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Is x^3-6x^2+11x-6≤0?

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Joined: 05 Feb 2013
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18 Apr 2013, 04:05
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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!
Unfortunately the official answer is A!
Source: Nova's Gmat Data Sufficiency Prep Course
Did I miss something?
[Reveal] Spoiler: OA

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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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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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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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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Joined: 05 Feb 2013
Posts: 29
Location: Ukraine
GMAT 1: 680 Q48 V35
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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
Attachments

how to find roots.pdf [91.3 KiB]

Re: Is x^3-6x^2+11x-6≤0?   [#permalink] 22 Apr 2013, 23:35
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