The function `f(x)=2x^(3)-3(a+b)x^(2)+6abx` has a local maximum at `x=a`, if

Options

A. `a gt b`
B. `a lt b`
C. `a gt 0`
D. `a lt 0`

Answer

To solve the problem, we need to find the conditions under which the function \( f(x) = 2x^3 - 3(a+b)x^2 + 6abx \) has a local maximum at \( x = a \).

### Step-by-Step Solution:

1. **Find the first derivative of \( f(x) \)**:
\[ f'(x) = \frac{d}{dx}(2x^3 - 3(a+b)x^2 + 6abx) \]
Using the power rule:
\[ f'(x) = 6x^2 - 6(a+b)x + 6ab \]

2. **Set the first derivative equal to zero at \( x = a \)**:
\[ f'(a) = 6a^2 - 6(a+b)a + 6ab = 0 \]
Simplifying this:
\[ 6a^2 - 6(a^2 + ab) + 6ab = 0 \]
\[ 6a^2 - 6a^2 - 6ab + 6ab = 0 \]
This simplifies to \( 0 = 0 \), which is always true. This means \( x = a \) is a critical point.

3. **Find the second derivative of \( f(x) \)**:
\[ f''(x) = \frac{d}{dx}(6x^2 - 6(a+b)x + 6ab) \]
Using the power rule again:
\[ f''(x) = 12x - 6(a+b) \]

4. **Evaluate the second derivative at \( x = a \)**:
\[ f''(a) = 12a - 6(a+b) \]
For a local maximum, we require:
\[ f''(a) < 0 \]
Thus:
\[ 12a - 6(a+b) < 0 \]
Simplifying this inequality:
\[ 12a - 6a - 6b < 0 \]
\[ 6a - 6b < 0 \]
Dividing by 6:
\[ a - b < 0 \]
Therefore:
\[ a < b \]

### Conclusion:
The function \( f(x) \) has a local maximum at \( x = a \) if \( a < b \).