Question

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begin{aligned} x ln x &=frac{x^{3}}{e^{2}} ln x cdot ln x &=ln left(frac{x^{3}}{e^{2}}right)=ln left(x^{3}right)-ln left(e^{2}right) (ln x)^{2}=& 3 ln (x)-2 text { tet } ln x quad text { be } & y therefore y^{2}-3 y+2 &=0 y=1,2 & ln (x)=1,8 therefore x &=e, e^{2} end{aligned}
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( begin{aligned} x_{1}=& e^{2}, quad x_{2}=e x_{1}=&left(x_{2}right)^{2} end{aligned} )

# pairs (x, y) Given x2 - xy + y2 = 4 (x + y - 4), where x, satisfying the equation is (A) only one + y - 4), where x, y both are real numbers. The number of pairs (y (D) None of these (B) only two (C) three (A) xy = 2x2 are the roots of the equation e2.x x = x3 with X1 > X2 then (D) x (B) x1 = x3 (C) 2x1 = x3 = x

Solution