Question

If \( \lim_{x \to \frac{\pi}{2}} \dfrac{b(1 - \sin x)(\pi - 2x)^2}{1} = \frac{1}{3} \), then \( \int_0^{3b-6} |x^2 + 2x - 3| \, dx \) is equal to:

• A. 3
• B. 4
• C. \(\frac{7}{4}\)
• D. 2

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
First, we solve the limit using L'Hôpital's Rule or expansion to find the value of the constant \(b\). Then, we substitute \(b\) into the upper limit of the integral and evaluate the definite integral of the absolute value function.

Step 2: Key Formula or Approach:
1. Limit: Let \(\pi/2 - x = \theta\). As \(x \to \pi/2, \theta \to 0\). \[ \lim_{\theta \to 0} \frac{b(1 - \cos \theta)}{(2\theta)^2} = \lim_{\theta \to 0} \frac{b(2\sin^2 \theta/2)}{4\theta^2} = \frac{2b(\theta^2/4)}{4\theta^2} = \frac{b}{8} \] Given \(b/8 = 1/3 \implies b = 8/3\).

Step 3: Detailed Explanation:
1. Upper limit of integral: \(3(8/3) - 6 = 8 - 6 = 2\). 2. The integral is \( \int_0^2 |x^2 + 2x - 3| dx \). 3. The expression inside the absolute value \(x^2 + 2x - 3 = (x+3)(x-1)\) changes sign at \(x=1\). - For \(0 \le x \le 1\), \(x^2 + 2x - 3\) is negative. - For \(1 \le x \le 2\), \(x^2 + 2x - 3\) is positive. 4. Splitting the integral: \[ \int_0^1 -(x^2 + 2x - 3) dx + \int_1^2 (x^2 + 2x - 3) dx \] \[ = \left[ -\frac{x^3}{3} - x^2 + 3x \right]_0^1 + \left[ \frac{x^3}{3} + x^2 - 3x \right]_1^2 \] \[ = (-1/3 - 1 + 3) + ( (8/3 + 4 - 6) - (1/3 + 1 - 3) ) \] \[ = 5/3 + ( 2/3 - (-5/3) ) = 5/3 + 7/3 = 12/3 = 4 \]

Step 4: Final Answer:
The value of the integral is 4.