Let `f(x) = ax^3 + bx^2 + cx + d sin x`. Find the condition that `f(x)` is always one-one function.

Answer

To determine the condition under which the function \( f(x) = ax^3 + bx^2 + cx + d \sin x \) is always one-one, we need to analyze its derivative. A function is one-one if it is either always increasing or always decreasing. This can be established by ensuring that the derivative \( f'(x) \) is either always positive or always negative.

### Step-by-Step Solution:

1. **Find the derivative of the function:**
\[ f'(x) = \frac{d}{dx}(ax^3 + bx^2 + cx + d \sin x) \]
Using the power rule and the derivative of sine:
\[ f'(x) = 3ax^2 + 2bx + c + d \cos x \]

2. **Set the condition for the function to be one-one:**
For \( f(x) \) to be one-one, \( f'(x) \) must be either always greater than zero or always less than zero:
\[ f'(x) > 0 \quad \text{or} \quad f'(x) < 0 \]

3. **Analyze the quadratic part of the derivative:**
The expression \( 3ax^2 + 2bx + c \) is a quadratic function. The sign of this quadratic function depends on the leading coefficient \( 3a \) and the discriminant \( D \):
\[ D = (2b)^2 - 4(3a)(c) = 4b^2 - 12ac \]

4. **Condition for the quadratic to be always positive:**
- If \( a > 0 \), for \( 3ax^2 + 2bx + c \) to be always positive, the discriminant must be less than zero:
\[ D < 0 \Rightarrow 4b^2 - 12ac < 0 \Rightarrow b^2 < 3ac \]

5. **Consider the contribution of \( d \cos x \):**
Since \( d \cos x \) oscillates between \( -|d| \) and \( |d| \), we need to ensure that even at its minimum value, \( f'(x) \) remains positive. Therefore, we must also consider:
\[ 3ax^2 + 2bx + c + d \cos x > 0 \quad \text{for all } x \]

6. **Combine conditions:**
The condition for \( f(x) \) to be one-one is:
\[ b^2 < 3ac \quad \text{and} \quad d \text{ must be such that } 3ax^2 + 2bx + c + d \cos x > 0 \text{ for all } x. \]

### Final Condition:
Thus, the condition for \( f(x) \) to be always one-one is:
\[ b^2 < 3ac \quad \text{and } \quad d \text{ must be sufficiently small to ensure positivity.} \]