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Is x^3-6x^2+11x-60?

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Sergiy
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Is x^3-6x^2+11x-60? [#permalink] New post   Updated on: 18 Apr 2013, 04:19
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

33% (01:53) correct 67% (01:27) wrong based on 3 sessions
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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?

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Official Answer
A

Originally posted by Sergiy on 18 Apr 2013, 04:05.
Last edited by Bunuel on 18 Apr 2013, 04:19, edited 2 times in total.
Edited the question and moved to DS forum.
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Re: Is x^3-6x^2+11x-60? [#permalink] New post   18 Apr 2013, 04:15
\(x^3-6x^2+11x-6\leq{0}\) 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
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Re: Is x^3-6x^2+11x-60? [#permalink] New post   18 Apr 2013, 13:06
Sergiy wrote:
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


\(x^3-6x^2+11x-6 = x^3-x-6(x^2-2x+1) = x(x-1)(x+1)-6(x-1)^2 = (x-1)[x(x+1) - 6(x-1)] = (x-1)(x-2)(x-3)\)

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.
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yezz
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Re: Is x^3-6x^2+11x-60? [#permalink] New post   22 Apr 2013, 16:12
Zarrolou wrote:
\(x^3-6x^2+11x-6\leq{0}\) 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


how did u get the roots if u didnt use factorisation ?
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Sergiy
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Re: Is x^3-6x^2+11x-60? [#permalink] New post   22 Apr 2013, 23:35
yezz wrote:
Zarrolou wrote:
\(x^3-6x^2+11x-6\leq{0}\) 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


how did u get the roots if u didnt use factorisation ?


Just pick up one root, in our case it can be 1, then polynomial x^3-6x^2+11x-6<0 divide on x-1 (we have 1 as a root).
Then use the following approach:
see file attached

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Re: Is x^3-6x^2+11x-60? [#permalink]
22 Apr 2013, 23:35
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Bunuel
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