Dhaka Secondary (Higher Secondary) 2011/5
Forum rules
Please don't post problems (by starting a topic) in the "Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
Please don't post problems (by starting a topic) in the "Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
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.
বড় ভালবাসি তোমায়,মা
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...
*to hide something write : [ h i d e ] text [ / h i d e ] (without spaces)
*to hide something write : [ h i d e ] text [ / h i d e ] (without spaces)
Every logical solution to a problem has its own beauty.
(Important: Please make sure that you have read about the Rules, Posting Permissions and Forum Language)
(Important: Please make sure that you have read about the Rules, Posting Permissions and Forum Language)
-
- Posts:125
- Joined:Mon Dec 13, 2010 12:05 pm
- Location:চট্রগ্রাম,Chittagong
- Contact:
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 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.
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.
- 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 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.
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 $0$.