If the domain of the function f(x)=x^2-25(4-x^2)+ _10(x^2+2 x-15) is (-, ) [, ), then ^2+^…

Question

If the domain of the function

$f(x)=\frac{\sqrt{x^2-25}}{\left(4-x^2\right)}+\log _{10}\left(x^2+2 x-15\right)$ is $(-\infty, \alpha) \cup[\beta, \infty)$, then $\alpha^2+\beta^3$ is equal to :

Accepted Answer

To find the domain of the function
$$f(x) = \frac{\sqrt{x^2-25}}{(4-x^2)}+\log_{10}(x^2+2x-15),$$
we need to consider the domain conditions for both the square root function and the logarithmic function.

The square root function $\sqrt{x^2-25}$ requires that the argument of the square root be non-negative, so
$$x^2 - 25 \geq 0.$$
This inequality is satisfied when
$$x \leq -5 \quad \text{or} \quad x \geq 5.$$

The denominator of the rational part of $f(x)$, $(4-x^2)$, cannot be zero, otherwise, the function will become undefined due to division by zero. Thus, we must have
$$4 - x^2 \neq 0.$$
This inequality is violated when
$$x = \pm2.$$

Combining these conditions gives us the domain for the rational part of the function:
$$x \in (-\infty, -5] \cup (5, \infty) \quad \text{and} \quad x \neq 2,-2.$$

Moving on to the logarithmic function, $\log_{10}(x^2+2x-15)$, the argument must be positive:
$x^2 + 2x - 15 > 0.$
This is a quadratic inequality, which we can factor to find the solution:
$(x+5)(x-3) > 0.$
From this, we see that the inequality is satisfied for
$$x < -5 \quad \text{or} \quad x > 3.$$

The overall domain of $f(x)$ is the intersection of the domains for each piece. Taking the intersection of the two sets gives us:
$$x \in (-\infty, -5) \cup (5, \infty),$$

Since the question states that the domain is of the form $(-\infty, \alpha) \cup [\beta, \infty)$, we can infer that
$$\alpha = -5 \quad \text{and} \quad \beta = 5.$$

We calculate $\alpha^2 + \beta^3$ as follows:
$$\alpha^2 + \beta^3 = (-5)^2 + 5^3 = 25 + 125 = 150.$$

So the correct answer, representing the sum of $\alpha^2$ and $\beta^3$, is:
Option D $150$.

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