Question

Given \(P(x)=x^{4}+ax^{3}+bx^{2}+cx+d\) such that \(x=0\) is the only real root of \(P^{\prime}(x)=0\). If \(P(-1)\)

• A. $P(-1)$ is the minimum but $P(1)$ is not the maximum of P
• B. $P(-1)$ is not minimum but $P(1)$ is the maximum of P
• C. neither $P(-1)$ is the minimum nor $P(1)$ is the maximum of P
• D. $P(-1)$ is the minimum and $P(1)$ is the maximum of P

Hint:

When an upward-opening quartic curve has only a single real derivative root, its graph behaves visually like a standard parabola with a single absolute global trough. Sketching a quick asymmetrical U-shape matching $P(-1) < P(1)$ reveals the boundary properties at a glance!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The problem asks us to analyze the behavior of a fourth-degree polynomial \( P(x) \) within a specific interval \( [-1, 1] \). The core information provided is about its derivative \( P'(x) \). A fourth-degree polynomial is a continuous and differentiable function everywhere on the real line. The roots of the derivative \( P'(x) = 0 \) are known as critical points, where the function can attain its local maxima or minima. If a polynomial of even degree (like 4) has a positive leading coefficient (which is 1 here) and only one real critical point, that point must necessarily be the global minimum of the function. This is because, for \( x^4 \), as \( x \) approaches positive or negative infinity, the function value grows indefinitely (\( P(x) \to \infty \)).
Step 2: Key Formula or Approach:
1. We differentiate \( P(x) \) to get \( P'(x) = 4x^3 + 3ax^2 + 2bx + c \).
2. Since \( x = 0 \) is a root of \( P'(x) \), we solve for the constant \( c \).
3. We analyze the monotonicity of the function: where the function increases (\( P'(x)>0 \)) and where it decreases (\( P'(x)<0 \)).
4. We evaluate the function at the boundaries and the critical point to find the extremum values in the interval \( [-1, 1] \).
Step 3: Detailed Explanation:
Given \( P(x) = x^4 + ax^3 + bx^2 + cx + d \).
The derivative is \( P'(x) = 4x^3 + 3ax^2 + 2bx + c \).
We are told \( x = 0 \) is the only real root of \( P'(x) = 0 \).
Substituting \( x = 0 \) into the derivative equation:
\[ 4(0)^3 + 3a(0)^2 + 2b(0) + c = 0 \implies c = 0 \]
Now, the derivative is \( P'(x) = x(4x^2 + 3ax + 2b) \).
Since \( x = 0 \) is the only real root, the quadratic part \( 4x^2 + 3ax + 2b \) must not have any real roots (its discriminant \( D<0 \)).
If a fourth-degree polynomial has a positive leading coefficient and only one critical point at \( x = 0 \), the function must be decreasing for all \( x<0 \) and increasing for all \( x>0 \).
This means:
- For \( x \in [-1, 0) \), \( P'(x)<0 \), so \( P(x) \) is strictly decreasing.
- For \( x \in (0, 1] \), \( P'(x)>0 \), so \( P(x) \) is strictly increasing.
- At \( x = 0 \), \( P(0) \) is the absolute minimum of the function on the interval \( [-1, 1] \).
Now we check the maximum. Since the function decreases from \( -1 \) to \( 0 \) and then increases from \( 0 \) to \( 1 \), the maximum value in the interval \( [-1, 1] \) must occur at one of the boundaries: either \( P(-1) \) or \( P(1) \).
The problem states that \( P(-1)<P(1) \).
Consequently, between the two boundary values, \( P(1) \) is the larger one.
Therefore, the maximum of \( P(x) \) on \( [-1, 1] \) is \( P(1) \).
As \( P(0) \) is the minimum, and \( 0 \in (-1, 1) \), \( P(-1) \) cannot be the minimum because \( P(0)<P(-1) \).
Step 4: Final Answer:
In the given interval, the global minimum is at \( x=0 \). Thus, \( P(-1) \) is not the minimum. Since \( P(1)>P(-1) \), \( P(1) \) is the maximum value. This corresponds to choice (B).