Question

$f\left(x\right)=\left\{\begin{array}{ll}\frac{\sqrt{1+px}-\sqrt{1-px}}{x},& -1\le x<0\\ \frac{2x+1}{x-2},& 0\le x\le 1\end{array}\right.$
is continuous in the interval [−1, 1], then p is equal to
(a) −1
(b) −1/2
(c) 1/2
(d) 1

Answer 1

(b) $-\frac{1}{2}$
Given: $f\left(x\right)=\left\{\begin{array}{c}\frac{\sqrt{1+px}-\sqrt{1-px}}{x},\mathrm{if}-1\le x<0\\ \frac{2x+1}{x-2},\mathrm{if}0\le x\le 1\\ \end{array}\right.$

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

$$\begin{gathered} \Rightarrow \underset{h\to 0}{\lim }f\left(-h\right)=\underset{h\to 0}{\lim }f\left(h\right) \\ \Rightarrow \underset{h\to 0}{\lim }\left(\frac{\sqrt{1-ph}-\sqrt{1+ph}}{-h}\right)=\underset{h\to 0}{\lim }\left(\frac{2h+1}{h-2}\right) \\ \Rightarrow \underset{h\to 0}{\lim }\left(\frac{\left(\sqrt{1-ph}-\sqrt{1+ph}\right)\left(\sqrt{1-ph}+\sqrt{1+ph}\right)}{-h\left(\sqrt{1-ph}+\sqrt{1+ph}\right)}\right)=\underset{h\to 0}{\lim }\left(\frac{2h+1}{h-2}\right) \\ \Rightarrow \underset{h\to 0}{\lim }\left(\frac{\left(1-ph-1-ph\right)}{-h\left(\sqrt{1-ph}+\sqrt{1+ph}\right)}\right)=\underset{h\to 0}{\lim }\left(\frac{2h+1}{h-2}\right) \\ \Rightarrow \underset{h\to 0}{\lim }\left(\frac{\left(-2ph\right)}{-h\left(\sqrt{1-ph}+\sqrt{1+ph}\right)}\right)=\underset{h\to 0}{\lim }\left(\frac{2h+1}{h-2}\right) \\ \Rightarrow \underset{h\to 0}{\lim }\left(\frac{\left(2p\right)}{\left(\sqrt{1-ph}+\sqrt{1+ph}\right)}\right)=\underset{h\to 0}{\lim }\left(\frac{2h+1}{h-2}\right) \\ \Rightarrow \left(\frac{\left(2p\right)}{\left(2\right)}\right)=\left(\frac{1}{-2}\right) \\ \Rightarrow p=\frac{-1}{2} \end{gathered}$$