IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

BuddyGuy

What is the range of the function...f(x)= x[x]??

chander_singla06

BuddyGuy

anit_sahu

the range is obviously from [0,infinity)

taruntanuj007

Range is all real numbers!!! i think sahu made a mistake thinking its mod sign or something

magiclko

I think range is [0, infinity)

cuz [x] value of a negative number will be a negative integer only... so range will be only positive real numbers!!!

amaron

No,No.anit_sahu is right.it obviously it can not be negative as if x<0 then obviously [x]<0 implies negative * negative =positive

therefore the function is always>=0

That solves the problem!

cbg2007

all postive values

visioniit

all positive values i.e. [0,infinity)

DON007

hey look.............

everybody...............

if u r saying that range is

from 0 to infinity........................

i think so it is wrong ..................absolutely wrong................

because............lets find 0.2 in the range.................

for this lets put it in the function...............

we get..............

y= 0.2 * 0 ( 0 is greatest integer of 0.2)

y=0..........

so............there cant be any value from (0,1)..................

in the range...............u can check it............

similarly there r many values that will not lie in the range......................

lets find 2.82...............

put x= 2.82 in function./............

we get y= 2.82* 2 =5.64...........

or take the underroot of 2.82..................

then.............

y= sqrt 2.82* 1 .............. ( 1 is greatest integer of sqrt 2.82)

= sqrt 2.82...............

so there is no particular value 4 its range ................

i think so.............

i m not against anybody ...............but please think abt it................

sorry ......if i m wrong.................

thanku

iitkgp_bipin

Consider any number a.b where a is its greatest integer and b is its fractional part.

x[x] = (a.b)[a.b] = a(a.b)
It varies from (a)(a) to (a)(a+1) i.e. [a²,a²+a)

Hence range is [a²,a²+a) where a = 0,1,2,3............

Similarly you can obtain for negative values.

apurvanaik

hey DON007,

consider -0.2.

now [-0.2]= -1 so (-1)*(-0.2)= 0.2

so the range is [0, infinity)

DON007

great work apurv.........

but what i m sying is that ........there r values in range 0 to infinity which cant lie.............

that is ................

eg .......2...........

if u take

x=2............then y= 2* 2= 4..........

x=-2 ..........then.....y= -2*-2=4

x=sqrt2......then y= sqrt2* 1= sqrt2.

x=-sqrt2...then y= -sqrt2*2= 2* sqrt2............

so this is only one example...........but there r many values.........

i may be wrong.............

okkkkkk

thanku

apurvanaik

hello DON007,

if you try to plot a graph of this function you will understand it better. When x takes values [0,1) the corresponding y is 0. when x takes values from [1,2) it is the x=y line. this line will end just before 2. Now for positive x, y doesnt take values between 0 and 1 but for x=(0,-1] it does. which is again x=y line in 2nd quadrant. similarly, take x=[2,3) now this will be x=2y line which starts from y=4. so for positive x y doesnt take values between [2,4). now consider values of x between (-1,-2]. this is x=2y line which starts from 2 and ends just before 4....

just go on doing this. i think u will get all the values.

n my name is apurva.

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