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$f(x)$ is continuous on interval [1,2], and its derivative exist on [1,2], and we have $f(1)=f(2)=0$. Prove there exists at least one point $a, a \in [1,2] $ such that

${f(a) \over a}=2007f^{'}(a)$.

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up vote 0 down vote accepted

Define $g(x)=x^{-1/2007}f(x)$ on $[1,2]$ and apply Rolle's theorem to $g$.

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Given the conditions, and $h(x)$ a continuous function then you can find $\xi$, such that $h(\xi)f(\xi)=f'(\xi)$ define then $g(x)=e^{-\int h(x)\mathrm{d}x}f(x)$ and because $f$ is zero at $1,2$ so is $g$ so you can apply rolle's theorem.

So given similar condition is if you can convert your equation in the previous form you can apply rolle's theorem. In the specific case $h(x)=\frac{1}{2007x}$

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