Key Concepts and Formulas

• Local Extrema: A differentiable function $f(x)$ has local maxima and minima at critical points where $f'(x) = 0$.
• Quadratic Equation Roots: For a quadratic equation $Ax^2 + Bx + C = 0$, the sum of roots is $x_1 + x_2 = -\frac{B}{A}$ and the product of roots is $x_1 x_2 = \frac{C}{A}$.
• Differentiation Rules: Power rule: $\frac{d}{dx}(x^n) = nx^{n-1}$. Constant rule: $\frac{d}{dx}(c) = 0$.

Step 1: Find the First Derivative

We are given the function $f(x) = 6x^3 - 45ax^2 + 108a^2x + 1$. To find the critical points where local maxima and minima occur, we need to find the first derivative $f'(x)$ and set it equal to zero.

$f'(x) = \frac{d}{dx}(6x^3 - 45ax^2 + 108a^2x + 1)$ Using the power rule and constant multiple rule, we have: $f'(x) = 6(3x^2) - 45a(2x) + 108a^2(1) + 0$ $f'(x) = 18x^2 - 90ax + 108a^2$

Step 2: Set the Derivative to Zero and Simplify

Local extrema occur when $f'(x) = 0$. So, we set the derivative we found in Step 1 to zero: $18x^2 - 90ax + 108a^2 = 0$ We can simplify this quadratic equation by dividing both sides by 18: $x^2 - 5ax + 6a^2 = 0$ The roots of this quadratic equation are $x_1$ and $x_2$, which correspond to the $x$-coordinates where the local maximum and minimum occur.

Step 3: Use the Given Condition to Find 'a'

We are given that $x_1 x_2 = 54$. For the quadratic equation $x^2 - 5ax + 6a^2 = 0$, the product of the roots is given by $\frac{C}{A}$, where $A = 1$ and $C = 6a^2$. Therefore, $x_1 x_2 = \frac{6a^2}{1} = 6a^2$ Since $x_1 x_2 = 54$, we have: $6a^2 = 54$ Divide both sides by 6: $a^2 = 9$ Taking the square root of both sides, we get $a = \pm 3$. Since we are given that $a > 0$, we must have: $a = 3$

Step 4: Calculate the Sum of the Roots

For the quadratic equation $x^2 - 5ax + 6a^2 = 0$, the sum of the roots is given by $-\frac{B}{A}$, where $A = 1$ and $B = -5a$. Therefore, $x_1 + x_2 = -\frac{-5a}{1} = 5a$ Since we found that $a = 3$, we have: $x_1 + x_2 = 5(3) = 15$

Step 5: Calculate the Final Result

We are asked to find the value of $a + x_1 + x_2$. We have $a = 3$ and $x_1 + x_2 = 15$. Therefore, $a + x_1 + x_2 = 3 + 15 = 18$

Common Mistakes & Tips

• Sign Errors: Be careful with signs when applying Vieta's formulas. Remember that the sum of the roots is $-\frac{B}{A}$.
• Constraint on 'a': Don't forget to use the given condition $a > 0$ to choose the correct value of $a$.
• Simplification: Always simplify the quadratic equation after finding the derivative to make the calculations easier.

Summary

We found the first derivative of the given function and set it to zero to find a quadratic equation whose roots are the x-coordinates of the local extrema. Using the given product of roots, we solved for 'a', and then used the sum of roots formula to find the value of $x_1 + x_2$. Finally, we added 'a' to the sum of the roots to obtain the desired result.

The final answer is $\boxed{18}$, which corresponds to option (D).