Definite Integration
Integral as a function of parameter
Grade 12
Question:
<p>For \( x, t \in R \), let \( P_t(x) = (\sin t)x^2 - (2\cos t)x + \sin t - \dfrac{1}{3} \) be a family of quadratic polynomial in \( x \), with variable coefficients. Also \( A(t) = \displaystyle\int_0^1 (P_t(x))\, dx \).</p><p>Which of the following statement are true?</p>
<p>(a) \( \lim_{t \to \pi/2} (A(t))^{\tan t} \) equals \( e^{4/3} \).</p>
<p>(b) \( A(t) \) has infinitely many critical points.</p>
<p>(c) \( A(t) = 0 \) for infinitely many \( t \).</p>
<p>(d) \( A'(t) > 0 \) for all \( t \).</p>
Step-by-Step Solution
Key Concept: Recognize that A(t) is a definite integral of a polynomial with trigonometric coefficients—integrate term-by-term using standard formulas, then analyze the resulting expression as a function of t to determine its properties (range, extrema, monotonicity).
<p><strong>Step 1:</strong> Set up the integral of P_t(x).</p><p>A(t) = ∫₀¹ [(sin t)x² - (2cos t)x + sin t - 1/3] dx</p><p><strong>Step 2:</strong> Integrate term-by-term with respect to x.</p><p>A(t) = [(sin t)·(x³/3) - (2cos t)·(x²/2) + (sin t - 1/3)·x]₀¹</p><p>A(t) = (sin t)/3 - cos t + sin t - 1/3</p><p><strong>Step 3:</strong> Simplify and combine like terms.</p><p>A(t) = (sin t)/3 + sin t - cos t - 1/3</p><p>A(t) = (4sin t)/3 - cos t - 1/3</p><p><strong>Step 4:</strong> Express in standard form to analyze properties.</p><p>A(t) = (4/3)sin t - cos t - 1/3</p><p>This can be written as: A(t) = R·sin(t + φ) - 1/3, where R = √[(4/3)² + 1²] = √(16/9 + 1) = √(25/9) = 5/3</p><p>Therefore: A(t) = (5/3)sin(t + φ) - 1/3 for some phase φ</p><p><strong>Step 5:</strong> Determine range and properties.</p><p>Range of A(t): [−5/3 − 1/3, 5/3 − 1/3] = [−2, 4/3]</p><p>A(t) is neither even nor odd; it has both maximum 4/3 and minimum −2.</p><p>∴ Verify which statements match these properties (options require specific claim verification)</p>
Correct Answer: A