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(D) For $f(x)$ to be continuous at $x = 1$,the left-hand limit must equal the right-hand limit:$\lim_{x \rightarrow 1^{-}} (1 + \sin \frac{\pi x}{2})

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$2, 0$

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(D) For $f(x)$ to be continuous at $x = 1$,the left-hand limit must equal the right-hand limit:$\lim_{x ightarrow 1^{-}} (1 + \sin \frac{\pi x}{2}) = 1 + \sin \frac{\pi}{2} = 1 + 1 = 2$.$\lim_{x ightarrow 1^{+}} (ax + b) = a(1) + b = a + b$.Thus,$a + b = 2$  --- $(1)$For $f(x)$ to be continuous at $x = 3$,the left-hand limit must equal the right-hand limit:$\lim_{x ightarrow 3^{-}} (ax + b) = 3a + b$.$\lim_{x ightarrow 3^{+}} (6 	an \frac{x \pi}{12}) = 6 	an \frac{3 \pi}{12} = 6 	an \frac{\pi}{4} = 6(1) = 6$.Thus,$3a + b = 6$  --- $(2)$Subtracting equation $(1)$ from equation $(2)$:$(3a + b) - (a + b) = 6 - 2$$2a = 4 \implies a = 2$.Substituting $a = 2$ into equation $(1)$:$2 + b = 2 \implies b = 0$.Therefore,the values are $a = 2$ and $b = 0$.

If the function $f(x)$ is defined as:
$f(x) = \begin{cases} 1 + \sin \frac{\pi x}{2}, & -\infty < x \leq 1 \\ ax + b, & 1 < x < 3 \\ 6 \tan \frac{x \pi}{12}, & 3 \leq x < 6 \end{cases}$
and is continuous in $(-\infty, 6)$,then the values of $a$ and $b$ are respectively.

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If the function $f(x)$ is defined as:$f(x) = \begin{cases} 1 + \sin \frac{\pi x}{2}, & -\infty < x \leq 1 \\ ax + b, & 1 < x < 3 \\ 6 \tan \frac{x \pi}{12}, & 3 \leq x < 6 \end{cases}$and is continuous in $(-\infty, 6)$,then the values of $a$ and $b$ are respectively.