Let `f(x)` be a non-positive continuous function and `F(x)=int_(0)^(x)f(t)dt AA x ge0` and `f(x) ge cF(x)` where `c lt 0` and let `g:[0, infty) to R` be a function such that `(dg(x))/(dx) lt g(x) AA x gt 0` and `g(0)=0`
The number of solution(s) of the equation
`|x^(2)+x-6|=f(x)+g(x)` is/are

Options

A. 2
B. 1
C. 0
D. 3

Explanation

To solve the problem, we need to analyze the given functions and the equation we are working with.

### Step 1: Understanding the Functions
We have:
- \( f(x) \) is a non-positive continuous function, meaning \( f(x) \leq 0 \) for all \( x \).
- \( F(x) = \int_0^x f(t) dt \) is the integral of \( f(x) \) from 0 to \( x \).
- The condition \( f(x) \geq cF(x) \) where \( c < 0 \) implies that \( f(x) \) is bounded below by a negative multiple of the area under the curve of \( f(t) \).

### Step 2: Analyzing the Function \( g(x) \)
The function \( g(x) \) satisfies:
- \( \frac{dg(x)}{dx} < g(x) \) for \( x > 0 \)
- \( g(0) = 0 \)

This implies that \( g(x) \) is a non-increasing function since its growth rate is less than its current value.

### Step 3: Analyzing the Equation
We need to analyze the equation:
\[ |x^2 + x - 6| = f(x) + g(x) \]

### Step 4: Behavior of \( |x^2 + x - 6| \)
The expression \( x^2 + x - 6 \) can be factored:
\[ x^2 + x - 6 = (x - 2)(x + 3) \]
The roots of this quadratic are \( x = 2 \) and \( x = -3 \). Since we are interested in \( x \geq 0 \), we have:
- For \( 0 \leq x < 2 \), \( |x^2 + x - 6| = 6 - x^2 - x \)
- For \( x \geq 2 \), \( |x^2 + x - 6| = x^2 + x - 6 \)

### Step 5: Analyzing \( f(x) + g(x) \)
Since \( f(x) \leq 0 \) and \( g(0) = 0 \) with \( g(x) \) being non-increasing, we can conclude:
- \( f(x) + g(x) \leq g(x) \leq 0 \) for all \( x \geq 0 \)

### Step 6: Comparing Both Sides
Now we compare the two sides of the equation:
1. For \( 0 \leq x < 2 \):
\[ 6 - x^2 - x \geq 0 \]
This is positive, while \( f(x) + g(x) \leq 0 \). Thus, no solutions exist in this interval.

2. For \( x \geq 2 \):
\[ x^2 + x - 6 \geq 0 \]
This is also positive, while \( f(x) + g(x) \leq 0 \). Thus, no solutions exist in this interval as well.

### Conclusion
Since in both cases \( |x^2 + x - 6| \) is positive while \( f(x) + g(x) \) is non-positive, there are no solutions to the equation.

The number of solutions of the equation \( |x^2 + x - 6| = f(x) + g(x) \) is **zero**.