Question

Let \(f(x) = \begin{cases} \frac{1}{3} & , x \le \frac{\pi}{2} \\ \frac{b(1 - \sin x)}{(\pi - 2x)^2} & , x>\frac{\pi}{2} \end{cases}\). If \(f\) is continuous at \(x = \frac{\pi}{2}\), then the value of \(\int_0^{3b-6} |x^2 + 2x - 3| \, dx\) is:

• A. 5
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For $f(x)$ to be continuous at $x = \pi/2$, the Left Hand Limit (LHL), Right Hand Limit (RHL), and the functional value must be equal. We use this to find the value of $b$, then evaluate the integral.
Step 2: Key Formula or Approach:
1. LHL = $f(\pi/2) = 1/3$.
2. RHL: $\lim_{x \to \frac{\pi}{2}^+} \frac{b(1 - \sin x)}{(\pi - 2x)^2}$. Use $x = \pi/2 + h$.
Step 3: Detailed Explanation:
1. Find $b$: Let $x = \frac{\pi}{2} + h$ as $x \to \frac{\pi}{2}^+, h \to 0^+$. RHL $= \lim_{h \to 0} \frac{b(1 - \sin(\frac{\pi}{2} + h))}{(\pi - 2(\frac{\pi}{2} + h))^2} = \lim_{h \to 0} \frac{b(1 - \cos h)}{(-2h)^2}$ Using $1 - \cos h \approx \frac{h^2}{2}$: RHL $= \lim_{h \to 0} \frac{b(h^2/2)}{4h^2} = \frac{b}{8}$. For continuity: $\frac{b}{8} = \frac{1}{3} \implies b = \frac{8}{3}$. 2. Evaluate the Integral: The upper limit is $3b - 6 = 3(\frac{8}{3}) - 6 = 8 - 6 = 2$. $I = \int_0^2 |x^2 + 2x - 3| dx$. The expression inside the modulus $x^2 + 2x - 3 = (x+3)(x-1)$ changes sign at $x=1$. For $x \in [0, 1]$, $|x^2 + 2x - 3| = -(x^2 + 2x - 3)$. For $x \in [1, 2]$, $|x^2 + 2x - 3| = x^2 + 2x - 3$. $I = \int_0^1 (3 - 2x - x^2) dx + \int_1^2 (x^2 + 2x - 3) dx$ $I = [3x - x^2 - \frac{x^3}{3}]_0^1 + [\frac{x^3}{3} + x^2 - 3x]_1^2$ $I = (3 - 1 - 1/3) + [(\frac{8}{3} + 4 - 6) - (\frac{1}{3} + 1 - 3)] = \frac{5}{3} + [\frac{2}{3} - (-\frac{5}{3})] = \frac{5}{3} + \frac{7}{3} = 4$. (Self-correction: Re-calculating sum leads to 4).
Step 4: Final Answer:
The value of the integral is 4.