1
$\begingroup$

Solve the equation:

$$\left \lfloor 3x-x^2 \right \rfloor = \left \lfloor x^2 + 1/2 \right \rfloor$$

In the solution it writes

We notice that $x^{2}+\frac{1}{2}> 0$ therfore $\left \lfloor x^2 - 1/2 \right \rfloor \geq 0$ . From there $\left \lfloor 3x-x^2 \right \rfloor = n \geq 0$.

So far all of this I understand but than it writes:

But $3x-x^2 \leq \frac{9}{4}$ and $\left \lfloor 3x-x^2 \right \rfloor< 3$.

I dont understand this last part how did they get $3x-x^2 \leq \frac{9}{4}$ ?

$\endgroup$
  • 4
    $\begingroup$ I fail to see step 1. Say $x=0$ then $x^2+1/2$ has floor zero so $x^2-1/2$ has floor $-1$ which isn't $\ge 0.$ $\endgroup$ – coffeemath Dec 20 '18 at 4:15
  • $\begingroup$ For $3x-x^2\le \frac 94$, note that $-(x-\frac32)^2$ is the underlying quadratic... $\endgroup$ – abiessu Dec 20 '18 at 4:18
  • $\begingroup$ I think the key observation is that $x^2+\frac{1}{2}$ and $3x-x^2$ are a convex and a concave parabola. $\endgroup$ – Mr.Robot Dec 20 '18 at 4:23
4
$\begingroup$

If we put $-x^2+3x$ into translated form we get:

$$-(x^2-3x)=-\left[\left(x-\frac 3 2\right)^2-\left(\frac32\right)^2\right]=-\left(x-\frac 3 2\right)^2+\frac94$$

We can now see that this is $x\mapsto x^2$ that has been dilated by a factor of -1 (graphically: reflected about the $x$ axis) and translated $\frac 3 2$ units right and $\frac 9 4$ units up. (In simple terms, the graph is open downwards and the vertex is at $\left(\frac 3 2, \frac 9 4\right)$.) Hence, for all $x$, $-x^2+3x\le \frac 9 4$.

Edit

I think there's a typo in the first part of your presented solution (see coffeemath's comment): it should be:

... therefore $\left\lfloor x^2+\frac 1 2\right\rfloor\ge0$. From there...

In any case, I think the solution is directing you to to identify potential cases to investigate: this equation can only hold for floors of 0, 1 and 2.

For example, investigating the floor 0 case:

$$\begin{align} \left\lfloor x^2+\frac 1 2\right\rfloor = 0&\Rightarrow 0\le x^2+\frac 1 2<1\\ x^2+\frac 1 2\ge 0&\Rightarrow x\in\mathbb R\\ x^2+\frac 1 2 < 1&\Rightarrow x\in\left]-\frac{\sqrt 2}{2},\frac{\sqrt2}{2}\right[\\\\ \left\lfloor-x^2+3x\right\rfloor=0&\Rightarrow0\le-x^2+3x<1\\ -x^2+3x\ge0&\Rightarrow x\in\left[0, 3\right]\\ -x^2+3x<1&\Rightarrow x\in\left]-\infty,\frac{3-\sqrt 5}{2}\right[\;\;\bigcup\;\;\left]\frac{3+\sqrt5}{2},\infty\right[ \end{align}$$

Taking the intersection of all these sets gives us part of the solution to the original problem:

$$x\in\left[0, \frac{3-\sqrt 5}{2}\right[$$

Similarly, investigating the floor 1 case gives us that the equation holds for $x\in\left[\frac{\sqrt 2}{2}, 1\right[$, and the floor 2 case gives us that the equation holds for $x\in\left[\frac{\sqrt6}{2},\frac{\sqrt{10}}{2}\right[$.

The solution set to the problem is thus:

$$\left[0, \frac{3-\sqrt 5}{2}\right[\;\;\bigcup\;\;\left[\frac{\sqrt 2}{2}, 1\right[\;\;\bigcup\;\;\left[\frac{\sqrt6}{2},\frac{\sqrt{10}}{2}\right[$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.