Q.

 For x>0,y>0 and xy1 , the solution of the system of equation xx+y=yx and yx+y=x2xyn is, where n>0

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a

y=1+4n1+8n2

b

x=1+1+8n2

c

x=1+1+8n2

d

y=1+4n+1+8n2

answer is A, C.

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Detailed Solution

xx+y=yn------(1)

yx+y=x2π·yn----(2)

 Multiplying (1)&(2)

(xy)x+γ=(xy)2π

x+y=2n(xy1)

substitue the above in (1) we get x2n=yn

x2yn=1 x2=yn>0 hence the quadratic equation x2+x-2n-0

x=1+1+8n2

 y=2nx =1+4n1+8n2

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