Question

A function $f\left(x\right)$ is continuous on its domain $[-2,2]$ where
$\begin{array}{cc}f\left(x\right)& =\frac{\sin ax}{x}+2,\text{for}-2\le x<0\\ & =3x+5\text{for}0\le x<1\\ & =\sqrt{x^2+8}-b,\text{for}1\le x⩽2\end{array}$
Show that $a+b+2=0$

Answer 1

$\underset{x\to 0^{-}}{lim}f\left(x\right)=$$\underset{x\to 0^{-}}{lim}\left(\frac{\sin ax}{x}\right)+2=$$\underset{x\to 0^{-}}{lim}\left(\frac{a\sin ax}{ax}\right)+2=a+2$............As $\underset{x\to 0}{lim}\frac{sinx}{x}=1$

$\underset{x\to 0^{+}}{lim}f\left(x\right)=$$\underset{x\to 0^{+}}{lim}(3x+5)=5$

Since $f\left(x\right)$ is continuous, so $\underset{x\to 0^{-}}{lim}f\left(x\right)=$$\underset{x\to 0^{+}}{lim}f\left(x\right)$

So, $a+2=5⟹a=3$

Now, $\underset{x\to 1^{-}}{lim}f\left(x\right)=$$\underset{x\to 1^{-}}{lim}(3x+5)=3+5=8$

$\underset{x\to 1^{+}}{lim}f\left(x\right)=$$\underset{x\to 1^{+}}{lim}(\sqrt{x^2+8}-b)=\sqrt{9}-b=3-b$

Since $f\left(x\right)$ is continuous

so, $\underset{x\to 1^{-}}{lim}f\left(x\right)=$$\underset{x\to 1^{+}}{lim}f\left(x\right)$

So, $8=3-b$

$⟹b=-5$

So, $a+b+2=3-5+2=0$

Hence proved.