Assertion-Reason

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.

10 Answers

39
Dr.House ·

i think c........................

11
rkrish ·

@mathemati..
You mean to say Statement 2 is wrong?????[7][7][7][7]

39
Dr.House ·

wats the answer dude????????

106
Asish Mahapatra ·

b555 y is stmnt 2 wrong?

39
Dr.House ·

SRY THAT WAS I MEAN. REASON IS OBVIOUSLY RITE NA, ANY IDIOT WOULD SAY DAT. SRY TO MISLEAD U GUYS

A is wrong

106
Asish Mahapatra ·

well 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)

39
Dr.House ·

that wa s even wat i got. once againn sry to mislead u all. anyways my brain is not working today.............

11
rkrish ·

@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).

106
Asish Mahapatra ·

well 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)

62
Lokesh Verma ·

LHS 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.

:)

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