Let $f(x)=x^2+2 b x+2 c^2$ and $g(x)=-x^2-2 c x+b^2 . x \in R$. If $b$ and $c$ are non-zero real numbers such that min $f(x)>\max g(x)$, then $\left|\frac{c}{b}\right|$ lies in the interval

If $f: R \rightarrow A$, defined by $f(x)=\cos x+\sqrt{3} \sin x-1$ is an onto function then $A=$

Let $g(x)=1+x-[x]$ and ${ }^{\prime}$

$$ f(x)= \begin{cases}-1, & x<0 \\ 0, & x=0,[x] \text { denotes the greatest integer less } \\ 1, & x>0\end{cases} $$

than or equal to $x$. Then for all $x, f(g(x))=$

1. Let [ $x$ ] represent the greatest integer less than or equal to $x,\{x\}=x-[x] \sqrt{2}=1.414$ and $\sqrt{3}=1.732$. If $f(x)=\left\{x+\left[\frac{x}{1+x^2}\right]\right\}$ is a real valued function, then $f(\sqrt{2})+f(-\sqrt{3})=$