Let a continous and differentiable function `f(x)` is such that `f(x)` and `(d)/(dx)f(x)` have opposite signs everywhere. Then,

Options

A. `f'(x)` is always increasing
B. `f(x)` is always increasing
C. `|f(x)|` is non - decreasing
D. `|f(x)|` is decreasing

Answer

To solve the problem, we need to analyze the given conditions about the function \( f(x) \) and its derivative \( f'(x) \).

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
We are told that \( f(x) \) and \( f'(x) \) have opposite signs everywhere. This means:
- If \( f(x) > 0 \), then \( f'(x) < 0 \) (the function is positive and decreasing).
- If \( f(x) < 0 \), then \( f'(x) > 0 \) (the function is negative and increasing).

This implies that \( f(x) \) cannot be zero at any point because if \( f(x) = 0 \), then \( f'(x) \) would also need to be zero, which contradicts the condition of having opposite signs.

2. **Considering the Absolute Value**:
We can consider the absolute value of the function, \( |f(x)| \). The behavior of \( |f(x)| \) will depend on whether \( f(x) \) is positive or negative:
- If \( f(x) > 0 \), then \( |f(x)| = f(x) \).
- If \( f(x) < 0 \), then \( |f(x)| = -f(x) \).

3. **Differentiating the Absolute Value**:
We differentiate \( |f(x)| \):
- For \( f(x) > 0 \):
\[ \frac{d}{dx}|f(x)| = f'(x) \]
- For \( f(x) < 0 \):
\[ \frac{d}{dx}|f(x)| = -f'(x) \]

4. **Analyzing the Derivative**:
- In the case where \( f(x) > 0 \), since \( f'(x) < 0 \), we have:
\[ \frac{d}{dx}|f(x)| = f'(x) < 0 \]
This means \( |f(x)| \) is decreasing.
- In the case where \( f(x) < 0 \), since \( f'(x) > 0 \), we have:
\[ \frac{d}{dx}|f(x)| = -f'(x) < 0 \]
This also means \( |f(x)| \) is decreasing.

5. **Conclusion**:
Since in both cases (whether \( f(x) \) is positive or negative), the derivative \( \frac{d}{dx}|f(x)| \) is negative, we conclude that \( |f(x)| \) is a decreasing function everywhere.

### Final Answer:
The correct option is that \( |f(x)| \) is decreasing.