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 ascertain the domain of the function \(f(x) = \frac{\sqrt{x^2 - 25}}{4 - x^2} + \log_{10}(x^2 + 2x - 15)\), each component must be evaluated.

1. The square root term \(\sqrt{x^2 - 25}\) requires \(x^2 - 25 \geq 0\).

This inequality simplifies to \(x^2 \geq 25\).

The solution is \(x \le -5\) or \(x \ge 5\).

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

\(x^2 eq 4\)

This excludes \(x = -2\) and \(x = 2\).

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

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

Factoring the quadratic \(x^2 + 2x - 15\) yields:

\((x + 5)(x - 3)\)

This inequality holds when:

• \(x>3\) or \(x<-5\).

Combining these restrictions determines the function's domain.

• The square root condition mandates \(x \le -5\) or \(x \ge 5\).
• The denominator restriction excludes \(x = -2\) and \(x = 2\).
• The logarithmic term requires \(x>3\) or \(x<-5\).

The intersection of these conditions yields the domain of \(f(x)\):

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

From this domain, we identify \(\alpha = -5\) and \(\beta = 5\).

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

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

The value of \(\alpha^2 + \beta^3\) is 150.