Question

If $f\left(x\right)=\left\{\begin{array}{cc}\frac{\sin (a+1)x+\sin x}{x},& x<0\\ c,& x=0\\ \frac{\sqrt{x+b{x}^2}-\sqrt{x}}{bx\sqrt{x}},& x>0\end{array}\right.$is continuous at x = 0, then
(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 ∈ R − {0}, c = $\frac{1}{2}$
(d) none of these

Answer 1

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

The given function can be rewritten as

$f\left(x\right)=\left\{\begin{array}{c}\frac{\sin \left(a+1\right)x+x\sin x}{x},\mathrm{for}x<0\\ c,\mathrm{for}x=0\\ \frac{\sqrt{x+b{x}^2}-\sqrt{x}}{b{x}^{{\displaystyle \frac{3}{2}}}},\mathrm{for}x>0\end{array}\right.$

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

We have
(LHL at x = 0) = $\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0-h\right)=\underset{h\to 0}{\lim }f\left(-h\right)$

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

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


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

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

Also, $f\left(0\right)=c$

If $f\left(x\right)$ is continuous at x = 0, then
​$\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)=f\left(0\right)$

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

Now, $\frac{\sqrt{1+bx}-1}{bx{\displaystyle }}$ exists only if $bx\ne 0\Rightarrow b\ne 0$.

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