1 vote

Best answer

Given that, $p \bigoplus q=\dfrac{p^{2}+q^{2}}{pq},p \bigodot q=\dfrac{p^{2}}{q}$

Performing these operations we get,

$x \bigoplus y = 2 \bigodot 2$

$\implies \dfrac{x^{2} + y^{2}}{xy} = \dfrac{2^{2}}{2}$

$\implies \dfrac{x^{2} + y^{2}}{xy} = 2$

$\implies x^{2} + y^{2} = 2xy$

$\implies x^{2} + y^{2} – 2xy = 0$

$\implies (x-y)^{2}= 0$

$\implies x = y$

So, the correct answer is $(B).$

Performing these operations we get,

$x \bigoplus y = 2 \bigodot 2$

$\implies \dfrac{x^{2} + y^{2}}{xy} = \dfrac{2^{2}}{2}$

$\implies \dfrac{x^{2} + y^{2}}{xy} = 2$

$\implies x^{2} + y^{2} = 2xy$

$\implies x^{2} + y^{2} – 2xy = 0$

$\implies (x-y)^{2}= 0$

$\implies x = y$

So, the correct answer is $(B).$