Rangpur Higher Secondary 2011/3

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BdMO
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Rangpur Higher Secondary 2011/3

Unread post by BdMO » Wed Feb 02, 2011 9:05 pm

Problem 3:
If $-3<f<4$ and $-2<g<1$, then what is the range of possible values of $fg$?

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leonardo shawon
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Re: Rangpur Higher Secondary 2011/3

Unread post by leonardo shawon » Wed Feb 02, 2011 11:45 pm

its $6<fg<4$
i did this way,, i multiple the upper limit and lower limit of f and g.
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Mehfuj Zahir
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Re: Rangpur Higher Secondary 2011/3

Unread post by Mehfuj Zahir » Thu Feb 03, 2011 12:08 am

THE INTERVAL OF fg should be [-8,6]
Last edited by Mehfuj Zahir on Thu Feb 03, 2011 2:38 am, edited 1 time in total.

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leonardo shawon
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Re: Rangpur Higher Secondary 2011/3

Unread post by leonardo shawon » Thu Feb 03, 2011 2:02 am

$-3<f<1$ and $-2<g<4$
if $f=-2.90$ and $g=-1.90$ their multiple $fg$ will give a positive answer!!!
thats why $6<fg<4$
Ibtehaz Shawon
BRAC University.

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Mehfuj Zahir
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Re: Rangpur Higher Secondary 2011/3

Unread post by Mehfuj Zahir » Thu Feb 03, 2011 2:27 am

if one of them is negative then?first time i was wrong the ans is (-8,6)

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*Mahi*
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Re: Rangpur Higher Secondary 2011/3

Unread post by *Mahi* » Fri Feb 04, 2011 6:05 pm

leonardo shawon wrote:$-3<f<1$ and $-2<g<4$
if $f=-2.90$ and $g=-1.90$ their multiple $fg$ will give a positive answer!!!
thats why $6<fg<4$
It's two ranges of $f$ and $g$ .It doesn't mean that if $f$ is $-3$ then $g$ is $-2$.
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leonardo shawon
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Re: Rangpur Higher Secondary 2011/3

Unread post by leonardo shawon » Fri Feb 04, 2011 11:42 pm

i got that later. My mistake! Sowry
Ibtehaz Shawon
BRAC University.

long way to go .....

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