Re: ISI 2012 ME - I Answer Key

Posted by Sinistral on Sep 24, 2013; 12:17pm
URL: http://discussion-forum.276.s1.nabble.com/ISI-2012-ME-I-Answer-Key-tp7515795p7584005.html

soumyab_sl wrote

so combining 1,2 and 3 we get f(x)  is real for x belonging to  [0,3]u[4,infinity)...............but the answer is only zero...............can u explain why???????????????????

So basically, u r trying to make the values under the root sign > 0 (all at the same time)
So, u need to take the intersection (NOT union) in the above case. but  it is evident that the values under the root sign cant be greater than zero at the same time (for a fixed value of x).

So, it is observed that x and x^2-4x become zero at x=0 and 3-x also doesn't become negative. So, {0} is a part of the domain. Now we need to check if anything else is also a part of the domain apart from 0.


After checking at various intervals,like for x<0 , x> 3, etc, which make the expression under the root sign negative; we find that not all 3 expression (the expressions under the root sign) can become < 0 at the same time. and for the remaining two (which are becoming negative for some value of x) it can never happen that the imaginary part of the two expressions get cancelled on addition for a fixed value of x. So basically we are only left with {0}.

Apologies for not coming out with a pure mathematical solution..