The problem asks to find the domain of the function $f(x) = \sqrt{1 - \frac{x+13}{x^2+4x+3}}$ and to discuss the continuity of $f(x)$ at $a=2$.

1. Problem Description

The problem asks to find the domain of the function $f(x) = \sqrt{1 - \frac{x+13}{x^2+4x+3}}$ and to discuss the continuity of $f(x)$ at $a=2$.

2. Solution Steps

a) Find the domain of $f(x)$.

For $f(x)$ to be defined, the expression inside the square root must be non-negative:

$1 - \frac{x+13}{x^2+4x+3} \geq 0$

First, factor the denominator: $x^2 + 4x + 3 = (x+1)(x+3)$.

So, $1 - \frac{x+13}{(x+1)(x+3)} \geq 0$.

Combine the terms on the left side into a single fraction:

$\frac{(x+1)(x+3) - (x+13)}{(x+1)(x+3)} \geq 0$

Simplify the numerator:

$\frac{x^2+4x+3 - x - 13}{(x+1)(x+3)} \geq 0$

$\frac{x^2+3x-10}{(x+1)(x+3)} \geq 0$

Factor the numerator:

$\frac{(x+5)(x-2)}{(x+1)(x+3)} \geq 0$

Now, we need to find the intervals where this inequality holds. The critical points are $x=-5, x=-3, x=-1, x=2$. We test the intervals determined by these critical points:

- $x < -5$: Choose $x=-6$. $\frac{(-1)(-8)}{(-5)(-3)} = \frac{8}{15} > 0$.

- $-5 < x < -3$: Choose $x=-4$. $\frac{(1)(-6)}{(-3)(-1)} = \frac{-6}{3} = -2 < 0$.

- $-3 < x < -1$: Choose $x=-2$. $\frac{(3)(-4)}{(-1)(1)} = \frac{-12}{-1} = 12 > 0$.

- $-1 < x < 2$: Choose $x=0$. $\frac{(5)(-2)}{(1)(3)} = \frac{-10}{3} < 0$.

- $x > 2$: Choose $x=3$. $\frac{(8)(1)}{(4)(6)} = \frac{8}{24} = \frac{1}{3} > 0$.

Therefore, the inequality holds when $x \in (-\infty, -5] \cup (-3, -1) \cup [2, \infty)$.

b) Discuss the continuity of $f(x)$ at $a=2$.

The function $f(x)$ is defined as $f(x) = \sqrt{\frac{(x+5)(x-2)}{(x+1)(x+3)}}$.

We found that the domain includes $x=2$.

Since $x=2$ is in the domain, we can evaluate $f(2) = \sqrt{\frac{(2+5)(2-2)}{(2+1)(2+3)}} = \sqrt{\frac{7 \cdot 0}{3 \cdot 5}} = \sqrt{0} = 0$.

We need to check if the limit exists at $x=2$. Since the function is only defined for $x \geq 2$ in a neighborhood around 2, we only need to check the right-hand limit.

$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \sqrt{\frac{(x+5)(x-2)}{(x+1)(x+3)}}$

Since the expression inside the square root is continuous at $x=2$, we have

$\lim_{x \to 2^+} \sqrt{\frac{(x+5)(x-2)}{(x+1)(x+3)}} = \sqrt{\frac{(2+5)(2-2)}{(2+1)(2+3)}} = \sqrt{\frac{7(0)}{3(5)}} = \sqrt{0} = 0$.

Since $\lim_{x \to 2^+} f(x) = f(2) = 0$, the function is right-continuous at $x=2$.

3. Final Answer

a) The domain of $f(x)$ is $(-\infty, -5] \cup (-3, -1) \cup [2, \infty)$.

b) $f(x)$ is right-continuous at $x=2$.