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

Since \(\mathrm{f}(x)\) is continuous in \([-2,2]\), it is continuous at \(x=0\) and \(x=1\).
\(\therefore \quad \lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 0^{+}} \mathrm{f}(x)\)
\(\Rightarrow \lim _{x \rightarrow 0^{-}}\left(\frac{\sin a x}{x}+3\right)=\lim _{x \rightarrow 0^{+}}(2 x+7)\)
\(\begin{aligned} & \Rightarrow a+3=0+7 \\ & \Rightarrow a=4\end{aligned}\)
Also, \(\lim _{x \rightarrow 1^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 1^{+}} \mathrm{f}(x)\)
\(\Rightarrow \lim _{x \rightarrow 1^{-}}(2 x+7)=\lim _{x \rightarrow 1^{+}}\left(\sqrt{x^2+8}-b\right)\)
\(\begin{aligned} & \Rightarrow 2(1)+7=\sqrt{1+8}-b \\ & \Rightarrow 9=3-b \\ & \Rightarrow b=-6\end{aligned}\)
\(\therefore \quad 2 a+3 b=8-18=-10\)