1x→[2npi ,(2n+1) pi)]-{2npi+npi/2} n is >=0
y→[0,1)
x→{2npi+npi/2}
y→[1,2)
x→{(2n+1)pi,(2n+2)pi }
y→(0,-1]
x→{-2npi,-(2n+1)pi}
y→(0,-1]
x→[-(2n+1)pi,-(2n+2)pi]-{-(2n+3/2)pi}
y→[0,1)
x→-(2n+3/2)pi
y→[1,2)
12 Answers
1
1@eureka wat u did in 3rd graph inpost#8 .....can u plz explain...sorry if its a silly doubt...but im not getting[2]
23kaustab the graph we r plotting isnt a function ....i.e for 1 value of x there can be more than 1 value of y ...
for eg: if u plot the graph of [|x|] + [|y|] = 3 ten u will understand every part of eureka's solution
1i mean to say i got uptil graph of y=[sinx]
now how u draw graphof [y]=[sinx]........sorry if its silly....but i m a novice
23see if 0 ≤sinx< 1 ...then [sinx] = 0 ok..
so [y] = 0
but [y]= 0 is satisfied by y lying between [0,1) so it will be like a shaded box ..... i think nishant sir will explain u betr