Question

If $f\left(x\right)=\{\begin{array}{ccc}\frac{\sin (a+1)x+2\sin x}{x}& ,& x<0\\ 2& ,& x=0\\ \frac{\sqrt{1+bx}-1}{x}& ,& x>0\end{array}$

is continuous at $x=0$, then find the values of $a$ and $b$.

Answer 1

Given:
The function $f\left(x\right)$ is continuous at $x=0$.

$\therefore \underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)=f\left(0\right)=2$ ................(1)

Now, $\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0-h)$
Thus,
$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(-h)$
= $\underset{h\to 0}{lim}\frac{\sin (a+1)(-h)+2\sin (-h)}{-h}$

= $\underset{h\to 0}{lim}\frac{-\sin (a+1)h-2\sin h}{-h}$

= $\underset{h\to 0}{lim}\frac{\sin (a+1)h}{h}+\underset{h\to 0}{lim}\frac{2\sin h}{h}$

= $(a+1)\underset{h\to 0}{lim}\frac{\sin (a+1)h}{(a+1)h}+2\underset{h\to 0}{lim}\frac{\sin h}{h}$

$=a+1+2$

$=a+3$ ....................(2)

From equation (1)

$\underset{x\to 0^{-}}{lim}f\left(x\right)=2\therefore a+3=2$
$\Rightarrow \begin{array}{c}a=-1\end{array}$
Now,
$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}(0+h)$

$\therefore \underset{x\to 0^{+}}{lim}\frac{\sqrt{1+bx}-1}{x}=\underset{h\to 0^{+}}{lim}\frac{\sqrt{1+bh}-1}{h}$

=$\underset{h\to 0}{lim}\frac{\sqrt{1+bh}-1}{h}\times \frac{\sqrt{1+bh}+1}{\sqrt{1+bh}+1}$

=$\underset{h\to 0}{lim}\left(\frac{(\sqrt{1+bh}{)}^2-(1{)}^2}{h(\sqrt{1+bh}+1)}\right)=\underset{h\to 0}{lim}\frac{1+bh-1}{h(\sqrt{1+bh}+1)}$

$=\underset{h\to 0}{lim}\frac{bh}{h\times [\sqrt{1+bh)}+1]}=\underset{h\to 0}{lim}\frac{b}{\sqrt{1+bh}+1}$

= $\frac{b}{\sqrt{1+b\times 0}+1}=\frac{b}{1+1}=\frac{b}{2}$

Now, again from equation (1)

$\underset{x\to 0^{+}}{lim}f\left(x\right)=f\left(0\right)=2$

$\therefore \frac{b}{2}=2$
$\Rightarrow \begin{array}{c}b=4\end{array}$

Hence, the values of $a=-1$ and $b=4$.