+1 vote

​​​​​​$\bigoplus$ and $\bigodot$ are two operators on numbers $\text{p}$ and $\text{q}$ such that $p \bigoplus q=\dfrac{p^{2}+q^{2}}{pq}$ and $p \bigodot q=\dfrac{p^{2}}{q}$;

If $x\bigoplus y=2\bigodot 2$, then $x=$

1. $\frac{y}{2}$
2. $y$
3. $\frac{3y}{2}$
4. $2y$

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+1 vote
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 $(B).$
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