$x\geq0$ in terms of $k$ that satisfy $\lfloor kx \rfloor = (k+1)\lfloor x \rfloor,$ where $k>0$ is an integer and $\lfloor r \rfloor$ is the greatest integer less than or equal to $r.$ You may wish to consider $k=2$ first.
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- Let
$f(x)=x^3 - 3x^2 - 3x + 1$and$g(x) = x+1.$For which values of$x$is$g(x) > f(x)?$ - Find the set of solutions for
$x$that satisfies$\lfloor \frac{x}{2} \rfloor = 4.$
- Let
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By rewriting
$x,$how can one simplify the equation? -
… in particular, let
$x = n+f,$where$n$is the integer part and$f$is the fractional part$(0\le f<1).$ -
Consider the relation between
$\lfloor n+f \rfloor$and$n.$ -
After simplifying, try considering the different possible cases for
$n.$ -
… and the corresponding values of
$f?$
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In this solution we use the notation
$x\in[a,b)$to mean$a \leq x < b.$For the general case, let
$x=n+f$where$n$is the integer part and$f\in[0,1)$is the fractional part. We have$\lfloor kn+kf \rfloor=(k+1)\lfloor n+f \rfloor$which gives$kn+\lfloor kf \rfloor=(k+1)n$and finally$\lfloor kf \rfloor=n.$This means that$n$can only take the values$n=0,1,\ldots,k-1.$Taking these in turn we get:- For
$n=0,$$f\in[0,\frac{1}{k})$and so$x\in[0,\frac{1}{k}).$ - For
$n=1,$$f\in[\frac{1}{k},\frac{2}{k})$and so$x\in[1+\frac{1}{k},1+\frac{2}{k}).$ $\cdots$- For
$n=r,$$f\in[\frac{r}{k},\frac{r+1}{k})$and so$x\in[r+\frac{r}{k},r+\frac{r+1}{k}).$ $\cdots$
In the end, we take the union of all these sets up to
$n=k-1,$which is formally written as$x\in\bigcup\limits_{n=0}^{k-1} \big[n+\frac{n}{k}, n+\frac{n+1}{k}\big).$Note: this notation is not required for full credit.
- For