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jatin_axn
Joined: 15 Jul 2005
Posts: 1
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 Posted: Fri Jul 15, 2005 12:34 am Post subject: algebra .... modulous |
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at what point following function is minimum
I x - 1 I + I x-2 I + .............. + I x- 99 I |
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tpsthegreat
Joined: 30 Jun 2005
Posts: 19
Location: ahmedabad
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| Posted: Sun Jul 17, 2005 5:55 pm Post subject: |
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hi ,
the func is minimum at x=50.
soln:
as we know that Ia+bI <= IaI + IbI
so minimum value of IaI + IbI is Ia+bI
so in ur question minimum value of func will be I99x-(1+2+3+......+99)I
= I99x-(99*100/2)I
= I99x-(99*50)I
the minimum value of this be 0 at x=50.
hence ans is 50.
bye |
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puppysatya
Joined: 22 Jul 2005
Posts: 2
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| Posted: Fri Jul 22, 2005 5:32 pm Post subject: |
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Hi tpsthegreat (i am afraid that i dont know your name),
that was a good solution. thanks for that |
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Salar
Joined: 24 Jan 2006
Posts: 6
Location: Chennai
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| Posted: Tue Jan 24, 2006 3:44 pm Post subject: algebra .... modulous |
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Boss,
There is a very simple way of doing it. Assume that F(x)= |x - 1 I + I x-2 I + .............. + I x- 99 I
Now remove the mod and differentiate the function. we would arrive at D(f(x)) as 50 and this will not change for any value of x(means its constant). So the function reaches minimum value of 50 for any value of x. Is it correct???
Salar |
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CZ@@@H@@@R
Serious about CAT
Joined: 14 Jan 2006
Posts: 64
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| Posted: Wed Jan 25, 2006 3:54 pm Post subject: Modulus |
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| tpsthegreat wrote: | hi ,
the func is minimum at x=50.
soln:
as we know that Ia+bI <= IaI + IbI
so minimum value of IaI + IbI is Ia+bI
so in ur question minimum value of func will be I99x-(1+2+3+......+99)I
= I99x-(99*100/2)I
= I99x-(99*50)I
the minimum value of this be 0 at x=50.
hence ans is 50.
bye |
Value of Mod >= 0 always. Since x not equal to 0. Hence Minimum value > 0
Let Sum = |x-1| + |x-2| + |x-3| + - - - - -- |x-99| 99 terms
Case 1: x<=1
Sum forms an A.P with first term |x-1| with common differece = +1 and Number of terms=99.
Hence Sum = 99/2 ( 2|x-1|+98 )
Minimum Sum (at |x-1|=0) = 4851
Case 2: 1<x<=99
Sum = x(x-1)/2 + (100-x)((100-x)-1)/2
Minimum Sum (at x=50) = 2450
Hence Function is minimum at x = 50 and Minimum Value = 2450
Please Correct it, If required. |
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CZ@@@H@@@R
Serious about CAT
Joined: 14 Jan 2006
Posts: 64
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| Posted: Wed Jan 25, 2006 4:16 pm Post subject: Re: algebra .... modulous |
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| Salar wrote: | Boss,
There is a very simple way of doing it. Assume that F(x)= |x - 1 I + I x-2 I + .............. + I x- 99 I
Now remove the mod and differentiate the function. we would arrive at D(f(x)) as 50 and this will not change for any value of x(means its constant). So the function reaches minimum value of 50 for any value of x. Is it correct???
Salar |
If Mod is removed F(x) = 99 x - 4950 (Eqn of Straight Line)
D(F(x)) = 99, > 0 Constant For all x
Minima or Maxima ??????
As far as F(x) is concerned
F(x) = |x-1|+|x-2|+ ----|x-99| = 99x-4950 For x>=99 |
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