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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63% (02:27) correct 
