Here since no value are give let us take

both x and y be positive integer such that x= 3 and y =2

From statement 1: \(x^2 \div (y+\frac{1}{y})\) = \(\frac{18}{5}\) = 3.6 (putting the value of x and y in the equation)

From Statement 2 : \(y^2 \div (x+\frac{1}{x})\) = \(\frac{12}{10}\)= 1.2 (putting the value of x and y in the equation)

From above we get statement 1 > Statement 2

Now if we consider negative integer such that x= -2 and y = -3 (since x>y)

From statement 1: \(x^2 \div (y+\frac{1}{y})\) = \(\frac{- 12}{10}\)= -1.2 (putting the value of x and y in the equation)

From Statement 2 : \(y^2 \div (x+\frac{1}{x})\) = \(\frac{- 18}{5}\) = -3.6(putting the value of x and y in the equation)

From above we have Statement 1 > Statement 2

But we have to look into every possibilities

Now if we consider negative integer such that x= 2 and y = -2 (since x>y)

From statement 1: \(x^2 \div (y+\frac{1}{y})\) = \(\frac{- 4}{5}\)= -0.8 (putting the value of x and y in the equation)

From Statement 2 : \(y^2 \div (x+\frac{1}{x})\) = \(\frac{4}{5}\) = 0.8(putting the value of x and y in the equation)

From above we have Statement 2 > Statement 1.

Hence the option is D
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