Consider the following functions

I. $f(x)= \begin{cases}\frac{1}{2}-x & , x<\frac{1}{2} \\ \left(\frac{1}{2}-x\right)^2 & , x \geq \frac{1}{2}\end{cases}$

II. $f(x)=|3 x-1|$

III. $f(x)=x|x|$

IV. $f(x)=|x|$

Then, on $[0,1]$ Lagrange's mean value theorem is applicable to the functions

$$ \mathop {\lim }\limits_{x \to \infty }\left[\frac{n+1}{n^2+1^2}+\frac{n+2}{n^2+2^2}+\frac{n+3}{n^2+3^2}+\ldots+\frac{n+2 n}{n^2+4 n^2}\right]= $$

Let $f(x)=\left\{\begin{array}{cl}1+\frac{2 x}{a}, & 0 \leq x \leq 1 \\ a x, & 1 < x \leq 2\end{array}\right.$.If $\lim _{x \rightarrow 1} f(x)$ exists, then the sum of the cubes of the possible values of $a$ is

Let $[P]$ denote the greatest integer $\leq P$. If $0 \leq a \leq 2$, then the number of integral values of ' $a$ ' such that $\lim \limits_{x \rightarrow a}\left(\left[x^2\right]-[x]^2\right)$ does not exist is