`f(x)=|x log_e x|` monotonically decreases in `(0,1/e)` (b) `(1/e ,1)` `(1,oo)` (d) `(1/e ,oo)`

Options

A. `(0,1//e)`
B. `(1//e,1)`
C. `(1.oo)`
D. `(1//e,oo)`

Explanation

To determine the intervals where the function \( f(x) = |x \log_e x| \) is monotonically decreasing, we need to analyze the derivative of the function. Here’s a step-by-step solution:

### Step 1: Define the function
The function is given as:
\[ f(x) = |x \log_e x| \]

### Step 2: Determine the intervals for \( f(x) \)
Since the logarithm function is defined only for \( x > 0 \), we will consider the intervals \( (0, 1) \) and \( (1, \infty) \).

### Step 3: Analyze the function in different intervals
1. **For \( x \in (0, 1) \)**:
- Here, \( \log_e x < 0 \), thus \( x \log_e x < 0 \).
- Therefore, \( f(x) = -x \log_e x \).

2. **For \( x \in (1, \infty) \)**:
- Here, \( \log_e x > 0 \), thus \( x \log_e x > 0 \).
- Therefore, \( f(x) = x \log_e x \).

### Step 4: Differentiate \( f(x) \)
Now we differentiate \( f(x) \) in both intervals.

1. **For \( x \in (0, 1) \)**:
\[ f(x) = -x \log_e x \]
Using the product rule:
\[ f'(x) = -\left( \log_e x + 1 \right) \]

2. **For \( x \in (1, \infty) \)**:
\[ f(x) = x \log_e x \]
Again using the product rule:
\[ f'(x) = \log_e x + 1 \]

### Step 5: Determine where \( f'(x) < 0 \)
1. **For \( x \in (0, 1) \)**:
\[ f'(x) = -(\log_e x + 1) < 0 \implies \log_e x + 1 > 0 \implies \log_e x > -1 \implies x > \frac{1}{e} \]
Thus, \( f(x) \) is decreasing in the interval \( \left( \frac{1}{e}, 1 \right) \).

2. **For \( x \in (1, \infty) \)**:
\[ f'(x) = \log_e x + 1 > 0 \text{ for } x > 1 \]
Thus, \( f(x) \) is increasing in the interval \( (1, \infty) \).

### Conclusion
The function \( f(x) \) is monotonically decreasing in the interval \( \left( \frac{1}{e}, 1 \right) \).

### Final Answer
The correct option is:
(b) \( \left( \frac{1}{e}, 1 \right) \)

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