Question

Let $ a>0 $. If the function $ f(x) = 6x^3 - 45ax^2 + 108a^2x + 1 $ attains its local maximum and minimum values at the points $ x_1 $ and $ x_2 $ respectively such that $ x_1x_2 = 54 $, then $ a + x_1 + x_2 $ is equal to:

• A. 15
• B. 18
• C. 24
• D. 13

Hint:

For finding the local maxima and minima, always set the derivative equal to zero and solve the resulting quadratic equation.

Answer 1

The Correct Option is B

The given function is \( f(x) = 6x^3 - 45ax^2 + 108a^2x + 1 \). 
To find the points where the function attains its local maxima and minima, we first find its first derivative: \[ f'(x) = 18x^2 - 90ax + 108a^2 \] Setting \( f'(x) = 0 \) to find critical points: \[ 18x^2 - 90ax + 108a^2 = 0 \] Dividing through by 18: \[ x^2 - 5ax + 6a^2 = 0 \] Solving this quadratic equation using the quadratic formula: \[ x = \frac{-(-5a) \pm \sqrt{(-5a)^2 - 4(1)(6a^2)}}{2(1)} = \frac{5a \pm \sqrt{25a^2 - 24a^2}}{2} = \frac{5a \pm a}{2} \] 
Thus, the critical points are: \[ x_1 = 2a \quad \text{and} \quad x_2 = 3a \] We are given that \( x_1x_2 = 54 \), so: \[ 2a \times 3a = 54 \] \[ 6a^2 = 54 \quad \Rightarrow \quad a^2 = 9 \quad \Rightarrow \quad a = 3 \] Now, \( x_1 = 2a = 6 \) and \( x_2 = 3a = 9 \), so: \[ a + x_1 + x_2 = 3 + 6 + 9 = 18 \] 
Thus, the correct answer is 18.

Answer 2

Step 1: Find the first derivative. To find the local maximum and minimum, we first compute the first derivative of \( f(x) \): \[ f'(x) = \frac{d}{dx} \left( 6x^3 - 45ax^2 + 108a^2x + 1 \right) = 18x^2 - 90ax + 108a^2 \] Set \( f'(x) = 0 \) to find the critical points: \[ 18x^2 - 90ax + 108a^2 = 0 \] Divide by 18: \[ x^2 - 5ax + 6a^2 = 0 \] Step 2: Solve the quadratic equation. This is a quadratic equation in \( x \). Using the quadratic formula: \[ x = \frac{-(-5a) \pm \sqrt{(-5a)^2 - 4(1)(6a^2)}}{2(1)} = \frac{5a \pm \sqrt{25a^2 - 24a^2}}{2} = \frac{5a \pm a}{2} \] Thus, the two roots are: \[ x_1 = \frac{5a + a}{2} = 3a \quad \text{and} \quad x_2 = \frac{5a - a}{2} = 2a \] Step 3: Use the given condition \( x_1x_2 = 54 \).We are given that \( x_1x_2 = 54 \), so: \[ (3a)(2a) = 54 \] \[ 6a^2 = 54 \quad \Rightarrow \quad a^2 = 9 \quad \Rightarrow \quad a = 3 \] Step 4: Calculate \( a + x_1 + x_2 \). Now that we know \( a = 3 \), we can find \( x_1 \) and \( x_2 \): \[ x_1 = 3a = 9 \quad \text{and} \quad x_2 = 2a = 6 \] Thus: \[ a + x_1 + x_2 = 3 + 9 + 6 = 18 \] Final Answer: \[ \boxed{18} \]