If $\lim\limits_{x \rightarrow 0} \frac{\mathrm{e}^{(\mathrm{a}-1) x}+2 \cos \mathrm{~b} x+(\mathrm{c}-2) \mathrm{e}^{-x}}{x \cos x-\log _{\mathrm{e}}(1+x)}=2$, then $\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2$ is equal to :

Let $[\cdot]$ denote the greatest integer function, and let $f(x)=\min \left\{\sqrt{2} x, x^2\right\}$.

Let $\mathrm{S}=\left\{x \in(-2,2)\right.$ : the function $\mathrm{g}(x)=|x|\left[x^2\right]$ is discontinuous at $\left.x\right\}$.

Then $\sum\limits_{x \in \mathrm{~S}} f(x)$ equals

Let $f: \mathbf{R} \rightarrow(0, \infty)$ be a twice differentiable function such that $f(3)=18, f^{\prime}(3)=0$ and $f^{\prime \prime}(3)=4$.

Then $\lim\limits _{x \rightarrow 1}\left(\log _e\left(\frac{f(2+x)}{f(3)}\right)^{\frac{18}{(x-1)^2}}\right)$ is equal to :

Given below are two statements:

Statement I: $ \lim\limits_{x \to 0} \left( \frac{\tan^{-1} x + \log_e \sqrt{\frac{1+x}{1-x}} - 2x}{x^5} \right) = \frac{2}{5} $

Statement II: $ \lim\limits_{x \to 1} \left( x^{\frac{2}{1-x}} \right) = \frac{1}{e^2} $

In the light of the above statements, choose the correct answer from the options given below: