Re: ISI 2004 ME - I Answer Key

Posted by Bankelal on Mar 26, 2014; 6:52pm
URL: http://discussion-forum.276.s1.nabble.com/ISI-2004-ME-I-Answer-Key-tp7515784p7585571.html

Hi n1,

Say x1 and x2 are the roots of eqn with X1>1 and X2<1.
Then (a^2)*((X2)^2) + 2b(X2) + c = 0

As we know that X2<1,
so (a^2)*((X2)^2)> 0 => 2b(X2) + c<0

or c/(-2b) < X2

and X2<1(given)

thus c/(-2b)< 1
hence (c)