$\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\frac{\sqrt{1+\mathrm{px}}-\sqrt{1-\mathrm{px}}}{\mathrm{x}} ; -1\le \mathrm{x}\le 0\\ \frac{2\mathrm{x}+1}{\mathrm{x}-2} ; 0\le \mathrm{x}\le 1\end{array}\right.$is continuous in $[-\mathrm{1,1}]$ then $2\mathrm{p}+1$ is

The given function is $\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\frac{\sqrt{1+\mathrm{px}}-\sqrt{1-\mathrm{px}}}{\mathrm{x}} ; -1\le \mathrm{x}\le 0\\ \frac{2\mathrm{x}+1}{\mathrm{x}-2} ; 0\le \mathrm{x}\le 1\end{array}\right.$

Given that the function is continuous in $\left[-1,1\right]$

Hence, the function is continuous at $x=0$

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

Left hand limit is 

$\begin{array}{rcl}\underset{x\to 0^{-}}{\lim }f\left(x\right)& =& \underset{x\to 0^{-}}{\lim }\frac{\sqrt{1+px}-\sqrt{1-px}}{x}\\ & =& \underset{x\to 0^{-}}{\lim }\frac{{\displaystyle \frac{p}{2\sqrt{1+px}}}+\frac{p}{2\sqrt{1-px}}}{1}\\ & =& p\end{array}$

Right hand limit is 

$\begin{array}{rcl}\underset{x\to 0^{+}}{\lim }f\left(x\right)& =& \underset{x\to 0^{-}}{\lim }\frac{2x+1}{x-2}\\ & =& -\frac{1}{2}\end{array}$

Hence, $p=-\frac{1}{2}\Rightarrow 2p+1=\boxed{\mathbf{0}}$