Anurag@Gurome wrote:gmatter2012 wrote:If A$B = A+B, if A > B and A$B = B-A, if A < B, then which of the followings is highest for (1/x$1/y)$(1/y$1/x)?
For positive numbers A and B and A > B, (A$B)$(B$A) = (A + B)$(A - B)
Now, (A + B) > (A - B)
Hence, (A + B)$(A - B) = (A + B) + (A - B) = 2A
And for positive numbers A and B and A < B, (A$B)$(B$A) = (B - A)$(B + A)
Now, (B - A) < (B + A)
Hence, (B - A)$(B + A) = (B + A) - (B - A) = 2A
Hence, the expression will attain highest value for highest value of A, i.e. highest value of 1/x, i.e. least value of x.
The correct answer is D.
Thank you Anurag, but some areas I couldn't understand, kindly bear with me.
For First case when A>B then
function is defined as A$B= A+B
but we do not have any definition of B$A but since A>B then I would assume B$A= A+B
How are we getting B$A= A-B when A>B
similarly when A<B then function is defined as A$B=B-A
but there is no definition for B$A but since A<B i would assume B$A = B-A
proceeding further for the first case:
(A+B)$(A+B) obviously I would get stuck here, as A+B = A+B and there is no definition for this condition.
same for the second case (B-A)$(B-A)..Dead end as both are equal
so my question is when A>B and A$B is defined as A+B how are we getting B$A= A-B
similarly when A<B and A$B= B-A then how are we deducing B$A = B+A
It seems that we are trying to keep the expression inside the brackets positive
example
For the first case when A>B then A$B = A+B( This of course is according to definition) and B$A= A-B( This will be positive as A>B)
when A<B
then A$B= B-A ( This again is according to definition ) but B$A is not defined but we are taking
B$A= B+A ( as this will be positive A<B)
but no where has it been mentioned that B$A is positive we know that A is positive and B is positive .
So please tell me how when A>B B$A= A -B
and when A<B then B$A = A+B
Am I missing something? or is my query justified, Kindly Bear with me.
