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 given and determine the conditions under which it has a local minimum at \( x = a \).

### Step 1: Differentiate the function
The function is given as:
\[ f(x) = \pi x^3 - \frac{3\pi}{2}(a + b)x^2 + 3\pi abx \]
We start by differentiating \( f(x) \) with respect to \( x \):
\[ f'(x) = \frac{d}{dx}(\pi x^3) - \frac{3\pi}{2}(a + b) \frac{d}{dx}(x^2) + 3\pi ab \frac{d}{dx}(x) \]
Calculating the derivatives:
\[ f'(x) = 3\pi x^2 - \frac{3\pi}{2}(a + b)(2x) + 3\pi ab \]
Simplifying this, we have:
\[ f'(x) = 3\pi x^2 - 3\pi(a + b)x + 3\pi ab \]
Factoring out \( 3\pi \):
\[ f'(x) = 3\pi \left( x^2 - (a + b)x + ab \right) \]

### Step 2: Set the first derivative to zero
For local extrema, we set \( f'(x) = 0 \):
\[ 3\pi \left( x^2 - (a + b)x + ab \right) = 0 \]
This gives us the quadratic equation:
\[ x^2 - (a + b)x + ab = 0 \]
Using the quadratic formula, the roots are:
\[ x = \frac{(a + b) \pm \sqrt{(a + b)^2 - 4ab}}{2} \]

### Step 3: Determine conditions for local minimum
For \( x = a \) to be a local minimum, it must be one of the roots of the quadratic equation. Thus, we set:
\[ a = \frac{(a + b) + \sqrt{(a + b)^2 - 4ab}}{2} \]
This simplifies to:
\[ 2a = a + b + \sqrt{(a + b)^2 - 4ab} \]
Rearranging gives:
\[ a - b = \sqrt{(a + b)^2 - 4ab} \]
Squaring both sides:
\[ (a - b)^2 = (a + b)^2 - 4ab \]
Expanding both sides:
\[ a^2 - 2ab + b^2 = a^2 + 2ab + b^2 - 4ab \]
This simplifies to:
\[ -2ab = -2ab \]
This condition is always true, indicating that \( x = a \) is indeed a root.

### Step 4: Second derivative test
Next, we find the second derivative \( f''(x) \):
\[ f''(x) = \frac{d}{dx}(3\pi (x^2 - (a + b)x + ab)) = 3\pi(2x - (a + b)) \]
For a local minimum at \( x = a \):
\[ f''(a) = 3\pi(2a - (a + b)) > 0 \]
This simplifies to:
\[ 3\pi(a - b) > 0 \]
Since \( 3\pi > 0 \), we conclude:
\[ a - b > 0 \implies a > b \]

### Conclusion
Thus, the values \( a \) and \( b \) can take are such that \( a > b \).