze the quadratic factor.

For the quadratic $4x^2 + 3ax + 2b$ to have no real roots, its discriminant must be negative. The discriminant, $D$, is given by: $D = (3a)^2 - 4(4)(2b) = 9a^2 - 32b$ So, we must have $9a^2 - 32b < 0$, which means $9a^2 < 32b$, or $b > \frac{9a^2}{32}$.

Step 3: Analyze the sign of $P'(x)$.

Since $4x^2 + 3ax + 2b$ has no real roots and its leading coefficient is positive, it is always positive. Therefore, the sign of $P'(x)$ is determined by the sign of $x$.

• If $x < 0$, then $P'(x) < 0$, so $P(x)$ is decreasing.
• If $x > 0$, then $P'(x) > 0$, so $P(x)$ is increasing.

This means that $x=0$ is a local minimum of $P(x)$.

Step 4: Use the given inequality $P(-1) < P(1)$.

We have $P(x) = x^4 + ax^3 + bx^2 + d$. Therefore, $P(-1) = (-1)^4 + a(-1)^3 + b(-1)^2 + d = 1 - a + b + d$ $P(1) = (1)^4 + a(1)^3 + b(1)^2 + d = 1 + a + b + d$ We are given that $P(-1) < P(1)$, so: $1 - a + b + d < 1 + a + b + d$ $-a < a$ $2a > 0$ $a > 0$

Step 5: Determine the behavior of $P(x)$ on the interval $[-1, 1]$.

Since $a > 0$, and $P'(x) = x(4x^2 + 3ax + 2b)$, we know that $P'(x) < 0$ for $x \in [-1, 0)$ and $P'(x) > 0$ for $x \in (0, 1]$. Thus, $P(x)$ is decreasing on $[-1, 0]$ and increasing on $[0, 1]$. This means $P(x)$ has a local minimum at $x=0$.

Since $P(x)$ is decreasing on $[-1, 0]$ and increasing on $[0, 1]$, $P(-1)$ is not the minimum value of $P(x)$ on $[-1,1]$ because $P(0)$ is the minimum. Now we need to check whether $P(1)$ is the maximum. Consider what happens as we increase from 0 to 1. The function is strictly increasing. So $P(1)$ is greater than $P(x)$ for all $x$ between 0 and 1. Similarly, $P(-1)$ is greater than $P(x)$ for all $x$ between -1 and 0. Since $P(-1) < P(1)$, we conclude that $P(1)$ is the maximum value of $P(x)$ on $[-1, 1]$.

Step 6: Conclusion

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

Common Mistakes & Tips

• Remember to consider the discriminant of the quadratic factor of $P'(x)$ to ensure it has no real roots.
• Don't forget that the sign of the leading coefficient of the quadratic factor determines whether it's always positive or always negative when it has no real roots.
• Paying close attention to the signs of derivatives is crucial for determining increasing/decreasing behavior.

Summary

We analyzed the derivative of the given polynomial, using the information that $x=0$ is the only real root of $P'(x)=0$. This allowed us to determine that $c=0$ and that the quadratic factor of $P'(x)$ has no real roots. Then, we used the given inequality $P(-1) < P(1)$ to find that $a > 0$. Finally, we concluded that $P(-1)$ is not the minimum, but $P(1)$ is the maximum on the interval $[-1, 1]$.

Final Answer The final answer is \boxed{P(-1) \text{ is not minimum but } P(1) \text{ is the maximum of } P}, which corresponds to option (A).