Let $$f: \rightarrow \mathbb{R} \rightarrow(0, \infty)$$ be strictly increasing function such that $$\lim _\limits{x \rightarrow \infty} \frac{f(7 x)}{f(x)}=1$$. Then, the value of $$\lim _\limits{x \rightarrow \infty}\left[\frac{f(5 x)}{f(x)}-1\right]$$ is equal to

If the function $$f:(-\infty,-1] \rightarrow(a, b]$$ defined by $$f(x)=e^{x^3-3 x+1}$$ is one - one and onto, then the distance of the point $$P(2 b+4, a+2)$$ from the line $$x+e^{-3} y=4$$ is :

$$\text { If } f(x)=\left|\begin{array}{ccc} x^3 & 2 x^2+1 & 1+3 x \\ 3 x^2+2 & 2 x & x^3+6 \\ x^3-x & 4 & x^2-2 \end{array}\right| \text { for all } x \in \mathbb{R} \text {, then } 2 f(0)+f^{\prime}(0) \text { is equal to }$$