If f(x) = begin{cases} frac{1-sin^3 x}{3 cos^ | vedclass

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(C) Since $f(x)$ is continuous at $x = \frac{\pi}{2}$,we have $LHL = RHL = f(\frac{\pi}{2})$.First,calculate $LHL$:$LHL = \lim_{x 	o \frac{\pi^-}{2}}

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$2$

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(C) Since $f(x)$ is continuous at $x = \frac{\pi}{2}$,we have $LHL = RHL = f(\frac{\pi}{2})$.First,calculate $LHL$:$LHL = \lim_{x 	o \frac{\pi^-}{2}} \frac{1-\sin^3 x}{3 \cos^2 x}$. This is a $\frac{0}{0}$ form.Using $L$'Hospital's Rule:$\lim_{x 	o \frac{\pi^-}{2}} \frac{-3 \sin^2 x \cos x}{3(2 \cos x)(-\sin x)} = \lim_{x 	o \frac{\pi^-}{2}} \frac{\sin x}{2} = \frac{1}{2}$.Thus,$\alpha = \frac{1}{2}$.Next,calculate $RHL$:$RHL = \lim_{x 	o \frac{\pi^+}{2}} \frac{\beta(1-\sin x)}{(\pi-2 x)^2} = \frac{1}{2}$.Let $x = \frac{\pi}{2} + h$,where $h 	o 0$. Then $\pi - 2x = \pi - 2(\frac{\pi}{2} + h) = -2h$.$\lim_{h 	o 0} \frac{\beta(1-\sin(\frac{\pi}{2} + h))}{(-2h)^2} = \lim_{h 	o 0} \frac{\beta(1-\cos h)}{4h^2} = \frac{1}{2}$.Using the limit formula $\lim_{h 	o 0} \frac{1-\cos h}{h^2} = \frac{1}{2}$:$\frac{\beta}{4} 	imes \frac{1}{2} = \frac{1}{2} \implies \frac{\beta}{8} = \frac{1}{2} \implies \beta = 4$.Therefore,$\alpha \beta = \frac{1}{2} 	imes 4 = 2$.

If $f(x) = \begin{cases} \frac{1-\sin^3 x}{3 \cos^2 x}, & x < \frac{\pi}{2} \\ \alpha, & x = \frac{\pi}{2} \\ \frac{\beta(1-\sin x)}{(\pi-2 x)^2}, & x > \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then $\alpha \beta =$

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