$$ \mathop {\lim }\limits_{n \to \infty } \frac{1}{n^3} \sum\limits_{k=1}^n k^2 x= $$

Let $f: R \rightarrow R$ be defined by

$$ f(x)=\left\{\begin{array}{cc} a-\frac{\sin [x-1]}{x-1}, & \text { if } x>1 \\ 1, & \text { if } x=1 \\ b-\left[\frac{\sin [x-1]-[x-1]}{([x-1])^3},\right. & \text { if } x<1 \end{array}\right. $$

where $[t]$ denotes the greatest integer less than or equal to $t$. If $f$ is continuous at $x=1$, then $a+b=$

If $g$ is the inverse of the function $f(x)$ and $g(x)=x+\tan x$, then $f^{\prime}(x)=$

If $\sqrt{x-x y}+\sqrt{y-x y}=1$, then $\frac{d y}{d x}=$