3x2-2[x]x+[x]2+x2-[x2]=1
2x2-2x[x]+[x]2-[x2]=1
x belongs (0,1) 2x2=1 so we have the x=1/√2
x belongs (0,-1) 2x2-2x=1 then we wil have the line x=1/2-√3/2
x belongs (1,√2)
the graph will be 2x2-2x=0 in the given domain we don have ne graph.
so we have to do the same way for other intervals.
1msp ,u are wrong....
when x belongs to (0,-1),,,it becomes 2x2+ 2x=1
n when x is in (1,√2),,,,it becomes...2x2 -2x=1
62no.. this one is very simple..
I din intend that one.. but i have a question for the next question of the day :D
1sir..........wont {x}2 + {x2} have the same value for digits uptu 1st decimal place......??
62it will
but then the range of this function will be in between 0 and 2..
so the graph will make sense .. because of the narrow range.. one digit could be a lot of difference.. ;)
1sir,graph will oscillate about the x axis i think........i dont think that anyone will be able to draw it.....
62yup i kind of agree.. to a great extent..
but also keep in mind that the Graph of the day exercise here is mostly a qualitative analysis..
I have sometimes ignored to check the convexity or other minute things (unless it is crucial to the graph)
1sir i think u should also start a "equations analyze karo" section..:-)
62@arshad.. the suggestion is good
but so far GOD that i give also serve the same purpose.. [1]
