If `f:R to R` is given by
`f(x)=x^(3)+(a+2)x^(2)+3ax+5a` if f(x) is one-one function, then a belong to

Explanation

To determine the values of \( a \) for which the function \( f(x) = x^3 + (a+2)x^2 + 3ax + 5a \) is a one-one function, we need to analyze its derivative. A function is one-one if its derivative does not change sign, meaning it is either always positive or always negative.

### Step-by-Step Solution:

1. **Find the Derivative**:
We start by differentiating the function:
\[ f'(x) = \frac{d}{dx}(x^3 + (a+2)x^2 + 3ax + 5a) \]
Using the power rule, we get:
\[ f'(x) = 3x^2 + 2(a+2)x + 3a \]

2. **Analyze the Derivative**:
For \( f(x) \) to be one-one, \( f'(x) \) must either be always positive or always negative. Since the leading coefficient (3) of \( f'(x) \) is positive, the parabola opens upwards. Therefore, \( f'(x) \) will be always positive if it has no real roots, which occurs when the discriminant is less than zero.

3. **Calculate the Discriminant**:
The discriminant \( D \) of the quadratic \( f'(x) = 3x^2 + 2(a+2)x + 3a \) is given by:
\[ D = b^2 - 4ac = [2(a+2)]^2 - 4 \cdot 3 \cdot 3a \]
Simplifying this:
\[ D = 4(a+2)^2 - 36a \]
\[ D = 4(a^2 + 4a + 4) - 36a \]
\[ D = 4a^2 + 16a + 16 - 36a \]
\[ D = 4a^2 - 20a + 16 \]

4. **Set the Discriminant Less Than Zero**:
We require:
\[ 4a^2 - 20a + 16 < 0 \]
Dividing the entire inequality by 4:
\[ a^2 - 5a + 4 < 0 \]

5. **Factor the Quadratic**:
Factoring gives us:
\[ (a - 1)(a - 4) < 0 \]

6. **Determine the Intervals**:
The critical points are \( a = 1 \) and \( a = 4 \). To find the intervals where the product is negative, we test the intervals:
- For \( a < 1 \): Choose \( a = 0 \) → \( (0-1)(0-4) = 4 > 0 \)
- For \( 1 < a < 4 \): Choose \( a = 2 \) → \( (2-1)(2-4) = -2 < 0 \)
- For \( a > 4 \): Choose \( a = 5 \) → \( (5-1)(5-4) = 4 > 0 \)

Thus, the inequality holds for:
\[ 1 < a < 4 \]

### Conclusion:
The values of \( a \) for which the function \( f(x) \) is one-one are:
\[ a \in (1, 4) \]