# Solve this

Question:

If f(x)=\left\{\begin{aligned} \frac{\sin (a+1) x+\sin x}{x}, & x<0 \\ c &, x=0 \text { is continuous at } x=0, \text { then } \\ \frac{\sqrt{x+b x^{2}}-\sqrt{x}}{b x \sqrt{x}}, & x>0 \end{aligned}\right.

(a) $a=-\frac{3}{2}, b=0, c=\frac{1}{2}$

(b) $a=-\frac{3}{2}, b=1, c=-\frac{1}{2}$

(c) $a=-\frac{3}{2}, b \in R-\{0\}, c=\frac{1}{2}$

(d) none of these

Solution:

(c) $a=\frac{-3}{2} \quad, b \in R-\{0\}, c=\frac{1}{2}$

The given function can be rewritten as

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

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

We have

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

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

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

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

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

Also, $f(0)=c$

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 $b x \neq 0 \Rightarrow b \neq 0$.

Thus, $b \in R-\{0\}$.