If $$\lim_\limits{x \rightarrow 0} \frac{e^{a x}-\cos (b x)-\frac{cx e^{-c x}}{2}}{1-\cos (2 x)}=17$$, then $$5 a^{2}+b^{2}$$ is equal to

Let $$f$$ and $$g$$ be two functions defined by

$$f(x)=\left\{\begin{array}{cc}x+1, & x < 0 \\ |x-1|, & x \geq 0\end{array}\right.$$ and $$\mathrm{g}(x)=\left\{\begin{array}{cc}x+1, & x < 0 \\ 1, & x \geq 0\end{array}\right.$$

Then $$(g \circ f)(x)$$ is :

Let $$f(x)=\left[x^{2}-x\right]+|-x+[x]|$$, where $$x \in \mathbb{R}$$ and $$[t]$$ denotes the greatest integer less than or equal to $$t$$. Then, $$f$$ is :

If $$\alpha > \beta > 0$$ are the roots of the equation $$a x^{2}+b x+1=0$$, and $$\lim_\limits{x \rightarrow \frac{1}{\alpha}}\left(\frac{1-\cos \left(x^{2}+b x+a\right)}{2(1-\alpha x)^{2}}\right)^{\frac{1}{2}}=\frac{1}{k}\left(\frac{1}{\beta}-\frac{1}{\alpha}\right), \text { then } \mathrm{k} \text { is equal to }$$ :