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Find the range of the function \[f(x)=\frac{\lceil 2x\rceil-2 \lfloor x \rfloor}{\lfloor 2x \rfloor-2\lceil x\rceil} \]
Here, $\lceil x\rceil$ represents the minimum integer greater than $x$ and $\lfloor x \rfloor$ represents the maximum integer less than $x$.

soln for number5(partial) in these cases ,i come up with only 3 solutions,so are they all?

Please write full solution so that someone can tell whether you are right or wrong or where is your fault...

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Tahmid Hasan wrote:soln for number5(partial)
[\hide]i came up with only 1 range $-1$.let's express $x$=$p+k$ where p is an integer and k is a fraction.now let's make some cases
csae1. $k=0$.(then x=p).but i find that there is no domain for x as an integer.
case2.$k=\frac{1}{2}$.(i had to do this because of multiplying $x$ by 2)
case3.$0<k<\frac{1}{2}$
case4.$\frac{1}{2}<k<1$
in these cases ,i come up with only 1 range $-1$.but i was informed that there are 2 more solutions.[\hide]

for case (¡¡): $(-1)$;
case (¡¡¡): $\frac{-1}{2}$;
case(¡v): $(-2)$...

One thing that you must remember here is that the symbols that looks like floors and ceilings are not the same as the usual floors and ceilings!

moon bhai eida ki bola lage?

ha ha...eita amre confuse kre disilo at first, that's why I posted the caution!

What about all real numbers 'R' ?

actually they were meant to be floors and ceilings, but when it came for describing the unpopular symbols, the words were not exact. that's why the -1 came in the range

BdMO wrote:Find the range of the function \[f(x)=\frac{\lceil 2x\rceil-2 \lfloor x \rfloor}{\lfloor 2x \rfloor-2\lceil x\rceil} \]
Here, $\lceil x\rceil$ represents the minimum integer greater than $x$ and $\lfloor x \rfloor$ represents the maximum integer less than $x$.

According to the definition if $a$ integer $\lceil a\rceil -\lfloor a\rfloor =2,$otherwise $1$then let $k=\dfrac {\lceil 2x\rceil -2\lfloor x\rfloor} {\lfloor 2x\rfloor -2\lceil x\rceil }$
If $x$ integer $k=1+\frac 6 {\lfloor 2x\rfloor-2\lfloor x\rfloor -4}=-1$
If $x$ not integer,let $x=\lfloor x\rfloor +j $ where $j $ denotes the fractional part of $x,$and then
$k=1+\frac 3 {\lfloor 2x\rfloor -2\lfloor x\rfloor -2}$
Now if $j\le .5,\lfloor 2x\rfloor =\lfloor 2\lfloor x\rfloor \rfloor =2\lfloor x\rfloor ,k=-\frac 1 2$
If $j>.5,\lfloor 2x\rfloor =2\lfloor x\rfloor +1,k=-2$
So $k\in {-1,-2,-\frac 1 2 }$