Applications of Derivatives Question - None

Question

The set of values of $p$ for which $f(x) = p^2x - \int 2^{4-x^2}\,dx$ is increasing for all $x \in R$, is:

Accepted Answer

Step 1: Find f'(x) using the Fundamental Theorem of Calculus.

f'(x) = p² - d/dx[∫ 2^(4-x²) dx] = p² - 2^(4-x²)

Step 2: For f to be increasing for all x ∈ ℝ, we need f'(x) ≥ 0 for all x ∈ ℝ.

p² - 2^(4-x²) ≥ 0 for all x ∈ ℝ

p² ≥ 2^(4-x²) for all x ∈ ℝ

Step 3: Find the maximum value of 2^(4-x²).

Since 4-x² ≤ 4 for all x ∈ ℝ (with equality when x = 0), we have 2^(4-x²) ≤ 2⁴ = 16.

The maximum value of 2^(4-x²) is 16 (achieved at x = 0).

Step 4: For p² ≥ 2^(4-x²) to hold for ALL x, we need p² to be at least as large as the maximum value of 2^(4-x²).

p² ≥ 16

|p| ≥ 4

p ∈ (-∞, -4] ∪ [4, ∞)

∴ Answer: A

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