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Question

If u=log(x3+y3+z33xyz) then (x+y+z)(ux+uy+uz)=

A
0
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B
xy+z
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C
2
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D
3
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Solution

Given:

u=log(x3+y3+z33xyz)

Differentiate with respect to 'x'

dudx=ddx(log(x3+y3+z33xyz)

[Since, ddx(logx)=1x,x>0,ddx(xn)=nxn1 and ddx(constant)=0]

dudx=3x2+0+03yzx3+y3+z33xyz

dudx=3x23yzx3+y3+z33xyz

i.e., ux=dudx=3x23yzx3+y3+z33xyz..........(1)

Similarly,

uy=dudy=3y23xzx3+y3+z33xyz..........(2)

uz=dudz=3z23xyx3+y3+z33xyz..........(3)

Add all the three equations

dudx+dudy+dudz


=3x23yzx3+y3+z33xyz+3y23xzx3+y3+z33xyz+3z23xyx3+y3+z33xyz


=3(x2+y2+z2xyyzzx)x3+y3+z33xyz


=3(x2+y2+z2xyyzzx)(x+y+z)(x2+y2+z2xyyzzx)

[Since, a3+b3+c33abc=(a+b+c)(a2+b2+c2abbcca)]

=3x+y+z

ux+uy+uz=3x+y+z

Now, (x+y+z)(ux+uy+uz)

=(x+y+z)(3x+y+z)

=3

(x+y+z)(ux+uy+uz)=3

Hence, Option A is correct.


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