If f(x) is continuous on its domain [-2,2],wh | vedclass

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(B) Since $f(x)$ is continuous on $[-2,2]$,it must be continuous at $x=0$ and $x=1$.For continuity at $x=0$,$\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+}

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$-10$

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(B) Since $f(x)$ is continuous on $[-2,2]$,it must be continuous at $x=0$ and $x=1$.For continuity at $x=0$,$\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$.$\lim_{x 	o 0^-} (\frac{\sin ax}{x} + 3) = a + 3$.$\lim_{x 	o 0^+} (2x + 7) = 7$.Equating them,$a + 3 = 7 \implies a = 4$.For continuity at $x=1$,$\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1)$.$\lim_{x 	o 1^-} (2x + 7) = 2(1) + 7 = 9$.$\lim_{x 	o 1^+} (\sqrt{x^2+8} - b) = \sqrt{1+8} - b = 3 - b$.Equating them,$9 = 3 - b \implies b = -6$.Therefore,$2a + 3b = 2(4) + 3(-6) = 8 - 18 = -10$.

If $f(x)$ is continuous on its domain $[-2,2]$,where $f(x) = \begin{cases} \frac{\sin ax}{x} + 3, & -2 \leq x < 0 \\ 2x + 7, & 0 \leq x \leq 1 \\ \sqrt{x^2+8} - b, & 1 < x \leq 2 \end{cases}$ then the value of $2a + 3b$ is

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