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(D) Given that $f$ is a strictly decreasing function on $\mathbb{R}$.For the equation $\frac{x^2}{f(a^2+5a+3)} + \frac{y^2}{f(a+15)} = 1$ to represent

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$(-\infty, -6) \cup (2, \infty)$

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(D) Given that $f$ is a strictly decreasing function on $\mathbb{R}$.For the equation $\frac{x^2}{f(a^2+5a+3)} + \frac{y^2}{f(a+15)} = 1$ to represent an ellipse with the major axis along the $y$-axis, the denominator of the $y^2$ term must be greater than the denominator of the $x^2$ term.Thus, $f(a+15) > f(a^2+5a+3)$.Since $f$ is a strictly decreasing function, $f(x_1) > f(x_2)$ implies $x_1 Therefore, $a+15 Rearranging the inequality, we get $a^2+4a-12 > 0$.Factoring the quadratic expression, we have $(a+6)(a-2) > 0$.Solving this inequality, we find that $a  2$.Thus, the interval for $a$ is $(-\infty, -6) \cup (2, \infty)$.

Let $f$ be a strictly decreasing function defined on $\mathbb{R}$ such that $f(x) > 0, \forall x \in \mathbb{R}$. Let $\frac{x^2}{f(a^2+5a+3)} + \frac{y^2}{f(a+15)} = 1$ be an ellipse with the major axis along the $y$-axis. The value of $a$ can lie in the interval$(s)$:

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