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

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New post 24 Apr 2018, 23:10
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A
B
C
D
E

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  65% (hard)

Question Stats:

62% (02:30) correct 38% (02:43) wrong based on 112 sessions

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

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New post 25 Apr 2018, 00:07
1
1
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]

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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]

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New post 25 Apr 2018, 02:14
2
1
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]

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New post 25 Apr 2018, 15:58
2
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]

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New post 25 Apr 2018, 21:06
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]

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New post 20 Jun 2019, 11:08
1
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] 20 Jun 2019, 11:08
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