$f(x)= \begin{cases}\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x}, & -1 \leq x<0 \\ ~~~\frac{2 x+1}{x-2}~~~~~ \quad, & 0 \leq x \leq 1\end{cases}$ is continuous in the interval [-1, 1], then p is equal to

-1/2

It is given that f(x) is continuous in [-1, 1]. Therefore, it is also continuous at x = 0.

∴  $\lim\limits_{x \rightarrow 0^{-}} f(x)=\lim\limits_{x \rightarrow 0^{+}} f(x)=f(0)$

$\Rightarrow \lim\limits_{x \rightarrow 0} \frac{\sqrt{1+p x}-\sqrt{1-p x}}{x}=\lim\limits_{x \rightarrow 0} \frac{2 x+1}{x-2}=\frac{2 \times 0+1}{0-2}$

$\Rightarrow \lim\limits_{x \rightarrow 0} \frac{2 p x}{x(\sqrt{1+p x}+\sqrt{1-p x})}=-\frac{1}{2}$

$\Rightarrow \lim\limits_{x \rightarrow 0} \frac{2 p}{\sqrt{1+p x}+\sqrt{1-p x}}=-\frac{1}{2} \Rightarrow p=-\frac{1}{2}$