$f(x)= \begin{cases}\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x}, & -1 \leq x<0 \\ \frac{2 x+1}{x-2} & , 0 \leq x \leq 1\end{cases}$
is continuous in the interval $[-1,1]$, then $p$ is equal to
(a) $-1$
(b) $-1 / 2$
(c) $1 / 2$
(d) 1
(b) $-\frac{1}{2}$
Given: $f(x)=\left\{\begin{array}{c}\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x}, \text { if }-1 \leq x<0 \\ \frac{2 x+1}{x-2}, \text { if } 0 \leq x \leq 1\end{array}\right.$
If $f(x)$ is continuous at $x=0$, then
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)$
$\Rightarrow \lim _{h \rightarrow 0}\left(\frac{\sqrt{1-p h}-\sqrt{1+p h}}{-h}\right)=\lim _{h \rightarrow 0}\left(\frac{2 h+1}{h-2}\right)$
$\Rightarrow \lim _{h \rightarrow 0}\left(\frac{(\sqrt{1-p h}-\sqrt{1+p h})(\sqrt{1-p h}+\sqrt{1+p h})}{-h(\sqrt{1-p h}+\sqrt{1+p h})}\right)=\lim _{h \rightarrow 0}\left(\frac{2 h+1}{h-2}\right)$
$\Rightarrow \lim _{h \rightarrow 0}\left(\frac{(1-p h-1-p h)}{-h(\sqrt{1-p h}+\sqrt{1+p h})}\right)=\lim _{h \rightarrow 0}\left(\frac{2 h+1}{h-2}\right)$
$\Rightarrow \lim _{h \rightarrow 0}\left(\frac{(-2 p h)}{-h(\sqrt{1-p h}+\sqrt{1+p h})}\right)=\lim _{h \rightarrow 0}\left(\frac{2 h+1}{h-2}\right)$
$\Rightarrow \lim _{h \rightarrow 0}\left(\frac{(2 p)}{(\sqrt{1-p h}+\sqrt{1+p h})}\right)=\lim _{h \rightarrow 0}\left(\frac{2 h+1}{h-2}\right)$
$\Rightarrow\left(\frac{(2 p)}{(2)}\right)=\left(\frac{1}{-2}\right)$
$\Rightarrow p=\frac{-1}{2}$