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bazaltina [42]
3 years ago
11

Let $f(x)$ be a function defined for all positive real numbers satisfying the conditions $f(x) > 0$ for all $x > 0$ and $f

(x - y) = \sqrt{f(xy) + 1}$ for all $x > y > 0$. Determine $f(2009)$.
Mathematics
2 answers:
algol [13]3 years ago
3 0
Suppose we choose x=1 and y=\dfrac12. Then

f(x-y)=\sqrt{f(xy)+1}\implies f\left(\dfrac12\right)=\sqrt{f\left(\dfrac12\right)+1}\implies f\left(\dfrac12\right)=\dfrac{1+\sqrt5}2


Now suppose we choose x,y such that

\begin{cases}x-y=\dfrac12\\\\xy=2009\end{cases}


where we pick the solution for this system such that x>y>0. Then we find

\dfrac{1+\sqrt5}2=\sqrt{f(2009)+1}\implies f(2009)=\dfrac{1+\sqrt5}2

Note that you can always find a solution to the system above that satisfies x>y>0 as long as x>\dfrac12. What this means is that you can always find the value of f(x) as a (constant) function of f\left(\dfrac12\right).
Crazy boy [7]3 years ago
3 0

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

    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}

Now, suppose

x=2009, y=0

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

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