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EAMCET Engineering 2003 Solved Paper

Section: Mathematics

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Question : 121 of 160

Marks: +1, -0

If

u(x, y)=y\log x+x\log y

u_x u_y-u_x\log x-u_y\log y+\log x\log y

0
-1
1
2

Solution:

41. We have,

u(x, y)=y\log x+x\log y

On differentiating partially w.r.t.

x

and

y

respectively,

u_x=\;\frac{y}{x}+\log y, u_y=\log x+\;\frac{x}{y}

Now,

u_x u_y-u_x\log x-u_y\log y+\log x\log y

\;\; = \left(\;\frac{y}{x}+\log y\right) \left(\log x+\;\frac{x}{y}\right) - \left(\;\frac{y}{x}+\log y\right) \log x

\;\; - \left(\log x+\;\frac{x}{y}\right) \log y+\log x\log y

\;\; =1+\;\frac{y}{x}\log x+\;\frac{x}{y}\log y+\log x\log y

\;\; -\;\frac{y}{x}\log x-\log x\log y-\log x\log y

\;\; -\;\frac{x}{y}\log y+\log x\log y

\;\; = 1

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