If `y=f(x)` is a curve and if there exists two points `A(x_(1),f(x_(1)) and B(x_(2),f(x_(2))` on it such that `f'(x_(1))=-(1)/(f'(x_(2)))=(f(x_(2))-f(x_(1)))/(x_(2)-x_(1))`, then the tangent at `x_(1)` is normal at `x_(2)` for that curve. Now, anwer the following questions.
Number of such lines on the curve `y=|lnx|,` is

Options

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

Answer

To solve the problem, we need to analyze the conditions given for the curve \( y = | \ln x | \) and find the number of such lines where the tangent at one point is normal at another point.

### Step-by-Step Solution:

1. **Identify the function and its derivative**:
The function is given as \( y = f(x) = |\ln x| \). We need to find the derivative \( f'(x) \).

- For \( x > 1 \): \( f(x) = \ln x \) and \( f'(x) = \frac{1}{x} \)
- For \( 0 < x < 1 \): \( f(x) = -\ln x \) and \( f'(x) = -\frac{1}{x} \)

2. **Set up the conditions**:
According to the problem, we have:
\[ f'(x_1) = -\frac{1}{f'(x_2)} = \frac{f(x_2) - f(x_1)}{x_2 - x_1} \]

This means:
\[ f'(x_1) = -\frac{1}{f'(x_2)} \]

3. **Substituting the derivatives**:
- If \( x_1 > 1 \): \( f'(x_1) = \frac{1}{x_1} \)
- If \( x_2 > 1 \): \( f'(x_2) = \frac{1}{x_2} \)

Then:
\[ \frac{1}{x_1} = -\frac{1}{\frac{1}{x_2}} \implies \frac{1}{x_1} = -x_2 \]

This leads to:
\[ x_1 x_2 = -1 \]

4. **Considering the case when \( x_1 < 1 \)**:
- If \( x_1 < 1 \): \( f'(x_1) = -\frac{1}{x_1} \)
- If \( x_2 < 1 \): \( f'(x_2) = -\frac{1}{x_2} \)

Then:
\[ -\frac{1}{x_1} = -\frac{1}{-\frac{1}{x_2}} \implies -\frac{1}{x_1} = x_2 \]

This leads to:
\[ x_1 x_2 = 1 \]

5. **Finding the number of solutions**:
- For \( x_1 x_2 = -1 \): This is not possible since \( x_1 \) and \( x_2 \) must be positive.
- For \( x_1 x_2 = 1 \): This is valid and can be satisfied by infinitely many pairs \( (x_1, x_2) \) such that \( x_1 = k \) and \( x_2 = \frac{1}{k} \) for any \( k > 0 \).

### Conclusion:
Thus, the number of such lines on the curve \( y = |\ln x| \) is infinite.