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If $\mathrm{f}(x)$ is continuous on its domain $[-2,2]$, where
$$
\mathrm{f}(x)= \begin{cases}\frac{\sin a x}{x}+3 & , \text { for }-2 \leq x < 0 \\ 2 x+7 & , \text { for } 0 \leq x \leq 1 \\ \sqrt{x^2+8}-\mathrm{b}, & \text { for } 1 < x \leq 2\end{cases}
$$
then the value of $2 a+3 b$ is
Options:
$$
\mathrm{f}(x)= \begin{cases}\frac{\sin a x}{x}+3 & , \text { for }-2 \leq x < 0 \\ 2 x+7 & , \text { for } 0 \leq x \leq 1 \\ \sqrt{x^2+8}-\mathrm{b}, & \text { for } 1 < x \leq 2\end{cases}
$$
then the value of $2 a+3 b$ is
Solution:
1920 Upvotes
Verified Answer
The correct answer is:
$-10$
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, $\lim _{x \rightarrow 1^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 1^{+}} \mathrm{f}(x)$
$\Rightarrow \lim _{x \rightarrow 1^{-}}(2 x+7)=\lim _{x \rightarrow 1^{+}}\left(\sqrt{x^2+8}-b\right)$
$\begin{aligned} & \Rightarrow 2(1)+7=\sqrt{1+8}-b \\ & \Rightarrow 9=3-b \\ & \Rightarrow b=-6\end{aligned}$
$\therefore \quad 2 a+3 b=8-18=-10$
$\therefore \quad \lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 0^{+}} \mathrm{f}(x)$
$\Rightarrow \lim _{x \rightarrow 0^{-}}\left(\frac{\sin a x}{x}+3\right)=\lim _{x \rightarrow 0^{+}}(2 x+7)$
$\begin{aligned} & \Rightarrow a+3=0+7 \\ & \Rightarrow a=4\end{aligned}$
Also, $\lim _{x \rightarrow 1^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 1^{+}} \mathrm{f}(x)$
$\Rightarrow \lim _{x \rightarrow 1^{-}}(2 x+7)=\lim _{x \rightarrow 1^{+}}\left(\sqrt{x^2+8}-b\right)$
$\begin{aligned} & \Rightarrow 2(1)+7=\sqrt{1+8}-b \\ & \Rightarrow 9=3-b \\ & \Rightarrow b=-6\end{aligned}$
$\therefore \quad 2 a+3 b=8-18=-10$
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