If the domain of the function \[f(x) = \frac{\sqrt{x^2 - 25}}{(4 - x^2)} + \log_{10}(x^2 + 2x - 15)\]is $(-\infty, \alpha) \cup [\beta, \infty)$, then $\alpha^2 + \beta^3$ is equal to:
- 140
- 175
- 150
- 125
The Correct Option is C
Solution and Explanation
To find the domain of the function, we consider the restrictions imposed by each term separately.
Step 1: Analyzing \( \frac{\sqrt{x^2 - 25}}{4 - x^2} \)
The term \( \sqrt{x^2 - 25} \) requires:
\[ x^2 - 25 \geq 0 \implies x^2 \geq 25 \implies x \leq -5 \text{ or } x \geq 5 \]
The term \( \frac{1}{4 - x^2} \) requires:
\[ 4 - x^2 \neq 0 \implies x^2 \neq 4 \implies x \neq \pm 2 \]
Combining these conditions:
\[ x \leq -5 \text{ or } x \geq 5 \]
Step 2: Analyzing \( \log_{10}(x^2 + 2x - 15) \)
For the logarithmic term to be defined:
\[ x^2 + 2x - 15 \(>\) 0 \]
Factoring the quadratic:
\[ (x + 5)(x - 3) \(>\) 0 \]
Using the sign chart for this inequality:
\[ x \in (-\infty, -5) \cup (3, \infty) \]
Step 3: Combining the Conditions
The overall domain of \( f(x) \) is given by the intersection of the two sets of conditions:
\[ x \in (-\infty, -5) \cup [5, \infty) \]
Thus, \( \alpha = -5 \) and \( \beta = 5 \).
Calculating \( \alpha^2 + \beta^3 \)
\[ \alpha^2 + \beta^3 = (-5)^2 + 5^3 = 25 + 125 = 150 \]
Conclusion: \( \alpha^2 + \beta^3 = 150 \).
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