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The operation £ is defined for all real numbers x and y by the equatio

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The operation £ is defined for all real numbers x and y by the equatio  [#permalink]

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New post 06 Feb 2019, 02:25
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

81% (01:31) correct 19% (02:16) wrong based on 26 sessions

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The operation £ is defined for all real numbers x and y by the equatio  [#permalink]

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New post Updated on: 06 Feb 2019, 05:08
Bunuel wrote:
The operation ⊙ is defined for all real numbers x and y by the equation x⊙y = x^2 + y^2 - xy. Each of the following must be true EXCEPT

A. x⊙x = x^2
B. x⊙y = y⊙x
C. x⊙1 = x
D. x⊙0 = x^2
E. x⊙y = (-x)⊙(-y)


check the function x⊙y = x^2 + y^2 - xy
for all given options
only option C stands our invalid
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Originally posted by Archit3110 on 06 Feb 2019, 04:48.
Last edited by Archit3110 on 06 Feb 2019, 05:08, edited 1 time in total.
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Re: The operation £ is defined for all real numbers x and y by the equatio  [#permalink]

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New post 06 Feb 2019, 05:05
Bunuel wrote:
The operation ⊙ is defined for all real numbers x and y by the equation x⊙y = x^2 + y^2 - xy. Each of the following must be true EXCEPT

A. x⊙x = x^2
B. x⊙y = y⊙x
C. x⊙1 = x
D. x⊙0 = x^2
E. x⊙y = (-x)⊙(-y)


Rachit, it has got nothing to do with multiplication.
It is an operation and what it means is given, so check all values by substituting the values..
It is given that x⊙y = x^2 + y^2 - xy.
A. x⊙x = x^2..... x⊙x= x^2 + x^2 - x*x=2x^2-x^2=x^2...True
B. x⊙y = y⊙x..... In x⊙y = x^2 + y^2 - xy, x and y are interchangeable and still the equation remains the same. So True
C. x⊙1 = x...... x⊙1= x^2 + 1^2 - x*1=x^2+1-x. This is not equal to x. Hence false.
D. x⊙0 = x^2.... x⊙0 = x^2 + 0^2 - x*0=x^2...True
E. x⊙y = (-x)⊙(-y).... -x⊙+y = (-x)^2 + (-y)^2 - (-x)(-y)=x^2+y^2-xy.

C
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Re: The operation £ is defined for all real numbers x and y by the equatio  [#permalink]

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New post 06 Feb 2019, 05:08
chetan2u wrote:
Bunuel wrote:
The operation ⊙ is defined for all real numbers x and y by the equation x⊙y = x^2 + y^2 - xy. Each of the following must be true EXCEPT

A. x⊙x = x^2
B. x⊙y = y⊙x
C. x⊙1 = x
D. x⊙0 = x^2
E. x⊙y = (-x)⊙(-y)


Rachit, it has got nothing to do with multiplication.
It is an operation and what it means is given, so check all values by substituting the values..
It is given that x⊙y = x^2 + y^2 - xy.
A. x⊙x = x^2..... x⊙x= x^2 + x^2 - x*x=2x^2-x^2=x^2...True
B. x⊙y = y⊙x..... In x⊙y = x^2 + y^2 - xy, x and y are interchangeable and still the equation remains the same. So True
C. x⊙1 = x...... x⊙1= x^2 + 1^2 - x*1=x^2+1-x. This is not equal to x. Hence false.
D. x⊙0 = x^2.... x⊙0 = x^2 + 0^2 - x*0=x^2...True
E. x⊙y = (-x)⊙(-y).... -x⊙+y = (-x)^2 + (-y)^2 - (-x)(-y)=x^2+y^2-xy.

C


understood thanks..
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Re: The operation £ is defined for all real numbers x and y by the equatio   [#permalink] 06 Feb 2019, 05:08
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