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![[Post New]](/templates/default/images/icon_minipost_new.gif) 9 Aug 2007 21:26:40 IST
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solve ___x-1______ <1 x2-4x+3
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 9 Aug 2007 21:29:23 IST
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(x-1)/(x-3)(x-1)<1
therefore 1/(x-3)<1 hence x-3>1 x>4................!!!!!!!!!!!! oh yes........ x<4 but x {1,3}
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The inevitable truth of life.....everyone in our life is going 2 hurt sooner or later......u just have 2 realise who is worth.....
the PAIN or the PERSON...!!! |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 9 Aug 2007 22:40:09 IST
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Hey, nivedh you are wrong. (x-3) can be greater than zero or less than it, there are two possibilities. If x-1=0 , then the inequality holds true and x=1. Else if x-1 is not equal to zero, x2-4x+3=(x-1)(x-3) and the (x-1) in the numerator and denominator cancel each other. So, weget 1/(x-3) < 1. [ineq 1] Now, if x-3>=0 i.e. x>=3, from [ineq 1] we get x-3>1, i.e. x>4. combining both we get x>4. Again if x-3<0, i.e. x<3 , from [ineq 1] we get x-3<1 i.e. x<4. combining both we get x<4. So the answer is (x<4)U(x=1)U(x>4) Please do rate me.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 Aug 2007 08:39:53 IST
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x cannot b equal 2 1 nor can it b equal 2 3................!!!!!!!!!!!
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The inevitable truth of life.....everyone in our life is going 2 hurt sooner or later......u just have 2 realise who is worth.....
the PAIN or the PERSON...!!! |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 Aug 2007 16:47:09 IST
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1/(x-3)<1 1/(x-3) - 1 < 0 (4 - x) / (x-3)<0 (x - 4) / ( x-3) > 0 Now apply wavy curve method and get ans as (-infinity,3)U(4,infinty) rate if this helps u...
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 Aug 2007 17:00:15 IST
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the ans should be ..
x= R - (3,4) - {1}
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Keep working....................Iam comming..
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Success!! |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 Aug 2007 22:26:39 IST
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Nivedh, why can't x be 1? put x=1. the numerator becomes zero. So,the whole L.H.S becomes zero, which is obviously less than 1. Got it????????!!!!!!!.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 Aug 2007 22:49:35 IST
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(x-1) / (x-3)(x-1) 1/(x-3)<1 provided x not = 1 we have cancelled out (x-1) 1/(x-3) - 1 < 0 (4 - x) / (x-3)<0 (x - 4) / ( x-3) > 0
x E (-inf, 3) U (4, +inf) ~ {1}
*PLZ RATE*
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- Gaurav Ragtah (spideyunlimited)
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 Aug 2007 22:50:21 IST
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metal!!!!!! u cant put x = 1... u will get 0 in denominator also... not defined buddy
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- Gaurav Ragtah (spideyunlimited)
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 Aug 2007 23:46:08 IST
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ans: x (- ,3) (4, )-{1} soln: bringing 1 2 the LHS -X^2+5X-4/(X-1)(X-3)<0 MULIPLY BY - AND REVERSE SIGN OF INEQUALITY (X^2-5X+4)/(X-1)(X-3)>0 (X-1)(X-4)/(X-1)(X-3)>0 X is NOT = 1 as fn will b undefined (x-3)(x-4)>0 x belongs to (-inf , 3) U(4, inf)-{1} P.S. X is same as x(pls dont mind the capital fonts)
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 Aug 2007 20:58:38 IST
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Got it, spidey,x is not = 1. Then the correct answer is (-infinity,4)U(4,infinity) please rate.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 Aug 2007 21:17:30 IST
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see on factorising we get
(x-1)/(x-1)(x-3)<1
so we get (1/(x-3))-1<0
on solving: (4-x)/(x-3)<0
taking -1 common we get x-4/x-3>0 so finally we get by wavy curve method
x belongs to (-inf,3)U(4, inf)
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 Aug 2007 21:44:21 IST
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metal!!! it is (-inf, 3) U (4, +inf) ~ {1}
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- Gaurav Ragtah (spideyunlimited)
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