Daily Graph 31-08-09

Draw the graph of

[y] = [sin x]

12 Answers

62
Lokesh Verma ·

no karna.. there are mistakes.

1
xYz ·

x→[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)

24
eureka123 ·

transformations will work here

62
Lokesh Verma ·

eureka.. the work is incomplete..

24
eureka123 ·

shading...

24
eureka123 ·

ok now??

62
Lokesh Verma ·

slightly wrong still...

Correct now :D

1
Kaustab Sarkar ·

@eureka wat u did in 3rd graph inpost#8 .....can u plz explain...sorry if its a silly doubt...but im not getting[2]

23
qwerty ·

kaustab 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

1
Kaustab Sarkar ·

i 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

23
qwerty ·

see 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

1
Kaustab Sarkar ·

ok ok i think i hav got it.....

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