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![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Nov 2006 00:11:04 IST
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modulus of sinx=modulus of x - k for wht range of k this will have no real solution?
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19 Nov 2006 22:29:03 IST
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Dear Sonali, We onticed that this query has been there for a long time and hasn't been answered yet. We willt ry to get it covered as soon as possible. thanks for your patience. ~forum administrator
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God is real, unless declared as Integer!!
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19 Nov 2006 23:47:21 IST
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|sinx| = |x-k|
since for non real values of the output
1<|sinx| < infinity
as the output of mod function is always positive or 0
so, 1<|x-k| case 1: if x>k
solving mod:
1 x+1
hope it is correct. pls.feel free to say if it is wrong
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19 Nov 2006 23:50:07 IST
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sinx| = |x-k|
since for non real values of the output
1<|sinx| < infinity
as the output of mod function is always positive or 0
so, 1<|x-k|< infinity
case 1: if x>k
solving mod:
1 or x-1
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BE THE CHANGE WHICH YOU WANT TO SEE IN THE WORLD......-M.K.GANDHI(THE ONLY HUMAN IN THIS WORLD) |
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20 Nov 2006 21:27:11 IST
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|sinx| = |x-k|
so this will have non real solutions only when
|sinx| > 1
implies |x-k| > 1
case 1
x - k >1
 k < ( x - 1 )
Case 2
k - x > 1
 k > ( x +1 )
so for ( x - 1 ) > k > ( x +1 ) the abolve equation will have non real solutions
cheres
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Rahul A
-- If you thing you can, you can. If you think you can't, you're right !! -- |
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21 Nov 2006 12:50:01 IST
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If | sin x | = | x - k| , then as 0 £ | sin x | £ 1, we have 0 £ | x - k| £ 1. To have no real solution If k £ x x - k ³ 1, then K £ x - 1; If k ³ x K - x ³ 1, then k ³ x + 1;
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22 Nov 2006 23:26:49 IST
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the modulus is on x and not on k
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24 Nov 2006 21:12:31 IST
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|sinx| = |x|-k Right???
For non-real solution,
|sinx| belongs to (1,infinity)
so 1< |x|-k < infinity
case1:
x is positive
1 1-x<-k x-1>k so, k
case 2:
x is negative
so,
1<-x-k 1+x<-k -1-x>k so, k<-(x+1)
the final answer is obtained by combining cases 1 & 2
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