Let f(x) be a non negative continuous and bounded function for all `xge0` .If `(cos x)f(x) lt (sin x- cosx)f(x) forall x ge 0`, then which of the following is/are correct?

Options

A. `f(6)+f(5)gt0`
B. `x^(2)-3x+2+f(7)=0` has 2 distinct solution
C. f(5)f(7)-f(5)=0
D. `underset(xrarr6)lim (f(x)-sin (pix))/(x-6)=1`

Explanation

To solve the problem, we need to analyze the inequality given in the question and the implications it has on the function \( f(x) \).

### Step-by-Step Solution:

1. **Understanding the Inequality**:
We start with the inequality:
\[ \cos(x) f(x) < \sin(x) - \cos(x) f(x) \quad \text{for all } x \geq 0 \]
Rearranging this gives:
\[ \cos(x) f(x) + \cos(x) f(x) < \sin(x) \]
or
\[ 2 \cos(x) f(x) < \sin(x) \]
This implies:
\[ f(x) < \frac{\sin(x)}{2 \cos(x)} \quad \text{for } \cos(x) \neq 0 \]

2. **Analyzing the Function**:
Since \( f(x) \) is non-negative, continuous, and bounded, we can infer that \( f(x) \) must be zero at points where the right side of the inequality becomes zero or negative.

3. **Finding Critical Points**:
The function \( \frac{\sin(x)}{2 \cos(x)} \) is undefined when \( \cos(x) = 0 \), which occurs at \( x = \frac{\pi}{2} + n\pi \) for \( n \in \mathbb{Z} \). At these points, \( \sin(x) \) is either 1 or -1, but since \( f(x) \) is non-negative, we focus on the intervals where \( \cos(x) > 0 \).

4. **Evaluating Specific Values**:
We can evaluate specific values of \( f(x) \) at critical points. For instance, at \( x = 0 \):
\[ f(0) < \frac{\sin(0)}{2 \cos(0)} = 0 \]
This implies \( f(0) = 0 \).

5. **Generalizing the Result**:
Since \( f(x) \) is continuous and non-negative, and we have established that \( f(0) = 0 \), we can conclude that \( f(x) \) must be zero at other critical points where \( \sin(x) - 2\cos(x)f(x) \) could potentially become zero or negative.

6. **Conclusion**:
By analyzing the inequality and the behavior of \( f(x) \), we can conclude that \( f(x) \) must be zero at specific points, leading us to determine the correctness of the options provided in the question.