The function `f(x)=pix^(3)-(3pi)/(2)(a+b)x^(2)+3piabx` has a local minimum at x = a, then the values a and b can take are

Options

A. `a=pi, b=e`
B. `a=e, b=pi`
C. `a=b=pi`
D. `a=b=e`

Answer

To solve the problem, we need to analyze the function \( f(x) = \pi x^3 - \frac{3\pi}{2}(a+b)x^2 + 3\pi abx \) and determine the conditions for it to have a local minimum at \( x = a \).

### Step 1: Differentiate the function
First, we find the first derivative of the function \( f(x) \):

\[ f'(x) = \frac{d}{dx} \left( \pi x^3 - \frac{3\pi}{2}(a+b)x^2 + 3\pi abx \right) \]

Using the power rule, we differentiate each term:

\[ f'(x) = 3\pi x^2 - 3\pi (a+b)x + 3\pi ab \]

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

\[ 3\pi x^2 - 3\pi (a+b)x + 3\pi ab = 0 \]

We can simplify this by dividing through by \( 3\pi \) (assuming \( \pi \neq 0 \)):

\[ x^2 - (a+b)x + ab = 0 \]

### Step 3: Factor the quadratic equation
Next, we factor the quadratic equation:

\[ (x - a)(x - b) = 0 \]

This gives us the critical points \( x = a \) and \( x = b \).

### Step 4: Determine the nature of the critical points
To determine whether \( x = a \) is a local minimum, we need to analyze the sign of the first derivative around \( x = a \).

1. For \( x < a \), \( f'(x) \) should be negative.
2. For \( x > a \), \( f'(x) \) should be positive.

### Step 5: Analyze the conditions
From the factorization \( (x - a)(x - b) = 0 \), we see that:

- If \( a < b \), then \( f'(x) < 0 \) for \( x < a \) and \( f'(x) > 0 \) for \( x > a \), indicating a local minimum at \( x = a \).
- If \( a > b \), then \( f'(x) > 0 \) for \( x < a \) and \( f'(x) < 0 \) for \( x > a \), indicating a local maximum at \( x = a \).

### Step 6: Conclusion
For \( f(x) \) to have a local minimum at \( x = a \), we must have \( a < b \).

### Final Answer
Thus, the values that \( a \) and \( b \) can take are such that \( a < b \).

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