Question

Let \( \frac{x^2}{f(a^2 + 2a + 7)} + \frac{y^2}{f(3a + 14)} = 1 \) represent an ellipse. The major axis of the given ellipse is the y-axis and \( f \) is a decreasing function. If the range of \( a \) is \( R - [\alpha, \beta] \), then \( \alpha + \beta \) is:

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For the equation $\dfrac{x^2}{A} + \dfrac{y^2}{B} = 1$ to be an ellipse with the major axis along the $y$-axis, we need $B>A>0$, i.e., the denominator under $y^2$ must be strictly greater than the denominator under $x^2$.
This means we need $f(3a+14)>f(a^2+2a+7)$.
Step 2: Using the Decreasing Nature of $f$:
Since $f$ is a decreasing function: \[ f(3a+14)>f(a^2+2a+7) \iff 3a+14<a^2+2a+7 \] (The inequality reverses because $f$ is decreasing.)
Step 3: Solving the Inequality:
\[ a^2 + 2a + 7 - 3a - 14>0 \] \[ a^2 - a - 7>0 \] Using the quadratic formula, the roots of $a^2 - a - 7 = 0$ are: \[ a = \frac{1 \pm \sqrt{1 + 28}}{2} = \frac{1 \pm \sqrt{29}}{2} \] Since the coefficient of $a^2$ is positive, $a^2 - a - 7>0$ outside the roots: \[ a \in \left(-\infty,\, \frac{1-\sqrt{29}}{2}\right) \cup \left(\frac{1+\sqrt{29}}{2},\, +\infty\right) \] In other words, $a \in \mathbb{R} - \left[\dfrac{1-\sqrt{29}}{2},\, \dfrac{1+\sqrt{29}}{2}\right]$.
Step 4: Computing $\alpha + \beta$:
\[ \alpha = \frac{1-\sqrt{29}}{2}, \quad \beta = \frac{1+\sqrt{29}}{2} \] \[ \alpha + \beta = \frac{1-\sqrt{29}}{2} + \frac{1+\sqrt{29}}{2} = \frac{2}{2} = 1 \] Step 5: Final Answer:
\[ \alpha + \beta = 1 \] The answer is Option (1).