Question

Find the number of elements in the range of
$f\left(x\right)=\left[x\right]+\left[2x\right]+\left[\frac{2}{3}x\right]+\left[3x\right]+\left[4x\right]+\left[5x\right]$ for $0\le x<3$
and hence when $xϵ[0,n]$, where $nϵN$ and [] denotes the greatest integer function.

Answer 1

Given, $f\left(x\right)=\left[x\right]+\left[2x\right]+\left[\frac{2}{3}x\right]+\left[3x\right]+\left[4x\right]+\left[5x\right]$
Since, [hx] changes its value at every integral multiple of $\frac{1}{k}$.
$\Rightarrow$ [x] will changes at every integral multiple of 1.
$\Rightarrow$ [2x] will changes at every integral multiple of $\frac{1}{2}$.
$\Rightarrow$ [3x] will changes at every integral multiple of $\frac{1}{3}$.
$\Rightarrow$ [4x] will changes at every integral multiple of $\frac{1}{4}$.
$\Rightarrow$ [5x] will changes at every integral multiple of $\frac{1}{5}$.
and $\left[\frac{2}{3}x\right]$ will changes at every integral multiple of $\frac{3}{2}$.
They would change all together at every multiple of LCM of
$\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{3}{2}\}=3$
ie, all will would change of every multiple of 3.
Now, number of total points at which f(x) will change its value in interval [0, 3)
will depend on the total number of different terms in the following cases:
[x]=0, 1, 2
[2x]=$0,\frac{1}{2},\frac{2}{2},\frac{3}{2},\frac{4}{2},\frac{5}{2}$
[3x]=$0,\frac{1}{3},\frac{2}{3},\frac{3}{3},\frac{4}{3},\frac{5}{3},\frac{6}{3},\frac{7}{3},\frac{8}{3}$
[4x]=$0,\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{4}{4},\frac{5}{4},\frac{6}{4},\frac{7}{4},\frac{8}{4},\frac{9}{4},\frac{10}{4},\frac{11}{4}$
[5x]=$0,\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},\frac{5}{5},\frac{6}{5},\frac{7}{5},\frac{8}{5},\frac{9}{5},\frac{10}{5},\frac{11}{5},\frac{12}{5},\frac{13}{5},\frac{14}{5}$
$\left[\frac{2}{3}x\right]=0,\frac{3}{2},\frac{6}{2}$
$\therefore$ f(x) will change its value in the intervals.
$0\le x<\frac{1}{5}$, $\frac{1}{5}\le x<\frac{1}{4}$, $\frac{1}{4}\le x<\frac{1}{3}$, ..., $\frac{14}{5}\le x<3$
$\Rightarrow$ total number of different terms in above equation=30
Hence, number of terms in the range of f(x) for $0\le x<3$ is 30.
If $xϵ[0,n]$, then
(i) If n=3k, number of terms in the range=(30k+1)
(ii) If n=3k+1, then number of terms in the range=30k+(3+4+5+1)-2=30k+11
(iii) If n=3k+2, then number of terms in the range=30k+(6+8+10+2)-(2+2)-2=30k+22.