The function $f(x)=\left\{\begin{array}{cc}\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x}, & -1 \leq x<0, \\ \frac{2 x+1}{x-2}, & 0 \leq x \leq 1,\end{array}\right.$ is continuous in the interval [-1, 1], then p is equal to

$-\frac{1}{2}$

$f(x)=\left\{\begin{array}{cc}\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x}, & -1 \leq x<0, \\ \frac{2 x+1}{x-2}, & 0 \leq x \leq 1,\end{array}\right.$

Checking at 0

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

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

so  $f(0)=\frac{2(0)+1}{0-2}=\frac{-1}{2}$

so  $p = \frac{-1}{2}$

Option: B