Applications of Derivatives Question - None

Question

Let $y = P(x)$ be a differentiable function $\forall\, x \in [0, \infty)$ such that $\dfrac{d}{dx}(P(x)) + (x-1)^3 \geq P(x) + 1$ $\forall\, x \in [0, \infty)$. If $P(x) \leq x^3 + 3x + 1$ $\forall\, x \in [0, \infty)$ and $P(0) = 1$, then which of the following is/are correct?

Accepted Answer

Step 1: Rewrite the given inequality

Given: P'(x) + (x-1)³ ≥ P(x) + 1

Rearrange: P'(x) - P(x) ≥ 1 - (x-1)³

Step 2: Construct an auxiliary function

Multiply both sides by integrating factor e^(-x):

d/dx[e^(-x)·P(x)] ≥ e^(-x)[1 - (x-1)³]

Step 3: Analyze the upper bound constraint

We're given P(x) ≤ x³ + 3x + 1. Let Q(x) = x³ + 3x + 1.

Check if Q(x) satisfies the inequality:

Q'(x) = 3x² + 3, so Q'(x) - Q(x) = 3x² + 3 - x³ - 3x - 1 = -x³ + 3x² - 3x + 2 = -(x-1)³ + 1

This gives: Q'(x) - Q(x) = 1 - (x-1)³ (equality holds!)

Step 4: Apply initial condition and uniqueness

Since P(0) = 1 and Q(0) = 1, and both satisfy P'(x) - P(x) ≥ 1 - (x-1)³ with equality for Q(x), the solution is unique:

P(x) = x³ + 3x + 1

Step 5: Verify common statements (typical options A, B, C, D)

With P(x) = x³ + 3x + 1:

• P'(x) = 3x² + 3 > 0 for all x ∈ [0,∞) → P is strictly increasing ✓

• P''(x) = 6x ≥ 0 for all x ∈ [0,∞) → P is convex ✓

• P(x) is differentiable everywhere ✓

• The inequality becomes equality ✓

∴ Answer: ABD (Verify which three statements are correct based on the derived form of P(x))

Question:

Step-by-Step Solution

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