Let frac{x^2}{f(a^2 + 7a + 3)} + frac{y^2}{f( | vedclass

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(B) For 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, and

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$40$

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(B) For 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, and both must be positive: $f(3a+15) > f(a^2+7a+3) > 0$.Since $f$ is a strictly decreasing function, $f(x_1) > f(x_2) \implies x_1 Therefore, $3a+15 Rearranging the inequality gives $a^2 + 4a - 12 > 0$.Factoring the quadratic expression, we get $(a+6)(a-2) > 0$.This inequality holds when $a \in (-\infty, -6) \cup (2, \infty)$.The set of values for $a$ is $R - [-6, 2]$.Comparing this with $R - [\alpha, \beta]$, we get $\alpha = -6$ and $\beta = 2$.Thus, $\alpha^2 + \beta^2 = (-6)^2 + (2)^2 = 36 + 4 = 40$.

Let $\frac{x^2}{f(a^2 + 7a + 3)} + \frac{y^2}{f(3a + 15)} = 1$ represent an ellipse with major axis along the y-axis, where $f$ is a strictly decreasing positive function on $R$. If the set of all possible values of $a$ is $R - [\alpha, \beta]$, then $\alpha^2 + \beta^2$ is equal to:

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