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(C) Given that the function $f(x)$ is continuous in the interval $(-\infty, 6)$,it must be continuous at the points $x = 1$ and $x = 3$.For continuity

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$2, 0$

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(C) Given that the function $f(x)$ is continuous in the interval $(-\infty, 6)$,it must be continuous at the points $x = 1$ and $x = 3$.For continuity at $x = 1$:$\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1)$$1 + \sin(\frac{\pi}{2}) = a(1) + b$$1 + 1 = a + b \implies a + b = 2$ ..... $(i)$For continuity at $x = 3$:$\lim_{x 	o 3^-} f(x) = \lim_{x 	o 3^+} f(x) = f(3)$$a(3) + b = 6 	an(\frac{3\pi}{12})$$3a + b = 6 	an(\frac{\pi}{4})$$3a + b = 6(1) \implies 3a + b = 6$ ..... $(ii)$Subtracting equation $(i)$ from equation $(ii)$:$(3a + b) - (a + b) = 6 - 2$$2a = 4 \implies a = 2$Substituting $a = 2$ into equation $(i)$:$2 + b = 2 \implies b = 0$Thus,the values are $a = 2$ and $b = 0$.

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

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