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Bunuel
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Is x^3 – 6x2 + 11x – 6 > 0? (1) 2 < x < 3 (2) 1 ≤ x < 2 12 Jul 2019, 02:25
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65% (hard)

Question Stats:

47% (01:47) correct 53% (02:27) wrong based on 57 sessions

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Is \(x^3 – 6x^2 + 11x – 6 > 0\)?

(1) \(2 < x < 3\)
(2) \(1 ≤ x < 2\)
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Kinshook
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Is x^3 – 6x2 + 11x – 6 > 0? (1) 2 < x < 3 (2) 1 ≤ x < 2 12 Jul 2019, 02:33
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Bunuel
Is \(x^3 – 6x2 + 11x – 6 > 0\)?

(1) \(2 < x < 3\)
(2) \(1 ≤ x < 2\)

Q: Is (x-1)(x-2)(x-3)>0?

#1: \(2 < x < 3\)
For 2<x<3 => (x-1)(x-2)(x-3)<0 => (x-1)(x-2)(x-3) is NOT >0
SUFFICIENT

#2:\(1 ≤ x < 2\)
For 1<x<3 => (x-1)(x-2)(x-3)>0
But for x=1 => (x-1)(x-2)(x-3)=0
NOT SUFFICIENT

IMO A
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Is x^3 – 6x2 + 11x – 6 > 0? (1) 2 < x < 3 (2) 1 ≤ x < 2 08 Sep 2019, 01:27
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Kinshook
Bunuel
Is \(x^3 – 6x2 + 11x – 6 > 0\)?

(1) \(2 < x < 3\)
(2) \(1 ≤ x < 2\)

Q: Is (x-1)(x-2)(x-3)>0?

#1: \(2 < x < 3\)
For 2<x<3 => (x-1)(x-2)(x-3)<0 => (x-1)(x-2)(x-3) is NOT >0
SUFFICIENT

#2:\(1 ≤ x < 2\)
For 1<x<3 => (x-1)(x-2)(x-3)>0
But for x=1 => (x-1)(x-2)(x-3)=0
NOT SUFFICIENT

IMO A

HiKinshook,
Thanks for the soln, However can you please explain how you broke \(x^3 – 6x^2 + 11x – 6\)> 0 into (x-1)(x-2)(x-3) thanks
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Is x^3 – 6x2 + 11x – 6 > 0? (1) 2 < x < 3 (2) 1 ≤ x < 2 08 Sep 2019, 03:40
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Kinshook
Bunuel
Is \(x^3 – 6x2 + 11x – 6 > 0\)?

(1) \(2 < x < 3\)
(2) \(1 ≤ x < 2\)

Q: Is (x-1)(x-2)(x-3)>0?

#1: \(2 < x < 3\)
For 2<x<3 => (x-1)(x-2)(x-3)<0 => (x-1)(x-2)(x-3) is NOT >0
SUFFICIENT

#2:\(1 ≤ x < 2\)
For 1<x<3 => (x-1)(x-2)(x-3)>0
But for x=1 => (x-1)(x-2)(x-3)=0
NOT SUFFICIENT

IMO A

HiKinshook,
Thanks for the soln, However can you please explain how you broke \(x^3 – 6x^2 + 11x – 6\)> 0 into (x-1)(x-2)(x-3) thanks


stne
If you put x = 1 in the equation
1-6+11-6 = 0
=> x = 1 is the root of the equation and (x-1) is a factor

\(x^3 – 6x2 + 11x – 6 > 0\)
\(x^2 (x-1) -5x(x-1) +6(x-1) > 0\)
\((x-1)(x^2-5x+6)>0\)
\((x-1)(x-2)(x-3)>0\)

This is how i broke the equation.
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Is x^3 – 6x2 + 11x – 6 > 0? (1) 2 < x < 3 (2) 1 ≤ x < 2 08 Sep 2019, 03:59
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Kinshook
stne


Thanks for the soln, However can you please explain how you broke \(x^3 – 6x^2 + 11x – 6\)> 0 into (x-1)(x-2)(x-3) thanks


stne
If you put x = 1 in the equation
1-6+11-6 = 0
=> x = 1 is the root of the equation and (x-1) is a factor

\(x^3 – 6x2 + 11x – 6 > 0\)
\(x^2 (x-1) -5x(x-1) +6(x-1) > 0\)
\((x-1)(x^2-5x+6)>0\)
\((x-1)(x-2)(x-3)>0\)

This is how i broke the equation.

Hi Kinshook,
Still not clear how you got \(x^2 (x-1) -5x(x-1) +6(x-1)\) after knowing that (x-1) is a factor.
Once I come to know that (x-1) is a factor of the eqn.\(x^3 – 6x^2 + 11x – 6 > 0\)
Then how do I get \(x^2 (x-1) -5x(x-1) +6(x-1)\) ?
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Kinshook
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Is x^3 – 6x2 + 11x – 6 > 0? (1) 2 < x < 3 (2) 1 ≤ x < 2 08 Sep 2019, 04:23
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stne
Kinshook
stne


Thanks for the soln, However can you please explain how you broke \(x^3 – 6x^2 + 11x – 6\)> 0 into (x-1)(x-2)(x-3) thanks


stne
If you put x = 1 in the equation
1-6+11-6 = 0
=> x = 1 is the root of the equation and (x-1) is a factor

\(x^3 – 6x2 + 11x – 6 > 0\)
\(x^2 (x-1) -5x(x-1) +6(x-1) > 0\)
\((x-1)(x^2-5x+6)>0\)
\((x-1)(x-2)(x-3)>0\)

This is how i broke the equation.

Hi Kinshook,
Still not clear how you got \(x^2 (x-1) -5x(x-1) +6(x-1)\) after knowing that (x-1) is a factor.
Once I come to know that (x-1) is a factor of the eqn.\(x^3 – 6x^2 + 11x – 6 > 0\)
Then how do I get \(x^2 (x-1) -5x(x-1) +6(x-1)\) ?

Please put \(x^3 – 6x2 + 11x – 6 > 0\) on top as a reference
To get x^3 term we need to multiply (x-1) by x^2
x^2(x-1) = x^3 - x^2
To get -6x^2 = -x^2 - 5x^2
Since -x^2 is already available, we need to get -5x^2.
We need to multiply (x-1) by -5x
-5x(x-1) = - 5x^2 + 5x
Now to get 11x = 5x + 6x
Since 5x is already available, we need to get 6x
We need to multiply (x-1) by 6
6(x-1) = 6x - 6

Final expression
\(x^3 – 6x2 + 11x – 6 > 0\)
\((x^2-5x+6)(x-1) = x^3 – 6x2 + 11x – 6 > 0\)

Another way is to check that x = 1,2,3 are roots is to put values of x=1, x=2, & x=3 and to check whether the expression becomes 0.
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stne
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Is x^3 – 6x2 + 11x – 6 > 0? (1) 2 < x < 3 (2) 1 ≤ x < 2 08 Sep 2019, 04:37
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Kinshook
stne


Hi Kinshook,
Still not clear how you got \(x^2 (x-1) -5x(x-1) +6(x-1)\) after knowing that (x-1) is a factor.
Once I come to know that (x-1) is a factor of the eqn.\(x^3 – 6x^2 + 11x – 6 > 0\)
Then how do I get \(x^2 (x-1) -5x(x-1) +6(x-1)\) ?

Please put \(x^3 – 6x2 + 11x – 6 > 0\) on top as a reference
To get x^3 term we need to multiply (x-1) by x^2
x^2(x-1) = x^3 - x^2
To get -6x^2 = -x^2 - 5x^2
Since -x^2 is already available, we need to get -5x^2.
We need to multiply (x-1) by -5x
-5x(x-1) = - 5x^2 + 5x
Now to get 11x = 5x + 6x
Since 5x is already available, we need to get 6x
We need to multiply (x-1) by 6
6(x-1) = 6x - 6

Final expression
\(x^3 – 6x2 + 11x – 6 > 0\)
\((x^2-5x+6)(x-1) = x^3 – 6x2 + 11x – 6 > 0\)

Another way is to check that x = 1,2,3 are roots is to put values of x=1, x=2, & x=3 and to check whether the expression becomes 0.

Thanks Kinshook
Seems as though question required a lot of effort
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