Applications of Derivatives
Increasing and Decreasing Functions
Grade 12
Question:
<p>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:</p>
<p>(a) \([-4, 4]\)</p>
<p>(b) \((-\infty, -16] \cup [16, \infty)\)</p>
<p>(c) \((-\infty, -4] \cup [4, \infty)\)</p>
<p>(d) \([-16, 16]\)</p>
Step-by-Step Solution
Key Concept: A function is increasing for all x ∈ ℝ iff f'(x) ≥ 0 for all x. You must find f'(x), recognize that 2^(4-x²) is always positive and decreasing, then determine when the coefficient p² ensures non-negative derivative everywhere.
<p><strong>Step 1:</strong> Find f'(x) using the Fundamental Theorem of Calculus.</p><p>f'(x) = p² - d/dx[∫2^(4-x²)dx] = p² - 2^(4-x²)</p><p><strong>Step 2:</strong> For f to be increasing for all x ∈ ℝ, we need f'(x) ≥ 0 for all x.</p><p>This requires: p² - 2^(4-x²) ≥ 0 for all x ∈ ℝ</p><p>Rearranging: p² ≥ 2^(4-x²) for all x ∈ ℝ</p><p><strong>Step 3:</strong> Find the maximum value of 2^(4-x²).</p><p>Since the exponent (4-x²) is maximized when x² is minimized (i.e., when x = 0), the maximum value is:</p><p>2^(4-0) = 2⁴ = 16</p><p><strong>Step 4:</strong> For the inequality p² ≥ 2^(4-x²) to hold for ALL x, we need:</p><p>p² ≥ 16</p><p>Therefore: |p| ≥ 4, which means p ∈ (-∞, -4] ∪ [4, ∞)</p><p>∴ Answer: A</p>
Correct Answer: A