Let S be the set of real values of parameter `lamda` for which the equation f(x) = `2x^(3)-3(2+lamda)x^(2)+12lamda` x has exactly one local maximum and exactly one local minimum. Then S is a subset of

Options

A. `(-4,oo)`
B. `(-3,3)`
C. `(3,8)`
D. `(-oo,-1)`

Answer

To solve the problem, we need to analyze the function \( f(x) = 2x^3 - 3(2 + \lambda)x^2 + 12\lambda x \) and determine the values of the parameter \( \lambda \) for which the function has exactly one local maximum and one local minimum.

### Step 1: Differentiate the function
To find the local maxima and minima, we first differentiate the function:

\[ f'(x) = \frac{d}{dx}(2x^3 - 3(2 + \lambda)x^2 + 12\lambda x) \]

Calculating the derivative gives:

\[ f'(x) = 6x^2 - 6(2 + \lambda)x + 12\lambda \]

### Step 2: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:

\[ 6x^2 - 6(2 + \lambda)x + 12\lambda = 0 \]

Dividing the entire equation by 6 simplifies it to:

\[ x^2 - (2 + \lambda)x + 2\lambda = 0 \]

### Step 3: Use the quadratic formula
The roots of this quadratic equation represent the critical points (local maxima and minima). We can use the quadratic formula:

\[ x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1 \), \( b = -(2 + \lambda) \), and \( c = 2\lambda \). Plugging in these values gives:

\[ x_{1,2} = \frac{2 + \lambda \pm \sqrt{(2 + \lambda)^2 - 8\lambda}}{2} \]

### Step 4: Simplify the expression under the square root
We simplify the expression under the square root:

\[ (2 + \lambda)^2 - 8\lambda = 4 + 4\lambda + \lambda^2 - 8\lambda = \lambda^2 - 4\lambda + 4 = (\lambda - 2)^2 \]

### Step 5: Find the critical points
Now substituting back into the formula for the roots:

\[ x_{1,2} = \frac{2 + \lambda \pm (\lambda - 2)}{2} \]

This gives us two roots:

1. \( x_1 = \frac{2 + \lambda + \lambda - 2}{2} = \frac{2\lambda}{2} = \lambda \)
2. \( x_2 = \frac{2 + \lambda - (\lambda - 2)}{2} = \frac{4}{2} = 2 \)

### Step 6: Conditions for one local maximum and one local minimum
For the function to have exactly one local maximum and one local minimum, the critical points must be distinct. This means that the discriminant of the quadratic must be zero, which we have already established occurs when:

\[ (\lambda - 2)^2 = 0 \]

This implies:

\[ \lambda - 2 = 0 \implies \lambda = 2 \]

### Step 7: Conclusion
Thus, for \( \lambda = 2 \), the function has exactly one local maximum and one local minimum. Therefore, the set \( S \) is a subset of:

\[ \{2\} \]