Let `f(x) = x^(2) - 3x + 2` be a function defined from `R rarr R` and `|x| = {{:(x,, x ge 0),(-x ,, x le 0):}`, then answer the following
Value of k for which equation `|f|x|| = k` has six solutions

Options

A. `(1)/(2)`
B. `(1)/(4)`
C. Zero
D. `(1)/(8)`

Answer

To solve the problem, we need to find the value of \( k \) for which the equation \( |f(|x|)| = k \) has six solutions, where \( f(x) = x^2 - 3x + 2 \).

### Step-by-Step Solution:

1. **Identify the function**:
We start with the function \( f(x) = x^2 - 3x + 2 \).

2. **Find the roots of \( f(x) \)**:
We can factor the quadratic:
\[ f(x) = (x-1)(x-2) \]
The roots are \( x = 1 \) and \( x = 2 \).

3. **Graph the function \( f(x) \)**:
The graph of \( f(x) \) is a parabola that opens upwards. It intersects the x-axis at \( x = 1 \) and \( x = 2 \), and the vertex can be found using the formula \( x = -\frac{b}{2a} = \frac{3}{2} \). The corresponding y-coordinate at the vertex is:
\[ f\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) + 2 = \frac{9}{4} - \frac{9}{2} + 2 = \frac{9 - 18 + 8}{4} = -\frac{1}{4} \]

4. **Analyze \( |f(|x|)| \)**:
Since we are considering \( |x| \), we need to evaluate \( f \) for both positive and negative values of \( x \). The function \( f(|x|) \) will be the same for \( x \geq 0 \) and \( x < 0 \) due to the absolute value. Thus, we only need to consider \( f(x) \) for \( x \geq 0 \).

5. **Graph \( |f(x)| \)**:
The graph of \( f(x) \) will be reflected above the x-axis for the parts where \( f(x) < 0 \). The minimum point of \( f(x) \) is at \( x = \frac{3}{2} \) with \( f\left(\frac{3}{2}\right) = -\frac{1}{4} \). Therefore, the graph of \( |f(x)| \) will touch the x-axis at \( x = 1 \) and \( x = 2 \) and will rise above the x-axis elsewhere.

6. **Determine the value of \( k \)**:
For the equation \( |f(|x|)| = k \) to have six solutions, the horizontal line \( y = k \) must intersect the graph of \( |f(x)| \) at six points. This occurs when the line is below the maximum point of the graph and above the minimum point of the graph.

The minimum value of \( |f(x)| \) occurs at the vertex, which is \( \frac{1}{4} \). Thus, for \( k \) to yield six solutions, it must be:
\[ 0 < k < \frac{1}{4} \]

7. **Conclusion**:
The specific value of \( k \) that allows for six intersections is \( k = \frac{1}{4} \).

### Final Answer:
The value of \( k \) for which the equation \( |f(|x|)| = k \) has six solutions is:
\[ k = \frac{1}{4} \]