13th_November_2008

Plot the graph of

[y]= e[x]

24 Answers

62
Lokesh Verma ·

Good work..

This is 100% :)

62
Lokesh Verma ·

because: (answer this correctly u will know why!)

When [x]=0 .. what is x?

1
skygirl ·

NOW MY DOUBT.... :P
wen i am already taking [x],,, y should i consider x=0.1, 0.5, 0.75 etc... ???????????? all of these will boil down to [x]=0..... plz explain.... :(

62
Lokesh Verma ·

[x]=0 means it covers all points x=[0,1)

also for each x(in the above interval),......... y takes values such that [y]=1 so y=[1,2)

So, for x=0, y=[1,2)

for x=0.1, y=[1,2) (so this will be a line above x=0.1
similarly for x=0.5, y=[1,2) (so this will be a line above x=0.5
for x=0.75, y=[1,2) (so this will be a line above x=0.75

(So, we have these red lines as shown!) for x=0.1, 0.5 and 0.75

now this will go for all x between 0 and 1

if we make the red lines for all the x's the whole blue region will get covered.

Now u need to tell me why we dont include the two lines x=1 and y=2?? (That part i leave for u to understand .. an easy on though )

62
Lokesh Verma ·

The solution of this problem is a bit tricky!

The first thing to note is that the LHS is an integer.

RHS = e[x]

so this will take values e-4,e-3, e-2, e-1, e0, e1, e2, e3, e4....... so on..

So, for this to be equal to LHS it has to be an integer.

The only way this can be an integer is when [x]=0 or e0 so LHS =1

so we know that [x]=0. Also, we know that [y]=1 (by substituting [x]=0 in the above equation!)

So we need to draw the equation of [y]=1 when [x]=0

The above graph is the representation of that part!

I will try to explain how this is the graph of [y]=1 when [x]=0!

62
Lokesh Verma ·

1
skygirl ·

am talking of this one....

1
vibhav roy ·

I did not carefully read abhishek's answer. now as I read it i see that he had mentioned that the abobe line segment is not a part of it.

1
vibhav roy ·

I got abhishek's point for 0≤x<1 [x]=0 so,[y]=e[x]1 and so at all these values of x y will lie b/w 1≤y<2. so the graph will be the whole sqare except the line x=1 but why will the line y=2 be included.

62
Lokesh Verma ·

@vaibhav
We dont have to plot
[y]=[e[x]].

But [y]=e[x]

The first graph will be relatively simpler!!!!!!!

33
Abhishek Priyam ·

its [y]=e[x]

so LHS is integer so RHS should be integer.....

got some hint from it.

Don't see the answer or u will be biassed.

1
vibhav roy ·

I didn't get abhishek. I got the graph of y=e[x] but I coudn't modify it to [y]=[e[x]]. please help.

1
skygirl ·

i THINK the graph is like this....

33
Abhishek Priyam ·

Oh yes right edge at x=1 is also not its part....

33
Abhishek Priyam ·

damn... what i have missed.... ok let me see....

62
Lokesh Verma ·

99% correct abhishek.!

33
Abhishek Priyam ·

is the curve only a square having vertices (0,1)(1,1)(1,2)(0,2) above line segment of square not a pat of it....

62
Lokesh Verma ·

first step is to find the graph of

y=e[x]

Then we will need to refine that graph suitably to reach the graph of
[y]=e[x]

1
ANKIT MAHATO ·

do we need to find the graph of y=[x]
and y=e[x] and combine these two suitably

1
voldy ·

I can't see any graph. above ? only the fn y= [e[x]] have you hidden it , if yes how do we see it?

62
Lokesh Verma ·

Yes srinath...
corrected above :)

1
voldy ·

y = e[x] is discontinous at all lnx such that x belong sto I , right?

62
Lokesh Verma ·

No sky.. this is not the answer...
There are a "few" mistakes

This (The one sky has given!) is graph of y=[e[x]]

1
skygirl ·

correct me, if wrong...

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