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Let $A = \{(x, y) : x > 0, y > 0, x^{2} + y^{2} = 1\}$ and let $B = \{(x, y) : x > 0, y > 0, x^{6} + y^{6} = 1\}$ Then $A \cap B$

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{(0,1),(1,0)}

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Correct option is C

x6+y6=x2+y2x4−x2y2+y4, so if (x,y)A∩B then 1=x4−x2y2+y4=x2+y22−3x2y2 =1−3x2y2⇒x2y2=0 x=0 or y=0 but x>0,y>0 so A∩B=ϕ

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Let A= (x, y) : x>0, y>0, x2+y2=1 and let B=(x,y):x>0,y>0,x6+y6=1 Then A∩B

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Let $A = \{(x, y) : x > 0, y > 0, x^{2} + y^{2} = 1\}$ and let $B = \{(x, y) : x > 0, y > 0, x^{6} + y^{6} = 1\}$ Then $A \cap B$