Question

Let \( \alpha \) and \( \beta \) (\( \alpha<\beta \)) be roots of \( 18x^2 - 9\pi x + \pi^2 = 0 \), \( f(x) = x^2, g(x) = \cos x \). Then
\[ \int_{\alpha}^{\beta} x (g \circ f(x)) \, dx = ? \]

• A. \( \frac{\sqrt{3} - 1}{4} \)
• B. \( \frac{\sqrt{3}}{4} \)
• C. \( \frac{2 + \sqrt{3}}{2} \)
• D. \( \frac{1}{2} \left( \sin \frac{\pi^2}{9} - \sin \frac{\pi^2}{36} \right) \)

Hint:

For integrals like \( x \cos(x^2) \), use substitution \( u = x^2 \) to simplify.

Answer 1

The Correct Option is D

We are given: \[ f(x) = x^2,\quad g(x) = \cos x \Rightarrow (g \circ f)(x) = \cos(x^2) \] We are to evaluate: \[ \int_{\alpha}^{\beta} x \cos(x^2) dx \] Let \( u = x^2 \Rightarrow du = 2x dx \Rightarrow x dx = \frac{1}{2} du \), and limits become \( \alpha^2, \beta^2 \).
So integral becomes: \[ \int_{\alpha}^{\beta} x \cos(x^2) dx = \frac{1}{2} \int_{\alpha^2}^{\beta^2} \cos u \, du = \frac{1}{2} [\sin u]_{\alpha^2}^{\beta^2} \] Now, solve quadratic: \[ 18x^2 - 9\pi x + \pi^2 = 0 \Rightarrow x = \frac{9\pi \pm \sqrt{(9\pi)^2 - 72\pi^2}}{2 \cdot 18} = \frac{\pi}{6},\ \frac{\pi}{3} \] So \( \alpha = \frac{\pi}{6}, \beta = \frac{\pi}{3} \Rightarrow \alpha^2 = \frac{\pi^2}{36}, \beta^2 = \frac{\pi^2}{9} \)
Final answer: \( \frac{1}{2} \left( \sin \frac{\pi^2}{9} - \sin \frac{\pi^2}{36} \right) \)