20250919 Junior Varsity Training

Problem 1

Find all the roots of equation $2x-2.8=\lfloor x\rfloor$.

Solution

$$\begin{array}{rl}2x-2.8=\lfloor x\rfloor & ⟹x-1<2x-2.8\le x\\ & ⟺1.8<x\le 2.8\\ & ⟹\lfloor x\rfloor =1\vee \lfloor x\rfloor =2\end{array}$$

If $\lfloor x\rfloor =1$,

$$\begin{array}{rl}2x-2.8=\lfloor x\rfloor & ⟺2x-2.8=1\\ & ⟺x=1.9\end{array}$$

If $\lfloor x\rfloor =2$,

$$\begin{array}{rl}2x-2.8=\lfloor x\rfloor & ⟺2x-2.8=2\\ & ⟺x=2.4\end{array}$$

Therefore,

$$2x-2.8=\lfloor x\rfloor ⟺x=1.9\vee x=2.4$$

The roots of equation $2x-2.8=\lfloor x\rfloor$ are $x=1.9$ and $x=2.4$.

Problem 2

Solve equation $x^2-\lfloor x\rfloor -2=0$.

Solution

$$\begin{array}{rl}{x}^2-\lfloor x\rfloor -2=0& ⟺x^2-2=\lfloor x\rfloor \\ & ⟹x-1<x^2-2\le x\\ & ⟺x-1<x^2-2\wedge x^2-2\le x\\ & ⟺(x<\frac{1-\sqrt{5}}{2}\vee x>\frac{1+\sqrt{5}}{2})\wedge (-1\le x\le 2)\\ & ⟺-1\le x<\frac{1-\sqrt{5}}{2}\vee \frac{1+\sqrt{5}}{2}<x\le 2\\ & ⟹\lfloor x\rfloor =-1\vee \lfloor x\rfloor =1\vee \lfloor x\rfloor =2\end{array}$$

If $\lfloor x\rfloor =-1$,

$$\begin{array}{rl}{x}^2-2=\lfloor x\rfloor & ⟺x^2-2=-1\\ & ⟺x=-1\end{array}$$

If $\lfloor x\rfloor =1$,

$$\begin{array}{rl}{x}^2-2=\lfloor x\rfloor & ⟺x^2-2=1\\ & ⟺x=\sqrt{3}\end{array}$$

If $\lfloor x\rfloor =2$,

$$\begin{array}{rl}{x}^2-2=\lfloor x\rfloor & ⟺x^2-2=2\\ & ⟺x=2\end{array}$$

Therefore,

$$x^2-\lfloor x\rfloor -2=0⟺x=-1\vee x=\sqrt{3}\vee x=2$$

The roots of equation $x^2-\lfloor x\rfloor -2=0$ are $x=-1$, $x=\sqrt{3}$ and $x=2$.

Problem 3

Solve equation $\lfloor \frac{5x-13}{3}\rfloor =\frac{3x+1}{5}$.

Solution

$$\begin{array}{rl}\lfloor \frac{5x-13}{3}\rfloor =\frac{3x+1}{5}& ⟺\exists k\in \mathbb{Z},\lfloor \frac{5x-13}{3}\rfloor =k\wedge \frac{3x+1}{5}=k\\ & ⟺\exists k\in \mathbb{Z},\lfloor \frac{5x-13}{3}\rfloor =k\wedge x=\frac{5k-1}{3}\\ & ⟺\exists k\in \mathbb{Z},\lfloor \frac{25k-44}{9}\rfloor =k\wedge x=\frac{5k-1}{3}\end{array}$$ $$\begin{array}{rl}\lfloor \frac{25k-44}{9}\rfloor =k& ⟺k\le \frac{25k-44}{9}<k+1\\ & ⟺\frac{11}{4}\le k<\frac{53}{16}\\ & ⟺k=3\end{array}$$

Therefore,

$$\begin{array}{rl}\lfloor \frac{5x-13}{3}\rfloor =\frac{3x+1}{5}& ⟺\exists k\in \mathbb{Z},k=3\wedge x=\frac{5k-1}{3}\\ & ⟺x=\frac{14}{3}\end{array}$$

The root of equation $\lfloor \frac{5x-13}{3}\rfloor =\frac{3x+1}{5}$ is $x=\frac{14}{3}$.

Problem 4

$$3x-3=\lfloor x\rfloor$$

Solution

$$\begin{array}{rl}3x-3=\lfloor x\rfloor & ⟹x-1<3x-3\le x\\ & ⟺1<x\le \frac{3}{2}\\ & ⟹\lfloor x\rfloor =1\end{array}$$

If $\lfloor x\rfloor =1$,

$$\begin{array}{rl}3x-3=\lfloor x\rfloor & ⟺3x-3=1\\ & ⟺x=\frac{4}{3}\end{array}$$

Therefore,

$$3x-3=\lfloor x\rfloor ⟺x=\frac{4}{3}$$

Problem 5

$$5-x^2=\lfloor x\rfloor$$

Solution

$$\begin{array}{rl}5-x^2=\lfloor x\rfloor & ⟹x-1<5-x^2\le x\\ & ⟺x-1<5-x^2\wedge 5-x^2\le x\\ & ⟺-3<x<2\wedge (x\le \frac{-1-\sqrt{21}}{2}\vee x\ge \frac{-1+\sqrt{21}}{2})\\ & ⟺-3<x\le \frac{-1-\sqrt{21}}{2}\vee \frac{-1+\sqrt{21}}{2}\le x<2\\ & ⟹\lfloor x\rfloor =-3\vee \lfloor x\rfloor =1\end{array}$$

If $\lfloor x\rfloor =-3$,

$$5-x^2=-3⟺x=-2\sqrt{2}$$

If $\lfloor x\rfloor =1$,

$$5-x^2=1$$

Therefore,

$$5-x^2=\lfloor x\rfloor ⟺x=-2\sqrt{2}$$

Problem 6

$$|x-2|=\lfloor x\rfloor$$

Solution

$$\begin{array}{rl}\lfloor x\rfloor \in \mathbb{Z}& ⟺|x-2|\in \mathbb{Z}\\ & ⟺x\in \mathbb{Z}\\ & ⟺\lfloor x\rfloor =x\end{array}$$

Therefore,

$$\begin{array}{rl}|x-2|=\lfloor x\rfloor & ⟺|x-2|=x\\ & ⟺x=1\end{array}$$

Problem 7

$$\frac{3x+2}{4}=\lfloor \frac{6x-1}{5}\rfloor$$

Solution

$$\begin{array}{rl}\frac{3x+2}{4}=\lfloor \frac{6x-1}{5}\rfloor & ⟺\exists k\in \mathbb{Z},\frac{3x+2}{4}=k\wedge \lfloor \frac{6x-1}{5}\rfloor =k\\ & ⟺\exists k\in \mathbb{Z},x=\frac{4k-2}{3}\wedge \lfloor \frac{6x-1}{5}\rfloor =k\\ & ⟺\exists k\in \mathbb{Z},x=\frac{4k-2}{3}\wedge \lfloor \frac{8k}{5}\rfloor =k+1\end{array}$$ $$\begin{array}{rl}\lfloor \frac{8k}{5}\rfloor =k+1& ⟺k+1\le \frac{8k}{5}<k+2\\ & ⟺k+1\le \frac{8k}{5}<k+2\\ & ⟺\frac{5}{3}\le k<\frac{10}{3}\\ & ⟺k=2\vee k=3\end{array}$$

Therefore,

$$\begin{array}{rl}\frac{3x+2}{4}=\lfloor \frac{6x-1}{5}\rfloor & ⟺\exists k\in \mathbb{Z},x=\frac{4k-2}{3}\wedge (k=2\vee k=3)\\ & ⟺x=2\vee x=\frac{10}{3}\end{array}$$

Problem 8

$$\{\begin{array}{l}y=2\lfloor x\rfloor +\sqrt{2}\\ x=3\lfloor y\rfloor -12.5\end{array}$$

Solution

$$\begin{array}{rl}y=2\lfloor x\rfloor +\sqrt{2}\wedge x=3\lfloor y\rfloor -12.5& ⟺\exists k\in \mathbb{Z},\lfloor x\rfloor =k\wedge y=2k+\sqrt{2}\wedge x=3\lfloor y\rfloor -12.5\\ & ⟺\exists k\in \mathbb{Z},\lfloor x\rfloor =k\wedge y=2k+\sqrt{2}\wedge x=3\lfloor 2k+\sqrt{2}\rfloor -12.5\\ & ⟺\exists k\in \mathbb{Z},\lfloor x\rfloor =k\wedge y=2k+\sqrt{2}\wedge x=6k-9.5\\ & ⟺\exists k\in \mathbb{Z},\lfloor 6k-9.5\rfloor =k\wedge y=2k+\sqrt{2}\wedge x=6k-9.5\end{array}$$ $$\begin{array}{rl}\lfloor 6k-9.5\rfloor =k& ⟺k\le 6k-9.5<k+1\\ & ⟺1.9\le k<2.1\\ & ⟺k=2\end{array}$$

Therefore,

$$\begin{array}{rl}y=2\lfloor x\rfloor +\sqrt{2}\wedge x=3\lfloor y\rfloor -12.5& ⟺\exists k\in \mathbb{Z},k=2\wedge y=2k+\sqrt{2}\wedge x=6k-9.5\\ & ⟺x=2.5\wedge y=4+\sqrt{2}\end{array}$$