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(D) For $f(x)$ to be continuous on $\mathbb{R}$, it must be continuous at $x=0, x=1,$ and $x=2$.At $x=0$: $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(

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$-1$

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(D) For $f(x)$ to be continuous on $\mathbb{R}$, it must be continuous at $x=0, x=1,$ and $x=2$.At $x=0$: $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) \implies \sin(0) = 0^2 + a^2 \implies a^2 = 0 \implies a = 0$.At $x=1$: $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) \implies 1^2 + a^2 = b(1) + 2 \implies 1 + 0 = b + 2 \implies b = -1$.At $x=2$: $\lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^+} f(x) \implies b(2) + 2 = 0 \implies 2(-1) + 2 = 0$, which is consistent.Thus, $a = 0$ and $b = -1$.Calculating $a+b+ab = 0 + (-1) + (0)(-1) = -1$.

If $f(x) = \begin{cases} \sin x, & \text{if } x \leq 0 \\ x^2+a^2, & \text{if } 0 < x < 1 \\ bx+2, & \text{if } 1 \leq x \leq 2 \\ 0, & \text{if } x > 2 \end{cases}$ is continuous on $\mathbb{R}$, then $a+b+ab = $

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