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Bunuel
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Is x^12 – 2x^11 negative? (1) x^2 < |x| (2) x^(-1) < -1 09 Dec 2019, 00:56
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51% (02:33) correct 49% (02:36) wrong based on 140 sessions

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Is \(x^{12}- 2x^{11}\) negative?


(1) \(x^2 < |x|\)

(2) \(x^{-1} < -1\)



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Is x^12 – 2x^11 negative? (1) x^2 < |x| (2) x^(-1) < -1 Updated on: 09 Aug 2020, 05:23
Originally posted by Archit3110 on 09 Dec 2019, 03:17.
Last edited by Archit3110 on 09 Aug 2020, 05:23, edited 1 time in total.
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given
\(x^{12}- 2x^{11}\) negative
#1
\(x^2 < |x|\)
x will be a fraction only possibility

but for x = + fraction we get yes and x = -ve we get no
insufficient
#2
\(x^{-1} < -1\)
is -ve ; possiblity ; x=-1/2,-1/3; we get NO
IMO B sufficient

Bunuel
Is \(x^{12}- 2x^{11}\) negative?


(1) \(x^2 < |x|\)

(2) \(x^{-1} < -1\)



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Is x^12 – 2x^11 negative? (1) x^2 < |x| (2) x^(-1) < -1 09 Dec 2019, 09:47
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We can factor out \(x^{11}\) to get
\(x^{11}*(x-2)\)

the only possibility for this expression to be negative is
when 0<x<2

1 stm --> gives us a range -1<x<1. Not sufficient
2 stm -- \(\frac{1}{x}<-1\) means x is negative. Sufficient

IMO
Ans: B
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Is x^12 – 2x^11 negative? (1) x^2 < |x| (2) x^(-1) < -1 09 Dec 2019, 10:23
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LevanKhukhunashvili
We can factor out \(x^{11}\) to get
\(x^{11}*(x-2)\)

the only possibility for this expression to be negative is
when 0<x<2

1 stm --> gives us a range -1<x<1. Not sufficient
2 stm -- \(\frac{1}{x}<-1\) means x is negative. Sufficient

IMO
Ans: B
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Actually, the first piece of information provided tells us that the range of x is between 0 and 1. When fractions are squared, they always get closer to 0, and for negative fractions, they will change signs and get closer to zero. So x must be positive for that to hold true. I could be wrong, but I believe that the answer is actually D.
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Is x^12 – 2x^11 negative? (1) x^2 < |x| (2) x^(-1) < -1 09 Dec 2019, 10:36
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edlc313
LevanKhukhunashvili
We can factor out \(x^{11}\) to get
\(x^{11}*(x-2)\)

the only possibility for this expression to be negative is
when 0<x<2

1 stm --> gives us a range -1<x<1. Not sufficient
2 stm -- \(\frac{1}{x}<-1\) means x is negative. Sufficient

IMO
Ans: B

Actually, the first piece of information provided tells us that the range of x is between 0 and 1. When fractions are squared, they always get closer to 0, and for negative fractions, they will change signs and get closer to zero. So x must be positive for that to hold true. I could be wrong, but I believe that the answer is actually D.
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IMO your reasoning about the first statement is not correct. if x=-1/2 that statement holds true, so the range is not 0<x<1
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Is x^12 – 2x^11 negative? (1) x^2 < |x| (2) x^(-1) < -1 13 Dec 2019, 09:20
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Quote:
Is \(x^{12}- 2x^{11}\) negative?


(1) \(x^2 < |x|\)

(2) \(x^{-1} < -1\)
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\(x^{12}- 2x^{11}<0…x^{11}(x- 2)<0\)
\(x<0:x-2>0…x>2=false…(x<0)\)
\(x>0:x-2<0…x<2…0<x<2\)


We need to find if \(0<x<2\)

(1) \(x^2 < |x|\) insufic

\(x^2 < |x|:-1<x<0…or…0<x<1\)

(2) \(x^{-1} < -1\) sufic

\(x^{-1} < -1…1/x<-1…1/x+1<0…(1+x)/x<0\)
\(x>0:1+x<0…x<-1=false…(x>0)\)
\(x<0:1+x>0…x>-1…-1<x<0\)

Ans (B)
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Is x^12 – 2x^11 negative? (1) x^2 < |x| (2) x^(-1) < -1 16 Dec 2019, 02:19
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[quote="Bunuel"]Is \(x^{12}- 2x^{11}\) negative?


(1) \(x^2 < |x|\)

(2) \(x^{-1} < -1\)


\(x^{12}- 2x^{11} < 0 \) <=> \(x^{11}*(x - 2) < 0 \) => check whether 0 < x < 2 or not

(1) -> -1 < x < 1. Insufficient as:
If -1 < x < 0: \(x^{11}*(x - 2) > 0 \)
If 0 < x < 1: \(x^{11}*(x - 2) < 0 \)

(2) -> -1 < x < 0 => \(x^{11}*(x - 2) > 0 \) => Sufficient

==> B
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Is x^12 – 2x^11 negative? (1) x^2 < |x| (2) x^(-1) < -1 04 Aug 2020, 09:23
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edlc313
LevanKhukhunashvili
We can factor out \(x^{11}\) to get
\(x^{11}*(x-2)\)

the only possibility for this expression to be negative is
when 0<x<2

1 stm --> gives us a range -1<x<1. Not sufficient
2 stm -- \(\frac{1}{x}<-1\) means x is negative. Sufficient

IMO
Ans: B

Actually, the first piece of information provided tells us that the range of x is between 0 and 1. When fractions are squared, they always get closer to 0, and for negative fractions, they will change signs and get closer to zero. So x must be positive for that to hold true. I could be wrong, but I believe that the answer is actually D.
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The range for which the first statement holds true is -1<x<1. Since we are looking for a 0<x<2 range, statement 1 isn't sufficient.
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