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(C) For the function to be continuous at $x = 0$, we must have $f(0) = \lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^-} f(x)$.Given $f(0) = a$.For $x > 0$, $f

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$\frac{3}{4}$

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(C) For the function to be continuous at $x = 0$, we must have $f(0) = \lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^-} f(x)$.Given $f(0) = a$.For $x > 0$, $f(x) = \frac{\sin x - \frac{1}{2} \sin 2x}{x^3} = \frac{\sin x - \sin x \cos x}{x^3} = \frac{\sin x(1 - \cos x)}{x^3} = \frac{\sin x}{x} \cdot \frac{2 \sin^2(x/2)}{x^2} = \frac{\sin x}{x} \cdot \frac{2 \sin^2(x/2)}{4(x/2)^2} = 1 \cdot \frac{1}{2} = \frac{1}{2}$.Thus, $a = \frac{1}{2}$.For $x So, $\lim_{x 	o 0^-} f(x) = b^2 \sin(\frac{\pi}{2} \cdot 1) = b^2$.Equating the limits, $b^2 = \frac{1}{2}$.Therefore, $a^2 + b^2 = (\frac{1}{2})^2 + \frac{1}{2} = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$.

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