Match the functions in Column I with their properties in Column II. In the following [ $x$ ] denotes the greatest integer less than or equal to $x$.
| Column I | Column II | ||
|---|---|---|---|
| A | $$ x|x| $$ |
I | Strictly increasing and continuous in $(-1,1)$ |
| B | $$ \sqrt{|x|} $$ |
II | Continuous but not differentiable in $(-1,1)$ |
| C | $$ x+[x] $$ |
III | Differentiable in $(-1,1)$ |
| D | $$ |x-1|+|x+1|+|x| $$ |
IV | Differentiable in $(-1,0) \cup(0,1)$ |
| V | Strictly increasing and not differentiable in $(-1,1)$ | ||
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
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