1
dont kno if this is correct[7]
6 Answers
3we just draw graph of y=1/2x ..... [simple 1 ...]
now we make it [y]=1/2x ... so we get lines only in places where 1/2x is an integer ....
then when we make it {x} ... the graph looks the same betw any two consecutive integer co ordinates on x axis
106{x}[y] = 1/2
as y is an integer. so if [y]=k, k is a natural no. then {x} = 1/2k
So consider [y] = 1 then {x} = 1/2
[y] = 2 {x} = 1/4
[y] = 3 {x} = 1/6
.
.
.
So graph will exist only in 1st and 2nd quadrant.
Its basically archana's graph in both 1st and 2nd quadrant but there also will be value of x in [-1,1] which she has not shown.