The exhaustive set of value of 'a' for which the function `f(x)=a/3(x^3+(a+2)x^2+(a-1)x+2` possess a negative point of minima is `(q,oo)`. The value of q is

Answer

To solve the problem, we will find the values of 'a' for which the function \( f(x) = \frac{a}{3}(x^3 + (a+2)x^2 + (a-1)x + 2) \) has a negative point of minima. We will follow these steps:

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

The function is given as:
\[ f(x) = \frac{a}{3}(x^3 + (a+2)x^2 + (a-1)x + 2) \]

To find the first derivative \( f'(x) \), we differentiate \( f(x) \):
\[ f'(x) = \frac{a}{3} \left( 3x^2 + 2(a+2)x + (a-1) \right) \]
Simplifying this gives:
\[ f'(x) = a \left( x^2 + \frac{2(a+2)}{3}x + \frac{(a-1)}{3} \right) \]

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

To find the critical points, we set \( f'(x) = 0 \):
\[ a \left( x^2 + \frac{2(a+2)}{3}x + \frac{(a-1)}{3} \right) = 0 \]
This implies either \( a = 0 \) or:
\[ x^2 + \frac{2(a+2)}{3}x + \frac{(a-1)}{3} = 0 \]

### Step 3: Find the second derivative of \( f(x) \)

Next, we find the second derivative \( f''(x) \):
\[ f''(x) = a \left( 2x + \frac{2(a+2)}{3} \right) \]

### Step 4: Determine conditions for minima

For \( f(x) \) to have a minimum at a critical point, we need \( f''(x) > 0 \). Thus:
\[ 2ax + 2(a+2) > 0 \]
This simplifies to:
\[ 2ax + 2a + 4 > 0 \]
Dividing by 2 gives:
\[ ax + a + 2 > 0 \]
Rearranging gives:
\[ ax > -a - 2 \]
Thus:
\[ x > -\frac{a + 2}{a} \quad \text{(for } a \neq 0\text{)} \]

### Step 5: Identify the conditions for a negative point of minima

For the point of minima to be negative, we require:
\[ -\frac{a + 2}{a} < 0 \]
This implies:
\[ a + 2 > 0 \quad \text{and} \quad a < 0 \]
From \( a + 2 > 0 \), we have:
\[ a > -2 \]

### Step 6: Combine the conditions

Thus, we have:
\[ -2 < a < 0 \]
This means the set of values for \( a \) that allow for a negative point of minima is \( (-2, 0) \).

### Conclusion

The exhaustive set of values of \( a \) for which the function possesses a negative point of minima is \( (q, \infty) \). From our analysis, we find that \( q = -2 \).

### Final Answer
The value of \( q \) is:
\[ \boxed{-2} \]