Key Concepts and Formulas

• Intermediate Value Theorem (IVT): If $f(x)$ is continuous on $[a, b]$ and $k$ is between $f(a)$ and $f(b)$, then there exists $c \in (a, b)$ such that $f(c) = k$. A common application is if $f(a)$ and $f(b)$ have opposite signs, there's a root in $(a, b)$.
• First Derivative Test: If $f'(x) > 0$ on an interval, $f(x)$ is strictly increasing. If $f'(x) < 0$, $f(x)$ is strictly decreasing.
• Odd Degree Polynomials: A polynomial of odd degree with real coefficients has at least one real root.

Step-by-Step Solution

1. Define the Function and Find its Derivative

Let $f(x) = 2x^5 + 5x^4 + 10x^3 + 10x^2 + 10x + 10$. We want to analyze the roots of this polynomial. First, we find the derivative to determine where the function is increasing or decreasing. $f'(x) = \frac{d}{dx}(2x^5 + 5x^4 + 10x^3 + 10x^2 + 10x + 10)$ $f'(x) = 10x^4 + 20x^3 + 30x^2 + 20x + 10$ We can factor out a 10: $f'(x) = 10(x^4 + 2x^3 + 3x^2 + 2x + 1)$

2. Analyze the Sign of the Derivative

We want to determine if $f'(x)$ is always positive, always negative, or changes sign. Let's focus on the expression inside the parenthesis: $x^4 + 2x^3 + 3x^2 + 2x + 1$. We can rewrite it to see if we can recognize a perfect square: $x^4 + 2x^3 + 3x^2 + 2x + 1 = x^2\left(x^2 + 2x + 3 + \frac{2}{x} + \frac{1}{x^2}\right)$ $= x^2\left(\left(x^2 + \frac{1}{x^2}\right) + 2\left(x + \frac{1}{x}\right) + 3\right)$ Let $y = x + \frac{1}{x}$. Then $y^2 = x^2 + 2 + \frac{1}{x^2}$, so $x^2 + \frac{1}{x^2} = y^2 - 2$. Substituting this into the expression, we get $= x^2\left((y^2 - 2) + 2y + 3\right)$ $= x^2(y^2 + 2y + 1)$ $= x^2(y + 1)^2$ Substituting back $y = x + \frac{1}{x}$, we have $= x^2\left(x + \frac{1}{x} + 1\right)^2$ $= x^2\left(\frac{x^2 + x + 1}{x}\right)^2$ $= (x^2 + x + 1)^2$ Therefore, $f'(x) = 10(x^2 + x + 1)^2$.

Now, let's analyze $x^2 + x + 1$. The discriminant is $b^2 - 4ac = 1^2 - 4(1)(1) = -3 < 0$. Since the leading coefficient is positive, $x^2 + x + 1 > 0$ for all real $x$. Thus, $(x^2 + x + 1)^2 > 0$ for all real $x$. Therefore, $f'(x) > 0$ for all real $x$.

3. Determine the Number of Real Roots

Since $f'(x) > 0$ for all $x$, $f(x)$ is strictly increasing. Also, $f(x)$ is a polynomial of degree 5, which is odd. Therefore, $f(x)$ has at least one real root. Since $f(x)$ is strictly increasing, it can have at most one real root. Thus, $f(x)$ has exactly one real root.

4. Locate the Real Root

We need to find an interval $(a, a+1)$ such that $f(a)$ and $f(a+1)$ have opposite signs. Let's evaluate $f(x)$ at some integer values: $f(0) = 10 > 0$. $f(-1) = 2(-1)^5 + 5(-1)^4 + 10(-1)^3 + 10(-1)^2 + 10(-1) + 10 = -2 + 5 - 10 + 10 - 10 + 10 = 3 > 0$. $f(-2) = 2(-2)^5 + 5(-2)^4 + 10(-2)^3 + 10(-2)^2 + 10(-2) + 10 = -64 + 80 - 80 + 40 - 20 + 10 = -34 < 0$.

Since $f(-2) < 0$ and $f(-1) > 0$, by the Intermediate Value Theorem, there is a root in the interval $(-2, -1)$.

5. Determine the Value of $|a|$

The problem states that all real roots lie in the interval $(a, a+1)$. Since we found the only real root lies in $(-2, -1)$, we have $(a, a+1) = (-2, -1)$. Thus, $a = -2$. Finally, $|a| = |-2| = 2$.

Common Mistakes & Tips

• Algebraic Manipulation: Be very careful with the algebraic manipulations, especially when completing the square or using substitutions. Double-check each step to avoid errors.
• Discriminant: Remember the discriminant test ($b^2-4ac$) to quickly determine if a quadratic is always positive or always negative.
• IVT Application: Make sure you understand the conditions required for the Intermediate Value Theorem to apply (continuity). Polynomials are always continuous.

Summary

We analyzed the given polynomial by finding its derivative and showing that it is always positive, implying the function is strictly increasing. This allowed us to conclude that the polynomial has exactly one real root. Using the Intermediate Value Theorem, we found that the root lies in the interval $(-2, -1)$. Thus, $a=-2$, and $|a|=2$.

Final Answer The final answer is \boxed{2}.