if the function `f: R rarr R ` given be `f(x)=x^3 + ax^2 + 5 x + sin 2x ` is invertible then

Options

A. `a in (-oo,-3)`
B. `a in (-3,3)`
C. `a in (3,oo)`
D. none of these

Explanation

To determine the conditions under which the function \( f(x) = x^3 + ax^2 + 5x + \sin(2x) \) is invertible, we need to analyze its derivative \( f'(x) \). A function is invertible if it is either strictly increasing or strictly decreasing over its entire domain. This means that the derivative \( f'(x) \) must not change signs; it should either be always positive or always negative.

### Step 1: Find the derivative of the function
The first step is to differentiate the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(x^3 + ax^2 + 5x + \sin(2x)) \]

Using the rules of differentiation, we get:

\[ f'(x) = 3x^2 + 2ax + 5 + 2\cos(2x) \]

### Step 2: Analyze the derivative
For \( f(x) \) to be invertible, we need \( f'(x) \) to be either always greater than zero or always less than zero for all \( x \).

### Step 3: Establish conditions for \( f'(x) > 0 \)
To ensure \( f'(x) > 0 \):

\[ 3x^2 + 2ax + 5 + 2\cos(2x) > 0 \]

The term \( 2\cos(2x) \) oscillates between -2 and 2. Therefore, we consider the minimum value of \( 2\cos(2x) \), which is -2. Thus, we have:

\[ 3x^2 + 2ax + 5 - 2 > 0 \]

This simplifies to:

\[ 3x^2 + 2ax + 3 > 0 \]

### Step 4: Establish conditions for \( f'(x) < 0 \)
Similarly, we analyze the case where \( f'(x) < 0 \):

\[ 3x^2 + 2ax + 5 + 2\cos(2x) < 0 \]

Considering the maximum value of \( 2\cos(2x) \), which is 2, we have:

\[ 3x^2 + 2ax + 5 + 2 < 0 \]

This simplifies to:

\[ 3x^2 + 2ax + 7 < 0 \]

### Step 5: Determine the conditions on \( a \)
To ensure that the quadratic \( 3x^2 + 2ax + 3 \) is always positive, we require that its discriminant is less than or equal to zero:

\[ D = (2a)^2 - 4 \cdot 3 \cdot 3 = 4a^2 - 36 \leq 0 \]

This gives:

\[ 4a^2 \leq 36 \implies a^2 \leq 9 \implies -3 \leq a \leq 3 \]

For the quadratic \( 3x^2 + 2ax + 7 \) to be always negative, we similarly require:

\[ D' = (2a)^2 - 4 \cdot 3 \cdot 7 = 4a^2 - 84 < 0 \]

This gives:

\[ 4a^2 < 84 \implies a^2 < 21 \implies -\sqrt{21} < a < \sqrt{21} \]

### Conclusion
Combining these conditions, we find that for \( f(x) \) to be invertible, \( a \) must satisfy:

\[ a \in (-\sqrt{21}, \sqrt{21}) \cap [-3, 3] \]

Thus, the final condition for \( a \) is:

\[ a \in [-3, 3] \]