1. Key Concepts and Formulas

• Domain of $\frac{1}{\sqrt{g(x)}}$: For this expression to be defined, the term inside the square root must be strictly positive, i.e., $g(x) > 0$.
• Domain of a Sum of Functions: If $f(x) = f_1(x) + f_2(x)$, the domain of $f(x)$ is the intersection of the domain of $f_1(x)$ and the domain of $f_2(x)$.
• Absolute Value Function: $|x|$ is defined as $x$ if $x \geq 0$ and $-x$ if $x < 0$.
• Solving Quadratic Inequalities: For a quadratic $ax^2+bx+c$, if $a>0$, the inequality $ax^2+bx+c < 0$ holds between its roots, and $ax^2+bx+c > 0$ holds outside its roots.

2. Step-by-Step Solution

The given function is $f(x)=\frac{1}{\sqrt{10+3 x-x^2}}+\frac{1}{\sqrt{x+|x|}}$. To find the domain of $f(x)$, we need to find the domain of each term and then take their intersection.

Step 1: Find the domain of the first term, $f_1(x) = \frac{1}{\sqrt{10+3x-x^2}}$

For $f_1(x)$ to be defined, the expression inside the square root in the denominator must be strictly positive: $10+3x-x^2 > 0$ To solve this quadratic inequality, we first rewrite it with a positive leading coefficient: $-(x^2 - 3x - 10) > 0$ Multiplying by $-1$ reverses the inequality sign: $x^2 - 3x - 10 < 0$ Next, we find the roots of the quadratic equation $x^2 - 3x - 10 = 0$. Factoring the quadratic, we look for two numbers that multiply to $-10$ and add to $-3$. These numbers are $-5$ and $2$. $(x-5)(x+2) < 0$ Since the parabola $y = x^2 - 3x - 10$ opens upwards, the inequality $(x-5)(x+2) < 0$ is satisfied when $x$ is between the roots $-2$ and $5$. Thus, the domain for the first term is $D_1 = (-2, 5)$.

Step 2: Find the domain of the second term, $f_2(x) = \frac{1}{\sqrt{x+|x|}}$

For $f_2(x)$ to be defined, the expression inside the square root in the denominator must be strictly positive: $x+|x| > 0$ We analyze this inequality by considering two cases based on the definition of the absolute value function:

• Case 1: $x \geq 0$ If $x \geq 0$, then $|x|=x$. The inequality becomes: $x+x > 0$ $2x > 0$ Dividing by $2$, we get $x > 0$. Since this case assumes $x \geq 0$, the combined condition is $x > 0$.

• Case 2: $x < 0$ If $x < 0$, then $|x|=-x$. The inequality becomes: $x+(-x) > 0$ $0 > 0$ This statement is false. Therefore, there are no values of $x < 0$ for which $x+|x| > 0$.

Combining both cases, the condition $x+|x| > 0$ is satisfied only when $x > 0$. Thus, the domain for the second term is $D_2 = (0, \infty)$.

Step 3: Find the overall domain of $f(x)$

The domain of $f(x)$ is the intersection of $D_1$ and $D_2$: Domain of $f(x) = D_1 \cap D_2 = (-2, 5) \cap (0, \infty)$. To find this intersection, we consider the numbers that are in both intervals. The numbers must be greater than $-2$ and less than $5$, AND greater than $0$. The common interval is where both conditions are met. This intersection is $(0, 5)$. The problem states that the domain of $f(x)$ is $(a, b)$. By comparing $(0, 5)$ with $(a, b)$, we find $a=0$ and $b=5$.

Step 4: Calculate the required expression $(1+a)^2+b^2$

Substitute the values $a=0$ and $b=5$ into the expression: $(1+a)^2+b^2 = (1+0)^2 + 5^2$ $= (1)^2 + 25$ $= 1 + 25$ $= 26$

3. Common Mistakes & Tips

• Strict Inequality for Denominators: Always remember that for expressions of the form $\frac{1}{\sqrt{g(x)}}$, the condition is $g(x) > 0$, not $g(x) \geq 0$.
• Absolute Value Cases: When dealing with $|x|$ in inequalities, carefully consider all possible cases for $x$ (e.g., $x \geq 0$ and $x < 0$) to ensure all valid solutions are found.
• Intersection of Intervals: Visualizing intervals on a number line can be very helpful to correctly determine their intersection.

4. Summary

The domain of the function $f(x)=\frac{1}{\sqrt{10+3 x-x^2}}+\frac{1}{\sqrt{x+|x|}}$ is determined by ensuring that the expressions under the square roots in the denominators are strictly positive. The first term requires $10+3x-x^2 > 0$, which leads to the domain $(-2, 5)$. The second term requires $x+|x| > 0$, which, by analyzing cases for $|x|$, leads to the domain $(0, \infty)$. The overall domain of $f(x)$ is the intersection of these two domains, which is $(0, 5)$. Therefore, $a=0$ and $b=5$. The expression $(1+a)^2+b^2$ evaluates to $(1+0)^2+5^2 = 1+25 = 26$.

The final answer is $\boxed{26}$.