# Determine the values of a, b, c for which the function

Question:

Determine the values of abc for which the function

$f(x)=\left\{\begin{array}{cl}\frac{\sin (a+1) x+\sin x}{x}, & \text { for } x<0 \\ c & , \quad \text { for } x=0 \text { is continuous at } x=0 . \\ \frac{\sqrt{x+b x^{2}}-\sqrt{x}}{b x^{3 / 2}}, & \text { for } x>0\end{array}\right.$

Solution:

The given function can be rewritten as:

$f(x)= \begin{cases}\frac{\sin (a+1) x+\sin x}{x}, & \text { for } x<0 \\ c & , \text { for } x=0 \\ \frac{\sqrt{x+b x^{2}}-\sqrt{x}}{b^{\frac{3}{2}}} & , \text { for } x>0\end{cases}$

$\Rightarrow f(x)= \begin{cases}\frac{b \sin (a+1) x+\sin x}{x}, & \text { for } x<0 \\ c \quad & , \text { for } x=0 \\ \frac{\sqrt{1+b x}-1}{b x}, & , \text { for } x>0\end{cases}$

We observe

$(\mathrm{LHL}$ at $x=0)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)$

$=\lim _{h \rightarrow 0}\left[\frac{-\sin (a+1) h-\sin (-h)}{h}\right]=\lim _{h \rightarrow 0}\left[\frac{-\sin (a+1) h}{h}-\frac{\sin h}{h}\right]$

$=-(a+1) \lim _{h \rightarrow 0}\left[\frac{\sin (a+1) h}{(a+1) h}\right]-\lim _{h \rightarrow 0} \frac{\sin h}{h}=-a-1$

$(\mathrm{RHL}$ at $x=0)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)$

$=\lim _{h \rightarrow 0}\left(\frac{\sqrt{1+b h}-1}{b h}\right)=\lim _{h \rightarrow 0}\left(\frac{b h}{b h(\sqrt{1+b h}+1)}\right)=\lim _{h \rightarrow 0}\left(\frac{1}{(\sqrt{1+b h}+1)}\right)=\frac{1}{2}$

And, $f(0)=c$

If $f(x)$ is continuous at $x=0$, then

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$

$\Rightarrow-a-1=\frac{1}{2}=c$

$\Rightarrow-a-1=\frac{1}{2}$ and $c=\frac{1}{2}$

$\Rightarrow a=\frac{-3}{2}, c=\frac{1}{2}$

Now, $\frac{\sqrt{1+b x}-1}{b x}$ exists only if $b x \neq 0 \Rightarrow b \neq 0$.

$\therefore b \in R-\{0\}$