Let `f(x) = 2x^3 + 3(1-3a)x^2 + 6(a^2-a)x +b` where `a, b in R`. Find the smallest integral value of a for which f(x) has a positive point of local maximum.

Answer

To solve the problem, we need to find the smallest integral value of \( a \) for which the function

\[ f(x) = 2x^3 + 3(1-3a)x^2 + 6(a^2-a)x + b \]

has a positive point of local maximum.

### Step 1: Find the first derivative \( f'(x) \)

To find the critical points where local maxima or minima may occur, we first need to compute the derivative of \( f(x) \):

\[ f'(x) = \frac{d}{dx}(2x^3 + 3(1-3a)x^2 + 6(a^2-a)x + b) \]

Calculating the derivative:

\[ f'(x) = 6x^2 + 6(1-3a)x + 6(a^2-a) \]

We can simplify this to:

\[ f'(x) = 6(x^2 + (1-3a)x + (a^2-a)) \]

### Step 2: Set the first derivative to zero to find critical points

To find the critical points, we set \( f'(x) = 0 \):

\[ 6(x^2 + (1-3a)x + (a^2-a)) = 0 \]

This simplifies to:

\[ x^2 + (1-3a)x + (a^2-a) = 0 \]

### Step 3: Apply the quadratic formula

The roots of the quadratic equation can be found using the quadratic formula:

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

Here, \( a = 1 \), \( b = (1-3a) \), and \( c = (a^2-a) \). Thus, we have:

\[ x = \frac{-(1-3a) \pm \sqrt{(1-3a)^2 - 4(1)(a^2-a)}}{2(1)} \]

### Step 4: Calculate the discriminant

The discriminant \( D \) must be non-negative for the roots to be real:

\[ D = (1-3a)^2 - 4(a^2-a) \]

Expanding this:

\[ D = (1 - 6a + 9a^2) - (4a^2 - 4a) \]
\[ D = 5a^2 - 2a + 1 \]

### Step 5: Conditions for local maxima

For \( f(x) \) to have a positive local maximum, both roots of the derivative must be positive. This requires:

1. The sum of the roots \( \alpha + \beta = -(1-3a) > 0 \)
2. The product of the roots \( \alpha \beta = a^2 - a > 0 \)

### Step 6: Analyze the conditions

1. From \( -(1-3a) > 0 \), we get:

\[ 1 - 3a < 0 \implies a > \frac{1}{3} \]

2. From \( a^2 - a > 0 \):

Factoring gives:

\[ a(a-1) > 0 \]

This inequality holds when \( a < 0 \) or \( a > 1 \).

### Step 7: Combine the conditions

We need to find the intersection of the conditions:

- From \( a > \frac{1}{3} \)
- From \( a < 0 \) or \( a > 1 \)

The valid range for \( a \) is:

\[ a > 1 \]

### Step 8: Find the smallest integral value

The smallest integral value of \( a \) that satisfies \( a > 1 \) is:

\[ \boxed{2} \]