Question

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

• A. 28
• B. 40
• C. 61
• D. 24

Answer 1

The Correct Option is B

For ellipse with major axis along \(y\)-axis, \[ f(3a+15)>f(a^2+7a+3) \] Since \(f\) is strictly decreasing, \[ 3a+15<a^2+7a+3 \] \[ a^2+4a-12>0 \] \[ (a+6)(a-2)>0 \] \[ a\in(-\infty,-6)\cup(2,\infty) \] Thus, \[ \mathbb{R}\setminus[\alpha,\beta]=\mathbb{R}\setminus[-6,2] \] So, \[ \alpha=-6,\qquad \beta=2 \] \[ \alpha^2+\beta^2=36+4=40 \] \[ \boxed{40} \]