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# For any operation ? that acts on two numbers x and y, the

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Manager
Joined: 29 Nov 2011
Posts: 76
For any operation ? that acts on two numbers x and y, the  [#permalink]

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03 Feb 2012, 20:12
6
00:00

Difficulty:

75% (hard)

Question Stats:

50% (01:09) correct 50% (01:16) wrong based on 112 sessions

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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
Math Expert
Joined: 02 Sep 2009
Posts: 52215

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04 Feb 2012, 03:31
2
2
Smita04 wrote:
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

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.
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Re: For any operation ? that acts on two numbers x and y, the  [#permalink]

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15 Jul 2018, 19:00
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Re: For any operation ? that acts on two numbers x and y, the &nbs [#permalink] 15 Jul 2018, 19:00
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