Definite Integration
Differentiation under integral sign
Grade 12
Question:
<p><strong>550.</strong> Suppose \(g'(x) < 0 \ \forall x \ge 0\) and \(\displaystyle\int_0^x tg'(t)\,dt \ \forall x \ge 0\). Which of the following statement(s) are <strong>correct</strong>?</p>
<p>(a) \(f\) is not increasing</p>
<p>(b) \(f\) is continuous \(\forall x > 0\)</p>
<p>(c) \(f(x) = xg(x) - \displaystyle\int_0^x g(t)\,dt\)</p>
<p>(d) \(f'(x)\) exists \(\forall x > 0\)</p>
Step-by-Step Solution
Key Concept: Given g'(x) < g(x) for all x > 0, we need to analyze the properties of f(x) = xg(x) - ∫₀ˣ g(t)dt using differentiation and the given inequality to determine which statements about f are correct.
<p><strong>Step 1: Find f'(x)</strong></p><p>Given: f(x) = xg(x) - ∫₀ˣ g(t)dt</p><p>Using the product rule and fundamental theorem of calculus:</p><p>f'(x) = g(x) + xg'(x) - g(x) = xg'(x)</p><p></p><p><strong>Step 2: Analyze whether f is increasing (Option A)</strong></p><p>From Step 1: f'(x) = xg'(x)</p><p>For x > 0: Since g'(x) < g(x) and we have no information that g'(x) > 0, we cannot conclude f'(x) > 0.</p><p>In fact, g'(x) could be negative, making f'(x) < 0 for x > 0.</p><p>Therefore, <strong>f is NOT necessarily increasing</strong>. Statement (a) is <strong>CORRECT</strong>.</p><p></p><p><strong>Step 3: Check continuity of f (Option B)</strong></p><p>f(x) = xg(x) - ∫₀ˣ g(t)dt</p><p>Since g is differentiable for x > 0, g is continuous for x > 0.</p><p>The product xg(x) is continuous for x > 0.</p><p>The integral ∫₀ˣ g(t)dt is continuous for x > 0 (by properties of definite integrals).</p><p>Therefore, f is continuous ∀x > 0. Statement (b) is <strong>CORRECT</strong>.</p><p></p><p><strong>Step 4: Verify the formula for f(x) (Option C)</strong></p><p>This is given as the definition of f(x) in the problem:</p><p>f(x) = xg(x) - ∫₀ˣ g(t)dt</p><p>By direct substitution, statement (c) is <strong>CORRECT</strong>.</p><p></p><p><strong>Step 5: Check existence of f'(x) (Option D)</strong></p><p>From Step 1, we computed:</p><p>f'(x) = xg'(x)</p><p>Since g'(x) exists for all x > 0 (given), and x is a differentiable function, f'(x) = xg'(x) exists for all x > 0.</p><p>Statement (d) is <strong>CORRECT</strong>.</p><p></p><p><strong>∴ Answer:</strong> A,B,C,D</p>
Correct Answer: A,B,C,D