Question

If the function $f\left(x\right)$ is continuous in the interval $[-2,2]$, find the values of $a$ and $b$ where
$f\left(x\right)=\{\begin{array}{c}\frac{\sin ax}{x}-2\text{for}-2\le x<0\\ 2x+1\text{for}0\le x\le 1\\ 2b\sqrt{x^2+3}-1\text{for}1<x\le 2\end{array}$

Answer 1

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}\frac{\sin ax}{x}-2$

$=\underset{x\to 0^{-}}{lim}(a\frac{\sin ax}{ax}-2)$

$=a\times 1-2=a-2[\because \underset{x\to 0}{lim}\frac{\sin x}{x}=1]$

Function is continuous at $x=0$

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

$\Rightarrow a-2=1$

$\Rightarrow a=3$

Again, $\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}(2x+1)$ $=3$

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

$=2b\sqrt{4}-1=4b-1$

Function is continuous at $x=1$

then $4b-1=3$

$\Rightarrow$ $4b=4$

$\Rightarrow$ $b=1$

$\therefore$ $a=3$ and $b=1$