Solving inequality involving floor function

Question

We have this inequality (over real numbers) : $$x^2-2x\le \frac{\sqrt{1-\lfloor x\rfloor^2}}{\lfloor x \rfloor + \lfloor -x \rfloor}$$

How we can solve it using both of algebraic and geometric methods ?

Answer 1

Hint: $x$ cannnot be integer, otherwise the denominator of the RHS would be $0$.

So $x$ is not integer, which gives $\lfloor x \rfloor+\lfloor -x \rfloor=-1$

Also $\lfloor x \rfloor\in\{-1,0,1\}$ since it must be integer and $1-\lfloor x\rfloor^2>0$ which is equivalent to $\lfloor x\rfloor^2<1$.

Then you solve on each interval $-1<x<0$, $0<x<1$, and $1<x<2$

EDIT: For a geometric solution, I would plot $x^2-2x$ and I would observe that the RHS is either $0$ or $-1$.

Answer 2

RHS is defined only for $-2< x<2$ and $x\ne0$..

If $-1<x<0$ or $0<x<1$, then RHS is $\frac{1}{-1}=-1$. Solve $x^2-2x+1\le0$ and only keep solutions in the initial interval.

If $1\le |x|<2$, RHS is $0$, solve $x^2-2x\le 0$ and only keep solutions in intervals $\left]-2,-1\right]$ and $\left[1,2\right[$.

Remember, when you encounter something like $\lfloor x\rfloor$ or $\left|x\right|$, try first to give it a value by making hypothesis on the positions of $x$.

Answer 3

$$\begin{cases}x= \lfloor x\rfloor+\{x\}\\-x= \lfloor -x\rfloor+1-\{x\}\end{cases}\Rightarrow\lfloor x\rfloor+\lfloor -x\rfloor=-1$$

$$\sqrt{1-\lfloor x\rfloor^2}=\begin{cases}0 \text{ if } -1\le x\lt0\\1\text{ if } 0\le x\lt1\\0 \text{ if } 1\le x\lt2\end{cases}$$ Hence $$\frac{\sqrt{1-\lfloor x\rfloor^2}}{\lfloor x\rfloor+\lfloor -x\rfloor}=\begin{cases}0 \text{ if } -1\le x\lt0\\-1\text{ if } 0\le x\lt1\\0 \text{ if } 1\le x\lt2\end{cases}$$

It follows that if $f(x)=x^2-2x$ then since $$f((-1,0))=(0,3)\\f([0,1))=[-1,0]\\f([1,2))=[-1,0)$$ the solution is $$1\le x\lt 2$$

Answer 4

I can't say anything more than Momo's answer. But since you asked for a geometric solution as well, this may also help: