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: Continuity at \(x=\pi/2\) For continuity, \[ \frac13=\lim_{x\to\pi/2^+}\frac{b(1-\sin x)}{(\pi-2x)^2} \] Let \(x=\frac\pi2+h\). Then, \[ 1-\sin x=1-\cos h \] \[ (\pi-2x)^2=4h^2 \] So, \[ \frac13=\frac{b}{4}\lim_{h\to0}\frac{1-\cos h}{h^2} \] Using \[ \lim_{h\to0}\frac{1-\cos h}{h^2}=\frac12 \] \[ \frac13=\frac{b}{8} \] \[ b=\frac83 \] Step 2: Evaluate integral \[ 3b-6=8-6=2 \] So, \[ I=\int_0^2|x^2+2x-3|dx \] Factor: \[ x^2+2x-3=(x+3)(x-1) \] Split at \(x=1\): \[ I=\int_0^1(3-2x-x^2)dx+\int_1^2(x^2+2x-3)dx \] \[ =\frac53+\frac73=4 \] \[ \boxed{4} \]