Let [t] denote the greatest integer less than or equal to t. If the function $$f(x) = \begin{cases} b^2 \sin\!\left(\dfrac{\pi}{2}\left[\dfrac{\pi}{2}(\cos x + \sin x)\cos x\right]\right), & x < 0 \\[10pt] \dfrac{\sin x - \dfrac{1}{2}\sin 2x}{x^3}, & x > 0 \\[10pt] a, & x = 0 \end{cases}$$ is continuous at x = 0,then $$a^{2} + b^{2}$$ is equal to

We need to find $$a^2 + b^2$$ given that f(x) is continuous at x = 0.

The right-hand limit as $$x \to 0^+$$ is

$$\lim_{x \to 0^+} \frac{\sin x - \tfrac{1}{2}\sin 2x}{x^3} \;=\; \lim_{x \to 0^+} \frac{\sin x - \sin x \cos x}{x^3} \;=\; \lim_{x \to 0^+} \frac{\sin x(1 - \cos x)}{x^3}$$

Using Taylor expansions: $$\sin x \approx x - \tfrac{x^3}{6}$$ and $$1 - \cos x \approx \tfrac{x^2}{2}$$, we get

$$= \lim_{x \to 0^+} \frac{x \cdot \tfrac{x^2}{2}}{x^3} = \tfrac{1}{2}$$

For continuity at 0 we must have $$a = \tfrac{1}{2}$$.

Next, the left-hand limit as $$x \to 0^-$$ is

$$\lim_{x \to 0^-} b^2 \sin\Bigl(\tfrac{\pi}{2}\bigl[\tfrac{\pi}{2}(\cos x + \sin x)\cos x\bigr]\Bigr).$$

As $$x \to 0^-$$, $$\cos x \to 1$$ and $$\sin x \to 0$$, so

$$\tfrac{\pi}{2}(\cos x + \sin x)\cos x \to \tfrac{\pi}{2}\cdot1\cdot1 = \tfrac{\pi}{2},$$

and the greatest integer function gives $$\bigl[\tfrac{\pi}{2}\bigr] = 1.$$ Hence the expression becomes

$$b^2 \sin\Bigl(\tfrac{\pi}{2}\cdot1\Bigr) = b^2 \sin\Bigl(\tfrac{\pi}{2}\Bigr) = b^2.$$

Continuity then requires $$b^2 = a = \tfrac{1}{2}$$.

Finally,

$$a^2 + b^2 = \Bigl(\tfrac{1}{2}\Bigr)^2 + \tfrac{1}{2} = \tfrac{1}{4} + \tfrac{1}{2} = \tfrac{3}{4}.$$

Therefore, $$a^2 + b^2 = $$ Option 2: $$\frac{3}{4}$$.