$$\begin{aligned} & \text { } f(x)=\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x} \text {, if } 1 \leq x<0 \\ & =\frac{2 x+1}{x-2} \quad \text {, if } 0 \leq x \leq 1 \\ \end{aligned}$$

is continuous in the interval $$[-1,1]$$, then $$p=$$

If $$\lim _\limits{x \rightarrow 5} \frac{x^k-5^k}{x-5}=500$$, then the value of $$k$$, where $$k \in N$$ is

$$\begin{aligned} & \text { If the function } \mathrm{f}(\mathrm{x})=1+\sin \frac{\pi}{2}, \quad-\infty<\mathrm{x} \leq 1 \\ & =\mathrm{ax}+\mathrm{b}, \quad 1<\mathrm{x}<3 \\ & =6 \tan \frac{x \pi}{12}, \quad 3 \leq x<6 \\ \end{aligned}$$

is continuous in $$(-\infty, 6)$$, then the values of $$\mathrm{a}$$ and $$\mathrm{b}$$ are respectively.

$$\lim _\limits{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3}=$$