Let $f(x)=2-|x-3|, 1 \le x \le 5$ and for rest of the values $f(x)$ can be obtained by using the relation $f(5x)=\alpha\, f(x)\forall\, x \in R$.
The value of $f(2007)$ taking $\alpha = 5$, is: 

For a function F, F(0) = 2, F(1) = 3, F(x + 2) = 2 F(x) - F(x + 1) for x $\geq$ 0, then F(5) is equal to-

If $f(x)=\cosh x+\sinh x$ then $f(x_{1}+x_{2}+.......+x_{n})=$

Let $f(x)=\left | x-x_{1} \right |+\left | x-x_{2} \right |$ where $x_{1} and x_{2}$ are distinct  real numbers. Then the number of points at which f(x) is minimum is: