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Question

If x,y,z are real and distinct, then x2+4y2+9z2āˆ’6yzāˆ’3zxāˆ’2xy=


A

Non-negative

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B

Non-positive

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C

Zero

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D

positive only

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Solution

The correct option is A

Non-negative


Given: x,y,zR and distinct.

Now, u=x2+4y2+9z26yz3zx2xy

=12(2x2+8y2+18z212yz6zx4xy)

=12{(x24xy+4y2)+(x26zx+9z2)+(4y212yz+9z2)}

=12{(x2y)2+(x3z)2+(2y3z)2}

Since it is sum of squares. u is always non-negative.


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