39i think c........................
1) x , y , z are +ve nos. such that x + y + z = 1
(A) : [ (1-2x)+(1-2y)+(1-2z) ]/3 ≥ [ (1-2x)(1-2y)(1-2z) ]1/3
(R) : For +ve nos. A.M. ≥ G.M.
A,B,C,D have the same usual meanings.
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10 Answers
39SRY THAT WAS I MEAN. REASON IS OBVIOUSLY RITE NA, ANY IDIOT WOULD SAY DAT. SRY TO MISLEAD U GUYS
A is wrong
106well at least one of (1-2x) or (1-2y) or (1-2z) has to be negative because, if all are positive then x,y,z≥1/2 ==> x+y+z≥3/2 which contradicts the given statement. hence the AM GM inequality cant be used here. Hence answer is (d)
39that wa s even wat i got. once againn sry to mislead u all. anyways my brain is not working today.............
11@ashish
You are right that AM-GM inequality cannot be used here BUT
are you sure that Statement 1 is false.
Well then,lets modify this ques. a little
Prove/Disprove:
For +ve x,y,z if x+y+z=1 ,then
[ (1-2x)+(1-2y)+(1-2z) ]/3 ≥ [ (1-2x)(1-2y)(1-2z) ]1/3
If you are able to prove it ,then the ans is (B), else (D).
106well it depends on how many of them i.e. (1-2x),(1-2y),(1-2z) are negative. if one term is negative then statement A is invalid because you cant be definite, if two of them are negative then it is correct.
.......... wait my brain is dazed.......just reached home after a full test........will think later and tell ...........(ignore wat ive written above)
62LHS will be 1/3 always.
When the 3 terms are +ve then RHS will be +ve and we can apply AM GM so there are no issues.
WHEN 1 of these terms is -ve then RHS will be -ve. Hence the INequality will satisfy. So we will have no issues at all..
So A is True and B is true.
:)