Solution:
It is given that, f(x) is a function such that, defined for all positive real numbers satisfying the conditions ,f(x) > 0 ,for all x > 0 , and also
[tex]f(x-y)=\sqrt{f(xy)+1}\\\\x>0,y>0\\\\for, x=1, \text{and} y=\frac{1}{2}\\\\f(1-\frac{1}{2})=\sqrt{f(1\times \frac{1}{2})+1}\\\\f(\frac{1}{2})^2=f(\frac{1}{2})+1\\\\f(\frac{1}{2})=\frac{1+\sqrt{5}}{2}[/tex]
Now, suppose
x=2009, y=0
[tex]f(2009-0)=\sqrt{f(2009*0)+1}\\\\f(2009)=\sqrt{f(0)+1}\\\\f(2009)=\sqrt{f(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}})+1}\\\\f(2009)=\sqrt{\sqrt{f(\frac{1}{\sqrt{2}}*\frac{1}{\sqrt{2}})+1}+1}\\\\f(2009)=\sqrt{\sqrt{f(\frac{1}{2})+1}+1}\\\\f(2009)=\sqrt{\sqrt{\frac{1+\sqrt{5}}{2}+1}+1}\\\\f(2009)=\sqrt\sqrt{\frac{3+\sqrt{5}}{2}}+1}\\\\f(2009)=\sqrt \sqrt{{5.236}{2}}+1}\\\\=\sqrt{3.6180}\\\\=1.9021[/tex]
