Let `f(x) = x^(3) + 3x^(2) + 9x + 6 sin x` then roots of the equation `(1)/(x-f(1))+(2)/(x-f(2))+(3)/(x-f(3))=0`, has

Options

A. no real root
B. one real root
C. two real roots
D. more than 2 real roots

Explanation

To solve the problem step by step, we start with the function \( f(x) = x^3 + 3x^2 + 9x + 6 \sin x \).

### Step 1: Find the derivative of \( f(x) \)
We need to determine if \( f(x) \) is an increasing or decreasing function. To do this, we calculate the derivative \( f'(x) \).

\[ f'(x) = \frac{d}{dx}(x^3 + 3x^2 + 9x + 6 \sin x) \]
\[ = 3x^2 + 6x + 9 + 6 \cos x \]

### Step 2: Analyze \( f'(x) \)
Now we analyze \( f'(x) \):

\[ f'(x) = 3x^2 + 6x + 9 + 6 \cos x \]

The term \( 3x^2 + 6x + 9 \) is a quadratic function that opens upwards (since the coefficient of \( x^2 \) is positive) and its minimum value occurs at the vertex:

\[ x = -\frac{b}{2a} = -\frac{6}{2 \cdot 3} = -1 \]

Calculating \( f'(-1) \):

\[ f'(-1) = 3(-1)^2 + 6(-1) + 9 + 6 \cos(-1) \]
\[ = 3 - 6 + 9 + 6 \cos(-1) \]
Since \( \cos(-1) \) is a positive value, \( f'(-1) \) will be greater than 0.

### Step 3: Show that \( f'(x) > 0 \)
The term \( 6 \cos x \) varies between -6 and 6. Therefore, the minimum value of \( f'(x) \) occurs when \( \cos x = -1 \):

\[ f'(x) \geq 3x^2 + 6x + 3 \]

The quadratic \( 3x^2 + 6x + 3 \) has a discriminant:

\[ D = b^2 - 4ac = 6^2 - 4 \cdot 3 \cdot 3 = 36 - 36 = 0 \]

This means it has a double root and does not cross the x-axis, indicating that \( f'(x) > 0 \) for all \( x \). Thus, \( f(x) \) is an increasing function for all \( x \).

### Step 4: Evaluate \( f(1), f(2), f(3) \)
Since \( f(x) \) is increasing, we have:

\[ f(1) < f(2) < f(3) \]

Let \( a = f(1), b = f(2), c = f(3) \) such that \( a < b < c \).

### Step 5: Analyze the equation
We need to analyze the equation:

\[ \frac{1}{x - a} + \frac{2}{x - b} + \frac{3}{x - c} = 0 \]

### Step 6: Define \( g(x) \)
Let:

\[ g(x) = \frac{1}{x - a} + \frac{2}{x - b} + \frac{3}{x - c} \]

### Step 7: Evaluate \( g(a), g(b), g(c) \)
Evaluate \( g(x) \) at the points \( a, b, c \):

1. \( g(a) = \frac{1}{a - a} + \frac{2}{a - b} + \frac{3}{a - c} \) → undefined (approaches \( +\infty \))
2. \( g(b) = \frac{1}{b - a} + \frac{2}{b - b} + \frac{3}{b - c} \) → undefined (approaches \( -\infty \))
3. \( g(c) = \frac{1}{c - a} + \frac{2}{c - b} + \frac{3}{c - c} \) → undefined (approaches \( +\infty \))

### Step 8: Conclusion
Since \( g(a) \to +\infty \) and \( g(b) \to -\infty \), there is at least one root between \( a \) and \( b \). Similarly, since \( g(b) \to -\infty \) and \( g(c) \to +\infty \), there is at least one root between \( b \) and \( c \).

Thus, the equation has **two real roots**.

### Final Answer
The roots of the equation have **two real roots**.