Dhaka Secondary (Higher Secondary) 2011/5

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BdMO
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Dhaka Secondary (Higher Secondary) 2011/5

Unread post by BdMO » Fri Jan 28, 2011 10:00 pm

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 $x$.

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Tahmid Hasan
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Re: Dhaka Secondary (Higher Secondary) 2011/5

Unread post by Tahmid Hasan » Sat Jan 29, 2011 1:12 pm

soln for number5(partial)
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 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

Unread post by Zzzz » Sat Jan 29, 2011 2:28 pm

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

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

Unread post by AntiviruShahriar » Sat Jan 29, 2011 4:53 pm

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): $(-2)$...

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

Unread post by Moon » Tue Feb 01, 2011 11:28 pm

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! ;)
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Tahmid Hasan
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Re: Dhaka Secondary (Higher Secondary) 2011/5

Unread post by Tahmid Hasan » Tue Feb 01, 2011 11:31 pm

moon bhai eida ki bola lage? :ugeek:
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Moon
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Re: Dhaka Secondary (Higher Secondary) 2011/5

Unread post by Moon » Tue Feb 01, 2011 11:56 pm

ha ha...eita amre confuse kre disilo at first, that's why I posted the caution! :)
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MUjtahid Akon
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Re: Dhaka Secondary (Higher Secondary) 2011/5

Unread post by MUjtahid Akon » Wed Feb 02, 2011 4:10 pm

What about all real numbers 'R' ?
Last edited by MUjtahid Akon on Wed Feb 02, 2011 5:08 pm, edited 1 time in total.

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Avik Roy
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Re: Dhaka Secondary (Higher Secondary) 2011/5

Unread post by Avik Roy » Wed Feb 02, 2011 4:36 pm

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

Unread post by Masum » Thu Feb 03, 2011 1:31 am

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 $k\in {-1,-2,-\frac 1 2 }$
One one thing is neutral in the universe, that is $0$.

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