\(f(x) \begin{cases}=\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x} & \text {, if } 1 \leq x < 0 \ =\frac{2 x+1}{x-2} \end{cases}\) \(\text {, if } 0 \leq x \leq 1\)
is continuous in the interval \([-1,1]\), then \(\mathrm{p}=\)

\(\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{\sqrt{1+p x}-\sqrt{1-p x}}{x} \)
\( =\lim _{x \rightarrow 0^{-1}} \frac{[(\sqrt{1+p x})-(\sqrt{1-p x})][(\sqrt{1+p x})+(\sqrt{1-p x})]}{x[(\sqrt{1+p x})+(\sqrt{1-p x})]} \)
\( =\lim _{x \rightarrow 0^{-}} \frac{[(1+p x)-(1-p x)]}{x[\sqrt{1+p x}+\sqrt{1-p x}]}=\lim _{x \rightarrow 0^{-}} \frac{2 p}{\sqrt{1+p x}+\sqrt{1-p x}} \)
\( =\frac{2 p}{2}=p \)
\( \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{2 x+1}{x-2}=\frac{1}{-2}\)
Since \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\), we get \(\mathrm{p}=\frac{-1}{2}\)