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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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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IMO D

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


Thanks.

Oh right! I see it now!!! thanks !!!!

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

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

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)

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

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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Thanks I appreciate your efforts and I get it

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