Question

In the following determine the value of ${}^{\prime }{p}^{\prime }$ so that the given function is continuous:
$f\left(x\right)=\{\begin{array}{c}\frac{\sqrt{1+px}-\sqrt{1-px}}{x},if-1\le x<0\\ \frac{2x+1}{x-2},if0\le x\le 1\end{array}$ .

Answer 1

Given,

$f\left(x\right)=\{\begin{array}{cc}\frac{\sqrt{1+px}-\sqrt{1-px}}{x},& \text{if}-1\le x<0\\ \frac{2x+1}{x-2},& \text{if}0\le x<1\end{array}$

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

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

$\Rightarrow \underset{h\to 0}{lim}f(-h)=\underset{h\to 0}{lim}f\left(h\right)$

$\Rightarrow \underset{h\to 0}{lim}\frac{\sqrt{1+p(-h)}-\sqrt{1-p(-h)}}{-h}=\underset{h\to 0}{lim}\frac{2h+1}{h-2}$

$\Rightarrow \underset{h\to 0}{lim}\frac{(\sqrt{1-ph}-\sqrt{1+ph})(\sqrt{1-ph}+\sqrt{1+ph})}{-h(\sqrt{1-ph}+\sqrt{1+ph})}=\frac{1}{-2}$

$\Rightarrow \underset{h\to 0}{lim}\frac{-2ph}{-h(\sqrt{1-ph}+\sqrt{1+ph})}=-\frac{1}{2}$

$\Rightarrow \underset{h\to 0}{lim}\frac{2p}{(\sqrt{1-ph}+\sqrt{1+ph})}=-\frac{1}{2}$

$\Rightarrow p=-\frac{1}{2}$