Let $$f: R \to R$$ be defined as $$f(x) = \begin{cases} 2\sin\left(-\frac{\pi x}{2}\right), & \text{if } x < -1 \\ |ax^2 + x + b|, & \text{if } -1 \leq x \leq 1 \\ \sin(\pi x), & \text{if } x > 1 \end{cases}$$
If $$f(x)$$ is continuous on $$R$$, then $$a + b$$ equals:

For $$f(x)$$ to be continuous on $$\mathbb{R}$$, it must be continuous at $$x = -1$$ and $$x = 1$$.

At $$x = 1$$: from the right, $$\displaystyle\lim_{x \to 1^+} \sin(\pi x) = \sin\pi = 0$$. From the middle piece, $$f(1) = |a + 1 + b|$$. For continuity, $$|a + 1 + b| = 0$$, so $$a + b = -1$$.

At $$x = -1$$: from the left, $$\displaystyle\lim_{x \to -1^-} 2\sin\left(-\dfrac{\pi x}{2}\right) = 2\sin\left(\dfrac{\pi}{2}\right) = 2$$. From the middle piece, $$f(-1) = |a - 1 + b| = |a + b - 1| = |-1 - 1| = |-2| = 2$$. This is consistent, confirming continuity at $$x = -1$$.

Therefore $$a + b = -1$$.