Consider f(x) = left[ frac{2(sin x - sin^3 x) | vedclass

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(A) Let $g(x) = \sin x - \sin^3 x = \sin x(1 - \sin^2 x) = \sin x \cos^2 x$.For $x \in (0, \pi)$,$\sin x > 0$ and $\cos^2 x \geq 0$. Thus,$g(x) \geq 0

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$f$ is continuous and differentiable at $x = \frac{\pi}{2}$

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(A) Let $g(x) = \sin x - \sin^3 x = \sin x(1 - \sin^2 x) = \sin x \cos^2 x$.For $x \in (0, \pi)$,$\sin x > 0$ and $\cos^2 x \geq 0$. Thus,$g(x) \geq 0$.Specifically,for $x 
eq \frac{\pi}{2}$,$g(x) > 0$.Since $g(x) > 0$ for $x \in (0, \pi) \setminus \{\frac{\pi}{2}\}$,we have $|g(x)| = g(x)$.Substituting this into the expression for $f(x)$:$f(x) = \left[ \frac{2g(x) + g(x)}{2g(x) - g(x)} ight] = \left[ \frac{3g(x)}{g(x)} ight] = [3] = 3$.Thus,$f(x) = 3$ for all $x \in (0, \pi) \setminus \{\frac{\pi}{2}\}$.Given $f(\frac{\pi}{2}) = 3$,the function is $f(x) = 3$ for all $x \in (0, \pi)$.$A$ constant function is continuous and differentiable everywhere in its domain.Therefore,$f$ is continuous and differentiable at $x = \frac{\pi}{2}$.

Consider $f(x) = \left[ \frac{2(\sin x - \sin^3 x) + |\sin x - \sin^3 x|}{2(\sin x - \sin^3 x) - |\sin x - \sin^3 x|} \right]$ for $x \in (0, \pi), x \neq \frac{\pi}{2}$,and $f(\frac{\pi}{2}) = 3$,where $[ \cdot ]$ denotes the greatest integer function. Then:

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