Question

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:

• A. 140
• B. 175
• C. 150
• D. 125

Answer 1

The Correct Option is C

To determine the domain of the given function \(f(x) = \frac{\sqrt{x^2 - 25}}{4 - x^2} + \log_{10}(x^2 + 2x - 15)\), we need to analyze each component of the function separately.

1. The square root function \(\sqrt{x^2 - 25}\) is defined if \(x^2 - 25 \geq 0\). This implies:

\(x^2 \geq 25\)

So, \(x \le -5\) or \(x \ge 5\).

1. The denominator \(4 - x^2\) should not be zero. Therefore:

\(x^2 \neq 4\)

Thus, \(x \neq -2\) and \(x \neq 2\).

1. The logarithmic function \(\log_{10}(x^2 + 2x - 15)\) is defined when:

\(x^2 + 2x - 15 > 0\)

To find the values of \(x\) that satisfy this inequality, we factor the quadratic:

\(x^2 + 2x - 15 = (x + 5)(x - 3)\)

This implies:

• \(x > 3\) or \(x < -5\)

Next, we combine all these restrictions to find the domain of the function.

• For \(\sqrt{x^2 - 25}\) we need \(x \le -5\) or \(x \ge 5\).
• The values \(x = -2\) and \(x = 2\) must be excluded because they make the denominator zero.
• For the term \(\log_{10}(x^2 + 2x - 15)\), we need \(x > 3\) or \(x < -5\).

Considering all these conditions together, the domain of \(f(x)\) is the intersection of the sets obtained from these conditions, resulting in:

\((-\infty, -5) \cup [5, \infty)\)

Thus, in the given problem statement, \(\alpha = -5\) and \(\beta = 5\).

Finally, we calculate \(\alpha^2 + \beta^3\):

• \(\alpha^2 = (-5)^2 = 25\)
• \(\beta^3 = 5^3 = 125\)
• The sum is: \(25 + 125 = 150\)

Thus, the value of \(\alpha^2 + \beta^3\) is 150.

Answer 2

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: \( \alpha^2 + \beta^3 = 150 \).