Too lengthy question:

case 1: a= 2 , b = 3

for (i) solution 2<x<3 when b > a

for (ii) a=b=2 solution x>2 or as in case 1 x> 3

similarly substitute the values for other options, and reject it as the conditions will not be satisfied.

so answer: A,B

First rewrite
x^2-(a+b)x+ab as (x-a)(x-b)

Option A: TRUE
(x-a)(x-b)<0
when a<x<b and b>a
will be true since in this case (x-a) would be positive and (x-b) would be negative

Option B: TRUE
(x-a)(x-b)>0
when x>b and b>=a
will be true since in this case (x-a) and (x-b) are both positive therefore the product is positive

Option C: False
(x-a)(x-b)>0
when a<x<b and b>=a
will be false because (x-a) will be positive and (x-b) will be negative therefore the product is negative

Option D:
(x-a)(x-b)<0
x<a and x>b and b>=a
is false because if x is less than a then (x-a) and (x-b) will be both negative and therefore the product is positive

Option E: False
(x-a)(x-b)>0
x<a<b where b<a
a<b and b>a is a contradiction therefore false


Final Answer: A,B


The real challenge is to do this in 1:45 tops.

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.