IIT JEE , AIEEE Forum - Mathematics , Algebra

akshay.khare91

find the domain of following function-

where[.] = greatest integer function

| | = modulus....

vnkt.swaroop

when x>0 it is x-1+7-x-6=0 which not defined.

So x<0 and it is [-x+1]+[7+x]-6

=[-x]+[1]+[7]+[x]-6

=-[x]-1+1+7+[x]-6

=1;so domain is (-infinity,0).

ankurgupta91

1) when 0<x<=1 , so, [-x]=-1

[1-x] + [7-x] - 6

= 1+ [-x] + 7 + [-x] -6

= 1-1+7-1-6

= 0

so no possible values in this interval

2) when 1<x<=7

[x-1] + [7-x] - 6

= [x] -1 +7 [-x] -6

= [x] +[-x]

=0 when x is integer or = -1 when x is non integer

so no possible values in this interval

3) when x>7

[x-1] + [x-7] -6

= [x] -1 +[x] -7 -6

= 2[x] - 14 > 0

= [x] > 7

x>=8

4) when x<= 0

[1-x] + [7-x] -6

= 1+[-x] +7 + [-x] -6

= 2[-x] + 2 >0

[-x] >-1

x<= 0

so the ans is x belongs to (-infinity , 0 ] union [8, infinity )

hope u gt it .......

little_genius

[/x-7/] + [/ x-1/ ]
for x>=7.....

=[ x-7]+[x-1]-6 which is >0.. since [x-1] >=6 and x>0.....
but for x=7 the function bcomes undefined ..
hence x>7 belonge to domain...

for 1 <=x<7

[/x-7/] + [/ x-1/ ] -6
=[7-x]+[x-1] -6 ..

now, we know [x]so [7-x]+[x-1] < 7-x+x-1 <6...

so [7-x]+[x-1] -6 <0 ... hence for 1<=x<7 the expression is undefined...

again , for 0< x<1 ..

[/7-x/]+[/x-1/] -6 ..=[7-x]+[1-x]-6<(7-x) +( 1-x ) -6<8-2x>0 since 0 also for x<0 its 8-2x-6>0..
so expression is real for -inf
hence DOMAIN= {-INF,1} UNION {7,INF}..........

sreeee

little genious....yahooooooo par :D

yes ur answer seems correct

ankurgupta91

little genius u r wrng
check by putting x= 7.3
it comes out to be zero
nd check by putting x =0.3
it again comes out zero (not possible)

hope u gt it ..........

akshay.khare91

hey all of u are wrong
the ans is R - [0,1) U {1,2,3,4,5,6} U [7,8).... (means these nos are excluded)

where R is the set of all real nos.

little_genius

@akshay khare ... what in my approach seems wrong?

ankurgupta91

hey akshay i think i hv done it rght
no value bt 1 nd 7 is satisfying the equation
check urself

if sqrt is nt given then ur answer is rght only

hope u gt it

little_genius

first of all look at my approach .... it maybe that i cud hv done some mistake somewhere...

ankurgupta91

@ little genius
see when 7<=x<8
[x-1] = 6 and [x-7] = 0
so , [x-1] + [x-7] - 6 =0
not possible
so no possible values btw 7 and 8
hope u gt it nw

vnkt.swaroop

reply to my answer.

akshay.khare91

i have already given the ans..
plss check wid that,.,..

akshay.khare91

heyy noone who can solve it correctly !!!!!!!!!

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