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If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd),

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Bunuel
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If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd), [#permalink] New post   24 Apr 2018, 23:10
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If \(0 < a < b < c < d\) and \(y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd)\), in which of the following intervals is y < 0?

I. \(a < x < b\)
II. \(b < x < c\)
III. \(c < x < d\)

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) II and III only
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D
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If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd), [#permalink] New post   25 Apr 2018, 00:07
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Ans: D
given that a,b,c,d are >0 and y = (x-a)(x-b)(x-c)(x-d) we need to find out is Y<0 or y is negative
now by given conditions:
I. y<0
II. y >0
III. y<0

I and III satisfy y<0
Ans : D
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Re: If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd), [#permalink] New post   25 Apr 2018, 02:14
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minatminat wrote:
B?

Im probably wrong but I will still give it a go...

y=(x^2−ax−bx+ab)(x^2−cx−dx+cd) y<0

(x^2−ax−bx+ab) = (x-a)(x-b)
(x^2−cx−dx+cd) = (x-c)(x-d)

For y <0
(x-a)(x-b) is +ve , then (x-c)(x-d) is - ve
So, x>a or x>b and x<c or x<d
So, the condition that matches y<0 is b<x<c


I think the problem is in the steps you are folowing..
for example
Statement 1 says x lies between a and b; a<x<b and we already know that a<b<c<d this mean b,c and d are greater than x. now by putting them in resolved given value of y = (x-a)(x-b)(x-c)(x-d) we get only one +ve and three -ve signs so y is negative.
Statement 2 says: b<x<c means c and d are both greater than x. so now we get two -ve and two +ve sings so y is +ve.
Statement 3 says: c<x<d means only d is bigger than x so one -ve and three +ve signs. which gives us y -ve.

:)
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Re: If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd), [#permalink] New post   13 Nov 2019, 05:39
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Bunuel wrote:
If \(0 < a < b < c < d\) and \(y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd)\), in which of the following intervals is y < 0?

I. \(a < x < b\)
II. \(b < x < c\)
III. \(c < x < d\)

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) II and III only


\(y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd)\)
\(y = (x-a)(x-b)(x-c)(x-d)\)

I. \(a < x < b:y = (x-a)(x-b)(x-c)(x-d)=[+][-][-][-]=[-][+]=[-]<0\)
II. \(b < x < c:y = (x-a)(x-b)(x-c)(x-d)=[+][+][-][-]=[+][+]=[+]>0\)
III. \(c < x < d:y = (x-a)(x-b)(x-c)(x-d)=[+][+][+][-]=[+][-]=[-]<0\)

Ans (D)
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If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd), [#permalink] New post   29 Mar 2021, 01:04
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aka14 wrote:
Can anyone show how you factor the y equation

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\(y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd)=\)

\(=(x(x - a) - b(x - a))(x(x - c) - d(x - c))=\)

\(=(x - a)(x - b)(x - c)(x - d)\).
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Re: If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd), [#permalink] New post   25 Apr 2018, 01:51
B?

Im probably wrong but I will still give it a go...

y=(x^2−ax−bx+ab)(x^2−cx−dx+cd) y<0

(x^2−ax−bx+ab) = (x-a)(x-b)
(x^2−cx−dx+cd) = (x-c)(x-d)

For y <0
(x-a)(x-b) is +ve , then (x-c)(x-d) is - ve
So, x>a or x>b and x<c or x<d
So, the condition that matches y<0 is b<x<c
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Re: If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd), [#permalink] New post   25 Apr 2018, 15:58
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IMO D

I used wavy line method by egmat to reach to conclusion.


Thanks.
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Re: If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd), [#permalink] New post   25 Apr 2018, 21:06
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dkumar2012 wrote:
minatminat wrote:
B?

Im probably wrong but I will still give it a go...

y=(x^2−ax−bx+ab)(x^2−cx−dx+cd) y<0

(x^2−ax−bx+ab) = (x-a)(x-b)
(x^2−cx−dx+cd) = (x-c)(x-d)

For y <0
(x-a)(x-b) is +ve , then (x-c)(x-d) is - ve
So, x>a or x>b and x<c or x<d
So, the condition that matches y<0 is b<x<c


I think the problem is in the steps you are folowing..
for example
Statement 1 says x lies between a and b; a<x<b and we already know that a<b<c<d this mean b,c and d are greater than x. now by putting them in resolved given value of y = (x-a)(x-b)(x-c)(x-d) we get only one +ve and three -ve signs so y is negative.
Statement 2 says: b<x<c means c and d are both greater than x. so now we get two -ve and two +ve sings so y is +ve.
Statement 3 says: c<x<d means only d is bigger than x so one -ve and three +ve signs. which gives us y -ve.

:)


Oh right! I see it now!!! thanks !!!!
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Re: If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd), [#permalink] New post   20 Jun 2019, 11:08
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D3N0 wrote:
Ans: D
given that a,b,c,d are >0 and y = (x-a)(x-b)(x-c)(x-d) we need to find out is Y<0 or y is negative
now by given conditions:
I. y<0
II. y >0
III. y<0

I and III satisfy y<0
Ans : D



Hi Could you explain how you got

1. y <0 for the first statement
2. y > 0 for the second statement
3. y < 0 for the third statement
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Re: If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd), [#permalink] New post   14 Nov 2019, 23:45
use wavy form approach so the interval you get is a<x<b and c<x<d since y<0.if y>0 it is b<x<c.
so the ans is Opt D 1 and 3
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Re: If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd), [#permalink] New post   29 Mar 2021, 00:57
Can anyone show how you factor the y equation

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Re: If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd), [#permalink] New post   29 Mar 2021, 01:06
Thanks I appreciate your efforts and I get it

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Re: If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd), [#permalink] New post   03 Aug 2022, 11:46
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Re: If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd), [#permalink]
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