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aimhigh715
- Senior | Next Rank: 100 Posts
- Posts: 41
- Joined: Thu Nov 08, 2007 3:22 pm
First Question
is x+1/x-3 < 0
stmt 1: -1<x<1
so both numerator i.e x+1 & denominator i.e x -3 will be -ve for all values of x incl 0 so the expr will be -ve < 0 SUFF
stmt 2: x^2 -4 <0
or x^2 <4 or -2<x<2
if x is greater than -1 say 1 then numerator would be +ve but denominator -ve so expr will be -ve <0
INSUFF
A
