Definite Integration
Definite Integration with Parameters
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 obtained by integrating a polynomial in x term-by-term, treating t as a parameter. The key is to evaluate the definite integral by treating coefficients as constants and finding bounds on A(t) using calculus or algebraic manipulation.
<p><strong>Step 1:</strong> Set up the integral by treating t as a parameter and x as the variable of integration:</p><p>$$A(t) = \int_0^1 \left[(\sin t)x^2 - (2\cos t)x + \sin t - \frac{1}{3}\right]dx$$</p><p><strong>Step 2:</strong> Integrate term-by-term:</p><p>$$A(t) = \left[\frac{(\sin t)x^3}{3} - (\cos t)x^2 + \left(\sin t - \frac{1}{3}\right)x\right]_0^1$$</p><p><strong>Step 3:</strong> Evaluate at bounds x=1 and x=0:</p><p>$$A(t) = \frac{\sin t}{3} - \cos t + \sin t - \frac{1}{3}$$</p><p>$$A(t) = \frac{4\sin t}{3} - \cos t - \frac{1}{3}$$</p><p><strong>Step 4:</strong> Analyze the expression. The maximum value of $\frac{4\sin t}{3} - \cos t$ occurs when we find critical points or use the amplitude form. This equals $\sqrt{\left(\frac{4}{3}\right)^2 + 1^2} = \sqrt{\frac{16}{9}+1} = \sqrt{\frac{25}{9}} = \frac{5}{3}$</p><p>Therefore: $A(t)_{\max} = \frac{5}{3} - \frac{1}{3} = \frac{4}{3}$ and $A(t)_{\min} = -\frac{5}{3} - \frac{1}{3} = -2$</p><p>∴ Answer: A</p>
Correct Answer: A