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39 Answers
62Zero scope.
This is only to make you guys think..
Sometimes even I dont know the graph fully. But I give them because it makes each of us THINK!
1
plz wo half half sare mei bhara hoga...i dont have the energy to fill d entire thing...
1nishant bhaiya give the answer none of us have the energy to work for that graph....
3how will be the area of the shaded portion will also be a part of the graph [x][y]=[xy]
11from (0,1) in x and in y in 1st quad the equality will be satisfied for all the values
same is the case wid the 3rd quad!
3at first i thot there will be no graph in the interval (0,1) bcos
[x]=0 [y]=[xy]/0 so confused abt that pt and when i come to the graph drawn it gives me a bit more idea.
11Very tough question bhaiya. No inspirations or bright ideas strking me right now.
1but genius we cant hav pts. wer either x or da y coordi 0<x/y<1 and da other coordi gr8er than 1 as LHS bcums "0" while RHS not equal to zero!!!!!!!
1yes u r correct .....
this graph seems to be complicated to draw than i assumed.....
ok lets any try other method...
oh.... this graph is really notmadeforiitjee .....
33for 0≤x<1
it is 0≤xy<1
for 1≤x<2
it is y(1-x)=I
I is integer...
for other x........
62Lets take an example.....
when [xy]=2
[x]=1, [y]=2 or viceversa will satisfy...
so we have to take the intersection of 2<=xy<3 (the hyperbolic region) with [x]=1 and [y]=2
so we have to take the intersection of 2<=xy<3 (the hyperbolic region) with [x]=2 and [y]=1
and the -ve values of [x] and [y]
Does this hint help some?
1the graph looks very beautiful.....
are the points on hyberbolae excluded??
give other infos...
13no....the points on hyperbola are included
this pattern will continue ...size of the hyperbola gets decreased
i m not sure yet[7]
1very easy graph.....wo kuch kuch lines se and x and y axis use hoke banega..........
right answer...bhaiya????