For function `f(x)=(lnx)/(x),` which of the following statements are true?

Options

A. (a)f(x) has horizontal tangent at x = e
B. (b)f(x) cuts the X-axis only at one point
C. (c)f(x) is many-one function
D. (d)f(x) has one vertical tangent

Answer

To analyze the function \( f(x) = \frac{\ln x}{x} \) and determine which statements are true, we will go through each option step by step.

### Step 1: Find the derivative of \( f(x) \)

To find the horizontal tangent, we first need to differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx}\left(\frac{\ln x}{x}\right) \]

Using the quotient rule, where \( u = \ln x \) and \( v = x \):

\[ f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2} \]

Calculating \( u' \) and \( v' \):

- \( u' = \frac{1}{x} \)
- \( v' = 1 \)

Substituting these into the quotient rule gives:

\[ f'(x) = \frac{x \cdot \frac{1}{x} - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2} \]

### Step 2: Determine where \( f'(x) = 0 \)

To find the points where the function has horizontal tangents, we set \( f'(x) = 0 \):

\[ \frac{1 - \ln x}{x^2} = 0 \]

This implies:

\[ 1 - \ln x = 0 \implies \ln x = 1 \implies x = e \]

**Conclusion for Option A:** \( f(x) \) has a horizontal tangent at \( x = e \) (True).

### Step 3: Find where \( f(x) \) cuts the x-axis

To find where \( f(x) \) cuts the x-axis, we set \( f(x) = 0 \):

\[ \frac{\ln x}{x} = 0 \]

This implies:

\[ \ln x = 0 \implies x = 1 \]

**Conclusion for Option B:** \( f(x) \) cuts the x-axis only at one point, \( x = 1 \) (True).

### Step 4: Determine if \( f(x) \) is a many-to-one function

To analyze the monotonicity of \( f(x) \), we check the sign of \( f'(x) \):

- For \( x < e \): \( \ln x < 1 \) implies \( f'(x) > 0 \) (increasing).
- For \( x > e \): \( \ln x > 1 \) implies \( f'(x) < 0 \) (decreasing).

Since \( f(x) \) is increasing on \( (0, e) \) and decreasing on \( (e, \infty) \), it is not one-to-one. Thus, it is a many-to-one function.

**Conclusion for Option C:** \( f(x) \) is a many-to-one function (True).

### Step 5: Check for vertical tangents

Vertical tangents occur where \( f'(x) \) is undefined or approaches infinity. Since \( f'(x) = \frac{1 - \ln x}{x^2} \), we check where the denominator is zero:

\[ x^2 = 0 \implies x = 0 \]

However, \( x = 0 \) is not in the domain of \( f(x) \) (since \( \ln x \) is undefined for \( x \leq 0 \)). Therefore, there are no vertical tangents.

**Conclusion for Option D:** \( f(x) \) does not have a vertical tangent (False).

### Final Summary of Statements:
- **A:** True
- **B:** True
- **C:** True
- **D:** False