Question:

<div class="solution"> <p><strong>Step 1:</strong> We are given a differentiable function \( y = P(x) \) defined for all \( x \in [0, \infty) \), and it satisfies the inequality \( \frac{d}{dx}(P(x)) + (x-1)^3 \geq P(x) + 1 \) for all \( x \in [0, \infty) \). Additionally, we know that \( P(x) \leq x^3 + 3x + 1 \) for all \( x \in [0, \infty) \) and \( P(0) = 1 \).</p> <p><strong>Step 2:</strong> To analyze the properties of \( P(x) \), let's first consider the given inequality \( \frac{d}{dx}(P(x)) + (x-1)^3 \geq P(x) + 1 \). We can rearrange this inequality to \( \frac{d}{dx}(P(x)) - P(x) \geq 1 - (x-1)^3 \). This form suggests that the derivative of \( P(x) \) is related to \( P(x) \) itself and a cubic term involving \( x \). To further understand \( P(x) \), we should examine its behavior over the interval \( [0, \infty) \) and utilize the given condition \( P(0) = 1 \) as an initial value.</p> <p><strong>Step 3:</strong> Let's define a new function \( Q(x) = P(x) - x^3 - 3x - 1 \). Since \( P(x) \leq x^3 + 3x + 1 \), we have \( Q(x) \leq 0 \) for all \( x \in [0, \infty) \). Moreover, \( Q(0) = P(0) - 1 = 0 \). We can use this function to simplify the given inequality and derive properties of \( P(x) \).</p> <p><strong>Step 4:</strong> By substituting \( P(x) = Q(x) + x^3 + 3x + 1 \) into the inequality \( \frac{d}{dx}(P(x)) + (x-1)^3 \geq P(x) + 1 \), we obtain \( \frac{d}{dx}(Q(x)) + 3x^2 + 3 + (x-1)^3 \geq Q(x) + x^3 + 3x + 2 \). Simplifying this inequality yields \( \frac{d}{dx}(Q(x)) - Q(x) \geq 0 \). This is a first-order linear differential inequality, which can be solved to understand the behavior of \( Q(x) \) and consequently \( P(x) \).</p> <p><strong>Step 5:</strong> Multiplying both sides of the inequality \( \frac{d}{dx}(Q(x)) - Q(x) \geq 0 \) by \( e^{-x} \), we get \( e^{-x}\frac{d}{dx}(Q(x)) - e^{-x}Q(x) \geq 0 \). The left-hand side can be recognized as the derivative of \( e^{-x}Q(x) \), thus \( \frac{d}{dx}(e^{-x}Q(x)) \geq 0 \). This implies that \( e^{-x}Q(x) \) is a non-decreasing function. Given that \( Q(0) = 0 \), we have \( e^{-0}Q(0) = 0 \), which means \( e^{-x}Q(x) \geq 0 \) for all \( x \in [0, \infty) \). Since \( e^{-x} \) is always positive, \( Q(x) \geq 0 \) for all \( x \in [0, \infty) \). However, from the definition of \( Q(x) \), we know \( Q(x) \leq 0 \), so \( Q(x) = 0 \) for all \( x \in [0, \infty) \), implying \( P(x) = x^3 + 3x + 1 \).</p> <p><strong>Step 6:</strong> With \( P(x) = x^3 + 3x + 1 \), we can now evaluate the options given. First, \( y = P(x) \) is indeed a monotonic function because its derivative \( P'(x) = 3x^2 + 3 \) is always positive for \( x \in