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03 Feb 2012, 21:12
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
53% (01:56) correct
47%
(02:05)
wrong
based on 192
sessions
History
Date
Time
Result
Not Attempted Yet
For any operation ? that acts on two numbers x and y, the commutator is defined as x?y – y?x. For which of the following operations is the commutator not equal to zero for some values of x and y?
I. x?y = xy
II. x?y = (x – y)²
III. x?y = x^3 + 3x²y + 3xy² + y^3
(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III
I. x?y = xy
II. x?y = (x – y)²
III. x?y = x^3 + 3x²y + 3xy² + y^3
(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III
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04 Feb 2012, 04:31
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Smita04
Let's check each option:
I. \(x?y=xy\) --> \(y?x=yx\) --> \(x?y-y?x=xy-yx=0\). Hence this option is ALWAYS equal to zero. Discard.
II. \(x?y=(x-y)^2\) --> \(y?x=(x-y)^2=(y-x)^2=(x-y)^2\)--> \(x?y-y?x=(x-y)^2-(x-y)^2=0\). Hence this option is ALWAYS equal to zero. Discard.
III. \(x?y=x^3+3x^2*y+3xy^2+y^3=(x+y)^3\) --> \(y?x=(y+x)^3\) --> \(x?y-y?x=(x+y)^3-(x+y)^3=0\). Hence this option is ALWAYS equal to zero. Discard.
Answer: ALL of the options equal to zero for ANY value of x and y. No correct answer among the answer choices.
If III were: \(x?y=x^3-3x^2*y+3xy^2-y^3=(x-y)^3\), then \(y?x=(y-x)^3\) --> \(x?y-y?x=(x-y)^3-(y-x)^3=(x-y)^3+(x+y)^3=2(x-y)^3\) --> this option equal to zero only if \(x=y\).
So, in this case the answer would be C (III only).
Hope it's clear.
18 Aug 2025, 23:22
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