The minimum value of $\frac{9 \cdot 3^{2x} + 6 \cdot 3^x + 4}{9 \cdot 3^{2x} - 6 \cdot 3^x + 4}$ is

  • A
    $-1$
  • B
    $\frac{1}{2}$
  • C
    $\frac{1}{4}$
  • D
    $\frac{1}{3}$

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Let $x$ be a real number. Match the following:
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)$ $3$
$(V)$ $4$

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$I$. $4$ divides $x$ or $4$ divides $y$.
$II$. $3$ divides $x+y$ or $3$ divides $x-y$.
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If the minimum value of $f(x)=2x^2+\alpha x+8$ is the same as the maximum value of $g(x)=-3x^2-4x+\alpha^2$, then $\alpha^2=$

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