## Dhaka Secondary (Higher Secondary) 2011/5

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### Dhaka Secondary (Higher Secondary) 2011/5

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 $x$.

Here, $\lceil x\rceil$ represents the minimum integer greater than $x$ and $\lfloor x \rfloor$ represents the maximum integer less than $x$.

- Tahmid Hasan
**Posts:**665**Joined:**Thu Dec 09, 2010 5:34 pm**Location:**Khulna,Bangladesh.

### Re: Dhaka Secondary (Higher Secondary) 2011/5

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

Last edited by Tahmid Hasan on Sat Jan 29, 2011 6:18 pm, edited 1 time in total.

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### Re: Dhaka Secondary (Higher Secondary) 2011/5

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

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### Re: Dhaka Secondary (Higher Secondary) 2011/5

for case (¡¡): $(-1)$;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]

case (¡¡¡): $\frac{-1}{2}$;

case(¡v): $(-2)$...

### Re: Dhaka Secondary (Higher Secondary) 2011/5

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!

"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please

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**install LaTeX fonts**in your PC for better looking equations,**learn****how to write equations**, and**don't forget**to read Forum Guide and Rules.- Tahmid Hasan
**Posts:**665**Joined:**Thu Dec 09, 2010 5:34 pm**Location:**Khulna,Bangladesh.

### Re: Dhaka Secondary (Higher Secondary) 2011/5

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

"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please

Please

**install LaTeX fonts**in your PC for better looking equations,**learn****how to write equations**, and**don't forget**to read Forum Guide and Rules.-
**Posts:**2**Joined:**Mon Jan 31, 2011 3:09 pm**Location:**Dhaka

### Re: Dhaka Secondary (Higher Secondary) 2011/5

What about all real numbers 'R' ?

Last edited by MUjtahid Akon on Wed Feb 02, 2011 5:08 pm, edited 1 time in total.

### Re: Dhaka Secondary (Higher Secondary) 2011/5

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

"Je le vois, mais je ne le crois pas!" - Georg Ferdinand Ludwig Philipp Cantor

### Re: Dhaka Secondary (Higher Secondary) 2011/5

According to the definition if $a$ integer $\lceil a\rceil -\lfloor a\rfloor =2,$otherwise $1$then let $k=\dfrac {\lceil 2x\rceil -2\lfloor x\rfloor} {\lfloor 2x\rfloor -2\lceil x\rceil }$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 $x$.

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 $k\in {-1,-2,-\frac 1 2 }$

One one thing is neutral in the universe, that is $0$.