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If the function $f(x)$ is continuous on its domain $[-2,2]$ where $f(x)=\frac{\sin a x}{x}+3,-2 \leq x < 0=x+5,0 \leq x \leq 1=\sqrt{x^2+8}-b, 1 < x \leq 2$, then $7 a+b+1$ is equal to
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Verified Answer
The correct answer is:
12
for continuity at $x=0, \lim _{x \rightarrow 0}-\frac{\sin a x}{x}+3=5$
$\begin{aligned}
& \Rightarrow a+3=5 \\
& \Rightarrow a=2
\end{aligned}$
for continuity at $x=1$,
$\begin{aligned}
& 6=\lim _{x \rightarrow 1}+\sqrt{x^2+8}-b \\
& \Rightarrow 6=\sqrt{1^2+8}-b \\
& \Rightarrow b=-3
\end{aligned}$
Now $7 a+b+1=7 \times 2-3+1=12$
$\begin{aligned}
& \Rightarrow a+3=5 \\
& \Rightarrow a=2
\end{aligned}$
for continuity at $x=1$,
$\begin{aligned}
& 6=\lim _{x \rightarrow 1}+\sqrt{x^2+8}-b \\
& \Rightarrow 6=\sqrt{1^2+8}-b \\
& \Rightarrow b=-3
\end{aligned}$
Now $7 a+b+1=7 \times 2-3+1=12$
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