For $x < 0$ : $ f(x) = b^2 \sin\left( \frac{\pi}{2} \left( \frac{\pi}{2} (\cos x + \sin x)\cos x \right) \right) $ For $x > 0$ : $ f(x) = \frac{\sin x - \frac{1}{2}\sin 2x}{x^3} $ For $x = 0$ : $ f(x) = a $

$ f(0) = a $ RHL $ = \lim_{x \to 0^+} \frac{\sin x (1 - \cos x)}{x^3} = \frac{1}{2} $ LHL $ = \lim_{x \to 0^-} \left[ b^2 \sin\left( \frac{\pi}{2} \left[ \frac{\pi}{2} (\sin x + \cos x) \cos x \right] \right) \right] = b^2 $ $ \therefore a = \frac{1}{2},; b^2 = \frac{1}{2} $ $ \Rightarrow a^2 + b^2 = \frac{1}{4} + \frac{1}{2} = \frac{3}{4} $