The number of ordered pairs (x, y) of real nu | vedclass

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(D) Given the system of equations:$x+y^2=12 \quad \dots(i)$$x^2+y=12 \quad \dots(ii)$Subtracting $(i)$ from $(ii)$ gives:$(x^2-x) + (y-y^2) = 0$$(x^2-

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$4$

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(D) Given the system of equations:$x+y^2=12 \quad \dots(i)$$x^2+y=12 \quad \dots(ii)$Subtracting $(i)$ from $(ii)$ gives:$(x^2-x) + (y-y^2) = 0$$(x^2-y^2) - (x-y) = 0$$(x-y)(x+y) - (x-y) = 0$$(x-y)(x+y-1) = 0$This implies $x=y$ or $x+y=1$.Case $1$: If $x=y$,then $x^2+x=12$ $\Rightarrow x^2+x-12=0$ $\Rightarrow (x+4)(x-3)=0$. Thus,$x=3, y=3$ and $x=-4, y=-4$. This gives $2$ solutions.Case $2$: If $x+y=1$,then $y=1-x$. Substituting into $(ii)$: $x^2+(1-x)=12 \Rightarrow x^2-x-11=0$. The discriminant $D = (-1)^2 - 4(1)(-11) = 1+44 = 45 > 0$. This quadratic equation has $2$ distinct real roots for $x$,each giving a corresponding real value for $y$. This gives $2$ more solutions.Total number of ordered pairs $(x, y)$ is $2+2=4$.

List-$I$	List-$II$
$(A)$ The minimum value of $2x^2 + 4x + 5$	$(I)$ $-1$
$(B)$ The maximum value of $\frac{x^2 + 4x + 1}{x^2 + x + 1}$	$(II)$ $1$
$(C)$ If $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$,$\forall x \in [a, b]$ then $b =$	$(III)$ $2$
$(D)$ If $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$,$\forall x \in [a, b]$ then $a =$	$(IV)$ $3$
	$(V)$ $4$

The number of ordered pairs $(x, y)$ of real numbers that satisfy the simultaneous equations $x+y^2=x^2+y=12$ is

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