Key Concepts and Formulas

• Critical Points: A critical point of a function $f(x)$ is a point $x_0$ where $f'(x_0) = 0$ or $f'(x_0)$ is undefined.
• Local Extrema: Local maxima and minima occur at critical points. The second derivative test can be used to determine the nature of a critical point: If $f''(x_0) > 0$, then $x_0$ is a local minimum; if $f''(x_0) < 0$, then $x_0$ is a local maximum.
• Three Distinct Real Roots: For a cubic equation of the form $f(x) = a$ to have three distinct real roots, $a$ must lie strictly between the local minimum and local maximum values of $f(x)$.

Step-by-Step Solution

1. Rearrange the Equation and Define the Function The given equation is $5x^3 - 15x - a = 0$. To determine the values of $a$ for which the equation has three distinct real roots, we isolate $a$: $a = 5x^3 - 15x$ Define the function $f(x) = 5x^3 - 15x$. We want to find the range of $a$ such that the horizontal line $y=a$ intersects the graph of $y=f(x)$ at three distinct points.

2. Find the Derivative of the Function To locate the critical points of $f(x)$, we find its derivative $f'(x)$: $f'(x) = \frac{d}{dx}(5x^3 - 15x) = 15x^2 - 15$

3. Determine Critical Points Critical points occur where $f'(x) = 0$. Thus, we solve for $x$: $15x^2 - 15 = 0$ $15(x^2 - 1) = 0$ $x^2 = 1$ $x = \pm 1$ So, the critical points are $x = -1$ and $x = 1$.

4. Calculate Local Maximum and Local Minimum Values We evaluate $f(x)$ at the critical points to find the local extrema:

• At $x = -1$: $f(-1) = 5(-1)^3 - 15(-1) = -5 + 15 = 10$ To determine if this is a local maximum or minimum, we use the second derivative test: $f''(x) = \frac{d}{dx}(15x^2 - 15) = 30x$ $f''(-1) = 30(-1) = -30 < 0$ Since $f''(-1) < 0$, $x = -1$ corresponds to a local maximum, and $f_{local\_max} = 10$.

• At $x = 1$: $f(1) = 5(1)^3 - 15(1) = 5 - 15 = -10$ $f''(1) = 30(1) = 30 > 0$ Since $f''(1) > 0$, $x = 1$ corresponds to a local minimum, and $f_{local\_min} = -10$.

5. Apply the Condition for Three Distinct Real Roots For the equation $f(x) = a$ to have three distinct real roots, $a$ must lie strictly between the local minimum and local maximum values of $f(x)$: $f_{local\_min} < a < f_{local\_max}$ $-10 < a < 10$ Thus, the set of all values of $a$ for which the equation has three distinct real roots is the interval $(-10, 10)$.

6. Identify $\alpha$ and $\beta$ and Calculate the Final Expression We are given that the interval is $(\alpha, \beta)$. Comparing this to our interval $(-10, 10)$, we have: $\alpha = -10$ $\beta = 10$ Now, we calculate $\beta - 2\alpha$: $\beta - 2\alpha = 10 - 2(-10) = 10 + 20 = 30$

Common Mistakes & Tips

• Strict Inequalities: Remember to use strict inequalities ($<$ and $>$) when finding the range of $a$ for distinct roots. Using $\le$ or $\ge$ would include cases where the cubic has a repeated root.
• Second Derivative Test: The second derivative test is a useful tool for determining whether a critical point is a local maximum or minimum. However, if the second derivative is zero, the test is inconclusive, and you would need to analyze the sign change of the first derivative.

Summary

To find the range of $a$ for which the cubic equation $5x^3 - 15x - a = 0$ has three distinct real roots, we express $a$ as a function of $x$, $f(x) = 5x^3 - 15x$, and find the local maximum and minimum values of $f(x)$. The value of $a$ must lie strictly between these two values. This gives us the interval $(-10, 10)$, so $\alpha = -10$ and $\beta = 10$. Therefore, $\beta - 2\alpha = 30$.

The final answer is \boxed{30}.