# Problem Of The Week #402 Jan 20th, 2020

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#### anemone

##### MHB POTW Director
Staff member
Here is this week's POTW:

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The set of real numbers $x$ for which

$\dfrac{1}{x-2009}+\dfrac{1}{x-2010}+\dfrac{1}{x-2011}\ge 1$

is the union of intervals of the form $a<x\le b$.

Find the sum of the lengths of these two intervals.

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#### anemone

##### MHB POTW Director
Staff member

You can find the suggested solution by other below.
Let $f(x)=\dfrac{1}{x-2009}+\dfrac{1}{x-2010}+\dfrac{1}{x-2011}$ .

Note that

$f(x)-f(y)=(y-x)\left(\dfrac{1}{(x-2009)(y-2009)}+\dfrac{1}{(x-2010)(y-2010)}+\dfrac{1}{(x-2011)(y-2011)}\right)$

If $x<y<2009$, then

$y-x>0$,

$\dfrac{1}{(x-2009)(y-2009)}>0,\,\dfrac{1}{(x-2010)(y-2010)}>0,\,\dfrac{1}{(x-2011)(y-2011)}>0$

Thus $f$ is decreasing on the interval $x<2009$, and because $f(x)<0$ for $x<0$, it follows that no values $x<2009$ satisfy $f(x)\ge 1$.

If $2009<x<y<2010$, then $f(x)-f(y)>0$ as before. Thus $f$ is decreasing in the interval $2009<x<2010$. Moreover $f\left(2009+\dfrac{1}{10}\right)=10-\dfrac{10}{9}-\dfrac{10}{19}>1$ and $f\left(2010-\dfrac{1}{10}\right)=\dfrac{10}{9}-10-\dfrac{10}{11}<1$. Thus there is a number $2009<x_1<2010$ such that $f(x)\ge 1$ for $2009<x\le x_1$ and $f(x)<1$ for $x_1<x<2010$.

Similarly, $f$ is decreasing on the interval $2010<x<2011$, $f\left(2010+\dfrac{1}{10}\right)>1$, and $f\left(2011-\dfrac{1}{10}\right)<1$. THus there is a number $2010<x_2<2011$ such that $f(x)\ge 1$ for $2010<x\le x_2$ and $f(x)<1$ for $x_2<x<2011$.

Finally, $f$ is decreasing on the interval $x>2011$, $f\left(2011+\dfrac{1}{10}\right)>1$ and $f(2014)=\dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{3}<1$. Thus there is a number $x_3>2011$ such that $f(x)\ge 1$ for $2011<x\le x_3$ and $f(x)<1$ for $x>x_3$.

The required sum of the lengths of these three intervals is

$x_1-2009+x_2-2010+x_3-2011=x_1+x_2+x_3-6020$

Multiplying both sides of the equation $\dfrac{1}{x-2009}+\dfrac{1}{x-2010}+\dfrac{1}{x-2011}=1$ by $(x-2009)(x-2010)(x-2011)$ and collecting terms on one side of the equation gives

$x^3-x^2(2009+2010+2011+1+1+1)+ax+b=0$

where $a$ and $b$ are real numbers. The three roots of this equation are $x_1,\,x_2$ and $x_3$. Thus $x_1+x_2+x_3=6020+3$ and consequently the required sum equals 3.

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