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25 Jun 2008 18:42:47 IST
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find the domain of following function-
where[.] = greatest integer function
| | = modulus....
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IMPOSSIBLES ARE OFTEN UNTRIED... |
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25 Jun 2008 23:33:36 IST
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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).
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26 Jun 2008 01:54:54 IST
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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 .......
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nobody is perfect......i m nobody.............. |
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26 Jun 2008 09:50:33 IST
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[/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}..........
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IIT GUWAHATI. |
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26 Jun 2008 10:10:46 IST
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little genious....yahooooooo par :D
yes ur answer seems correct
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26 Jun 2008 12:36:19 IST
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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 ..........
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nobody is perfect......i m nobody.............. |
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26 Jun 2008 20:35:09 IST
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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.
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26 Jun 2008 20:39:07 IST
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@akshay khare ... what in my approach seems wrong?
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IIT GUWAHATI. |
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26 Jun 2008 20:39:45 IST
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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
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nobody is perfect......i m nobody.............. |
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26 Jun 2008 20:42:57 IST
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first of all look at my approach .... it maybe that i cud hv done some mistake somewhere...
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IIT GUWAHATI. |
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26 Jun 2008 20:49:44 IST
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@ 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
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nobody is perfect......i m nobody.............. |
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26 Jun 2008 22:18:49 IST
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reply to my answer.
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27 Jun 2008 06:35:22 IST
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i have already given the ans..
plss check wid that,.,..
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3 Jul 2008 20:54:01 IST
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heyy noone who can solve it correctly !!!!!!!!!
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3 Jul 2008 21:41:33 IST
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akshay, there's a slight mistake in your answer. u say that x cannot be 0, but putting x = 0, the denominator comes out to be \sqrt{2}
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Chirag Sangani
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My solution:
Let us find the values of x for which the denominator is 0
So, we have
Now, both [|x-1|] and [|7-x|] must be non-negative integers.
Case 1:
Suppose, [|x-1|] = 6. The x which satisfy this equation are 
Correspondingly, [|7-x|]=0 The x satisfying this eqn are 
Their intersection is  (i)
Case 2:
Suppose, [|x-1|] = 5. The x which satisfy this equation are 
Correspondingly, [|7-x|]=1. The x which satisfy this equation are 
Their intersection is 
Similarly, continuing the cases, we will get  (ii)
Now,
Consider [|x-1|] = 0. The x which satisfy this equation are 
Correspondingly, [|7-x|]=6. The x which satisfy this eqn are 
Their intersection is  (iii)
Taking union of (i), (ii) and (iii), we get
which also shows symmetry and the ends of this range, as it should.
So, the domain would be 
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Chirag Sangani
www.chiragsangani.com |
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3 Jul 2008 22:38:48 IST
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hey yaar why r u nt getting it
sqrt is given there
so, [ |x-1| ] + [ |7-x| ] - 6 > 0
and btw 1 to 6 at non integral values it becomes (-1)
which is nt possible
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nobody is perfect......i m nobody.............. |
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damn.. the sqrt :( i forgot
so the answer would be simply 
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Chirag Sangani
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3 Jul 2008 23:24:06 IST
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hey astatine is right i think coz the
value coming out of modulus in always positive
hence no need to put that condition that denominator should be positive
just we should apply that denominator should not be 0.
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3 Jul 2008 23:28:27 IST
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no actually ankur has a valid point. the denominator is  . Even though [|x-1|] and [|7-x|] are always non-negative, if their sum is less than 6 then the quantity in the root will be less zero.
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Chirag Sangani
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