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A
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33% (01:53) correct
67%
(01:27)
wrong
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Hi Guys!
Can someone help me?
I have found interesting question and thing that official explanation contains a mistake:
Is x^3-6x^2+11x-6≤0?
(1) 2<x<3
(2) 2≤X<3
The roots of the inequality are: 1, 2 and 3.
See below:
1) First statement refers to positive sector from 1 to 2 exclusively, so the data tell us that inequality is positive – sufficient!
2) Second statement tells us that in region from 2 to 3 exclusively the inequality is negative, in point 2 it is equal to 0. Since we have “≤” both is acceptable – sufficient!
Hence, my answer is D.
Unfortunately the official answer is A!
Source: Nova's Gmat Data Sufficiency Prep Course
Did I miss something?
Can someone help me?
I have found interesting question and thing that official explanation contains a mistake:
Is x^3-6x^2+11x-6≤0?
(1) 2<x<3
(2) 2≤X<3
The roots of the inequality are: 1, 2 and 3.
See below:
1) First statement refers to positive sector from 1 to 2 exclusively, so the data tell us that inequality is positive – sufficient!
2) Second statement tells us that in region from 2 to 3 exclusively the inequality is negative, in point 2 it is equal to 0. Since we have “≤” both is acceptable – sufficient!
Hence, my answer is D.
Unfortunately the official answer is A!
Source: Nova's Gmat Data Sufficiency Prep Course
Did I miss something?
This Question is Locked Due to Poor Quality
Hi there,
The question you've reached has been archived due to not meeting our community quality standards. No more replies are possible here.
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18 Apr 2013, 04:15
Kudos
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\(x^3-6x^2+11x-6\leq{0}\) roots 1, 2 and 3
\((x-1)(x-2)(x-3)\leq{0}\)
the equation is \(\leq{0}\) in two intervals : \(x\leq{1}\) and \(2\leq{x}\leq{3}\)
(1) 2<x<3
(2) 2≤X<3
Both are sufficient. You're right. But we can check: pick \(2\) => \(8-6*4+11*2-6=0\) and \(0\) is \(\leq{0}\)
If the question were \(x^3-6x^2+11x-6<0\) no =
Than A would be the answer
Hope this helps, let me know
\((x-1)(x-2)(x-3)\leq{0}\)
the equation is \(\leq{0}\) in two intervals : \(x\leq{1}\) and \(2\leq{x}\leq{3}\)
(1) 2<x<3
(2) 2≤X<3
Both are sufficient. You're right. But we can check: pick \(2\) => \(8-6*4+11*2-6=0\) and \(0\) is \(\leq{0}\)
If the question were \(x^3-6x^2+11x-6<0\) no =
Than A would be the answer
Hope this helps, let me know
18 Apr 2013, 13:06
Kudos
Bookmarks
Sergiy
\(x^3-6x^2+11x-6 = x^3-x-6(x^2-2x+1) = x(x-1)(x+1)-6(x-1)^2 = (x-1)[x(x+1) - 6(x-1)] = (x-1)(x-2)(x-3)\)
From F.S 1 , we know that the above expression will always give a negative value for 2<x<3. Sufficient.
From F.S 2, we know that x=2 OR 2<x<3. In either case, the expression is 0 OR negative ; i.e <0. Sufficient.
D.
