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

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

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

If $$a=\lim _\limits{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}$$ and $$b=\lim _\limits{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots+n^2}{n^3}$$, then

If $$f(x) = {{{4^{x - \pi }} + {4^{x - \pi }} - 2} \over {{{(x - \pi )}^2}}}$$, for $$x \ne \pi $$, is continuous at $$x=\pi$$, then k =