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Let $f$ be a real valued function such that$f(x)$+ $2f \left(\dfrac{2002}{x}\right) =3x$, find $f(x)$

Attempt:

Substituting $x=1$ and $x=2002$ and solving the simultaneous equations obtained, I got:

$f(2002)= -2000$ and $f(1)= 4003$

Now, $f(1)+f(2002)= 2003$

Also, there are $2003-1$ integers between $1$ and $2002$ (inclusive).

How do I proceed? Any hints?

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2 Answers 2

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We have

$$f\left(\frac{2002}{x}\right)+2f(x)=3\left(\frac{2002}{x}\right)$$

Solving,

$$f(x)=\frac{4004}{x}-x$$

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Enter in $\frac{2002}{x}$ giving

$f(\frac{2002}{x})+2f(x)=3 \cdot \frac{2002}{x}$


$f(\frac{2002}{x})+2f(x)=\frac{6006}{x}$

$f(x)+2f(\frac{2002}{x})=3x$

So you have system of equations to solve

$a+2b=\frac{6006}{x}$

$b+2a=3x$

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  • $\begingroup$ Replacing $x$ in $f(\frac{2002}{x})$ with $\frac{2002}{x}$ gives $f(\frac{2002}{\frac{2002}{x}})=f(2002 \cdot \frac{x}{2002})=f(x)$ $\endgroup$
    – randomgirl
    Apr 10, 2018 at 17:08

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