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

Concept: The derivative of a polynomial function determines its stationary critical points and interval monotonicity profile. For a fourth-degree polynomial function $P(x)$ with a positive leading coefficient, the behavior at the absolute limits dictates that $P(x) \rightarrow \infty$ as $x \rightarrow \pm\infty$. If its derivative $P'(x) = 0$ possesses exactly one unique real root, that critical coordinate point must represent the absolute global minimum of the functional curve. Step 1: Analyze the critical point and monotonicity.
Differentiating the given polynomial function with respect to $x$: $$P'(x) = 4x^3 + 3ax^2 + 2bx + c$$ We are given that $x = 0$ is the only real root of $P'(x) = 0$. This implies that the derivative transforms its sign exclusively at $x = 0$. Since the polynomial opens upwards globally ($4 > 0$), the function must be:
• Strictly decreasing on the interval $(-\infty, 0)$
• Strictly increasing on the interval $(0, \infty)$

Step 2: Evaluate interval boundary conditions on $[-1, 1]$.
Because the curve strictly transitions from decreasing to increasing at the origin, the value $P(0)$ constitutes the absolute minimum of the function on any interval surrounding zero. Therefore, within the closed domain interval $[-1, 1]$:
• The true minimum is located at $x = 0$, meaning $P(-1)$ cannot be the minimum value.
• Since the function increases continuously as $x$ moves rightward from $0 \rightarrow 1$, we have $P(1) > P(x)$ for all $x \in [0, 1)$.

Step 3: Reconcile with the given boundary inequality.
The problem states that $P(-1) < P(1)$. Since the function decreases monotonically on $[-1, 0]$, every single functional value in that negative section satisfies $P(-1) \ge P(x) \ge P(0)$. Combining our two boundary bounds shows that $P(1)$ is strictly greater than all other values distributed inside the set: $$P(1) > P(-1) \ge P(x) \quad \forall x \in [-1, 1)$$ Thus, $P(1)$ is definitively the absolute maximum element of the function on this interval, while $P(-1)$ is not the minimum, which perfectly aligns with option (B).