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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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24 Apr 2018, 22:10
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Question Stats:

63% (02:25) correct 37% (02:43) wrong based on 161 sessions

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

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24 Apr 2018, 23:07
2
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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Joined: 11 Jan 2018
Posts: 19
Re: If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd),  [#permalink]

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25 Apr 2018, 00:51
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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25 Apr 2018, 01:14
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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25 Apr 2018, 14:58
2
IMO D

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

Thanks.
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Joined: 11 Jan 2018
Posts: 19
Re: If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd),  [#permalink]

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25 Apr 2018, 20:06
1
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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Joined: 07 May 2018
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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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20 Jun 2019, 10:08
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]

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13 Nov 2019, 04:39
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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Re: If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd),  [#permalink]

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14 Nov 2019, 22:45
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
Re: If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd),   [#permalink] 14 Nov 2019, 22:45
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