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Is x<0 ? [#permalink]
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10 Dec 2011, 17:52
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Is x<0 ? (1) x^3(1x^2)<0 (2) x^21<0
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Re: OG 11 [#permalink]
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11 Dec 2011, 16:32
IMO it is C. (1) By itself is not sufficient. (2) tells us x is a +ve or ve fraction. Not sufficient. (Together) If x is +ve fraction it does not satisfy st. (1) (plug in x = 1/2 for testing) If x is ve fraction it satifies (1). so x is a ve fraction and hence x < 0. SUFFICIENT.
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Re: OG 11 [#permalink]
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12 Dec 2011, 13:26
agree C
stmt 1: x<0 or x>1 stmt 2: x>1 x<1
Together: x>0 and x>1



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Re: OG 11 [#permalink]
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15 Dec 2011, 03:29
How does 1) X^3(1X^2)<0 led to x<0 or x>1 eg: X=2 2^3 (1(2)^2) = 8*3 = 24 eg: X=1 , result 0<0 , not true. eg:X=0 result 0<0, not true X=1 , 0<0, not true. i feel it is only when X>1 , X^3(1X^2)<0
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Re: OG 11 [#permalink]
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15 Dec 2011, 04:25
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Lucky2783 wrote: How does 1) X^3(1X^2)<0 led to x<0 or x>1 eg: X=2 2^3 (1(2)^2) = 8*3 = 24 eg: X=1 , result 0<0 , not true. eg:X=0 result 0<0, not true X=1 , 0<0, not true.
i feel it is only when X>1 , X^3(1X^2)<0 Actually just rewrite with factors: x^3(1x^2) < 0 So: (x^3)(1x)(1+x) < 0 so either all three or negative or any one can be negative for the expression to be true. If (x^3) is negative, x < 0 If (1x) is negative, x > 1 If (1+x) is negative, x < 1 When you combine statement 1 and 2, the only overlap is x < 1 so we know for sure that x is negative Hope this answers your question.
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Re: OG 11 [#permalink]
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15 Dec 2011, 04:52
i think we need to consider the three cases in parallel eg: if X<0 what will happen to other two expressions If (x^3) is negative, x < 0 say X is 0.5, X^3 is ive , 1X is +ive and 1+X too is positive. expression holds. Now say X=2, X^3 is ive , 1X is +ive but 1+X is ive . expression doesnt hold . so we cannot say that if X<0 , X^3 (1X) (1+X) < 0 1+X will be ive only when X <1 , in that case 1X will be +ive and X^3 will be ive , over all expression will be +ive so this set of values of X also ruled out. now 1X will be ive for all values of X greater than 1 and the other two X^3 and 1+X will be positive in this case. we can safely say that for all X>1 the expression will hold. may be i am wrong though in my thought process , pls advise.
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Re: OG 11 [#permalink]
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15 Dec 2011, 17:45
Is X<0 ?
1) X^3(1X^2)<0 2) X^21<0
C
From 1)
1 < x < 0 and 1 < x
From 2)
1 < x < 1
Combining,
1 < x < 0



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Re: OG 11 [#permalink]
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15 Dec 2011, 19:53
Study1 you sound promising...
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Re: OG 11 [#permalink]
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06 Jul 2012, 23:43
ashiima wrote: Is X<0 ?
1) X^3(1X^2)<0 2) X^21<0 Question asks if X is negative Statement1 If X is ve: x^3 will be ve, then (1x^2) has to be positive, can't say If X is +ve: x^3 will be +ve then (1x^2) has to be negative, can't say Insufficient Statement2 X^21<0, this is possible only when x is fraction. But fraction can be negative or positive, Insufficient Both statements together we get when x is ve, x^3 is negative, and 1x^2 is positive ( as x is a fraction now eg. 1/2) when x is +ve, not valid Hence x is ve, when both statements combined together. Hence C Hope I am clear.
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Re: Is x<0 ? [#permalink]
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07 Jul 2012, 03:01
Is x<0 ? (1) x^3(1x^2)<0 > \(x^3(1x)(1+x)<0\) > the roots are 1, 0, and 1 (equate the expressions to zero to get the roots and list them in ascending order), this gives us 4 ranges: \(x<1\), \(1<x<0\), \(0<x<1\), and \(x>1\). Now, test some extreme value: for example if \(x\) is very large number then \(x\) and \(1+x\) are positive, while \(1x\) is negative, which gives negative product for the whole expression, so when \(x>1\) the expression is negative. Now the trick: as in the 4th range expression is negative then in 3rd it'll be positive, in 2nd it'll be negative again and finally in 1st it'll be positive: +  + . So, the ranges when the expression is negative are: \(1<x<0\) and \(x>1\). So, \(x\) could be negative as well as positive. Not sufficient. Or: \(x^3(1x^2)<0\) > \(x(1x^2)<0\) > \(x<x^3\) > \(1<x<0\) or \(x>1\). Not sufficient. (2) x^21<0 > \(x^2<1\) > \(1<x<1\). Not sufficient. (1)+(2) Intersection of the ranges from (1) and (2) is \(1<x<0\), hence the answer to the question whether \(x<0\) is YES. Sufficient. Answer: C. Solving inequalities: x24x94661.html#p731476inequalitiestrick91482.htmleverythingislessthanzero108884.html?hilit=extreme#p868863xyplane71492.html?hilit=solving%20quadratic#p841486Hope it helps.
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Re: Is x<0 ? [#permalink]
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30 Jan 2014, 06:57
Answer is C.
(1): x³(1x²) < 0 > x³ < 0 or 1x² < 0. If x³ < 0 then x < 0, BUT x² < 1 x could be both. IS.
(2): Clearly IS. x² < 1, x could be < or > 0.
Together: From 1 we know EITHER x³ <0 OR 1x² < 0. From 2 we know that 1x² < 0 hence x³ is > 0, thus x is > 0.



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