Case Study

Let

$f(x) = \{x\} + \dfrac 13$

The graph of $\left\{ x \right\}$ is a discontinuous graph with slope $1$ whenever defined and is bounded in $0 \le \{x\} < 1$ . If we add ${1}/{3}$ our graph shifts ${1}/{3}$ units up and the bound for $f$ is $\dfrac 13 \le f(x) < \dfrac 43$ .

Let

$g(x) = \dfrac{1}{\lfloor x \rfloor} + \dfrac{1}{\lfloor 2x \rfloor}$

The graph of $\lfloor x \rfloor$ is a discontinuous graph with slope $0$ whenever defined, thus for a certain interval it assumes a certain constant value. The term $2x$ tells us that the graph will assume a set of constant values for the intervals

• $\displaystyle \bigcup_{n \in \mathbb{Z} \setminus 0} \left[ n , n + \dfrac 12 \right)$ ; and
• $\displaystyle \bigcup_{n \in \mathbb{Z} \setminus 0} \left[ n + \dfrac 12 , n + 1 \right)$

Note that for $g(x) = f(x)$ , we must have $n \ge 1$ as $f(x)$ is positive for all values of $x$ .

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Case 1:

When $x \in \displaystyle \bigcup_{n \in \mathbb{Z} , n \ge 1} \left[ n , n + \dfrac 12 \right)$ , we have $\lfloor x \rfloor = n$ and $\lfloor 2x \rfloor = 2n$ , thus

$\dfrac{1}{\lfloor x \rfloor} + \dfrac{1}{\lfloor 2x \rfloor} = \dfrac{1}{n} + \dfrac{1}{2n} = \dfrac{3}{2n}$

Also in this interval, we have

$0 \le \{x\} < \dfrac 12 \implies \dfrac 13 \le \{x\} + \dfrac 13 < \dfrac 56$

Thus

$\dfrac 13 \le \dfrac{3}{2n} < \dfrac 56$

Solving both inequalities (keeping in mind that $n \ge1$ ) gives

$\dfrac 95 < n \le \dfrac 92$

As $n$ is an integer therefore, $\boxed{n = 2,3,4}$ .

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Case 2:

When $x \in \displaystyle \bigcup_{n \in \mathbb{Z} , n \ge 1} \left[ n + \dfrac 12 , n + 1 \right)$ , we have $\lfloor x \rfloor = n$ and $\lfloor 2x \rfloor = 2n+1$ , thus

$\dfrac{1}{\lfloor x \rfloor} + \dfrac{1}{\lfloor 2x \rfloor} = \dfrac 1n + \dfrac{1}{2n+1} = \dfrac{3n+1}{n(2n+1)}$

Also in this interval

$\dfrac 12 \le \{x\} < 1 \implies \dfrac 56 \le \{x\} + \dfrac 13 < \dfrac 43$

Thus

$\dfrac 56 \le \dfrac{3n+1}{n(2n+1)} < \dfrac 43$

Solving first inequality (keeping in mind that $n \ge1$ ) gives

$1 \le n < \dfrac{13 + \sqrt{409}}{20} \implies 1 \le n < 1.66$

and solving second inequality gives

$n > \dfrac{5 + \sqrt{37}}{2} \implies n > 5.541$

As $n$ is an integer therefore $n = 1$ or $n \ge 6$ .

But observe that for $n=1$ , we have $\dfrac{3n+1}{2n^2 + n} = \dfrac 43$ which does not obey the condition that $\dfrac{3n+1}{2n^2 + n} < \dfrac 43$ .

Also for $n \ge 6$ , we have $\dfrac{3n+1}{2n^2 + n} < \dfrac{3(6)+1}{2(6)^2+6} = \dfrac{19}{78} < \dfrac 56$ which does not obey the condition that $\dfrac{3n+1}{2n^2 + n} \ge \dfrac 56$ .

Thus the given case will not give any solutions for $n$ .

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Hence there are only $\boxed{3}$ possible solutions (from the first case) which lie

• in the interval $\left[ 2 , \dfrac 52 \right)$ when $n = 2$ ;
• in the interval $\left[ 3 , \dfrac 72 \right)$ when $n = 3$ ;
• in the interval $\left[ 4 , \dfrac 92 \right)$ when $n = 4$ .