Consider the function f(x) = `{{:( x^(2) - 1"," , -1 le x le 1) , ( lnx "," , 1 lt x le e):}`
Let `f_(1) (x) = f (|x|)`
`f_(2) (x) = |f(|x|)|`
`f_(3) (x) = f (-x)`
Now answer the question
Number of integral solution of the equation `f_(1) (x) = f_(2) (x) ` is (are )

Options

A. 4
B. 3
C. 2
D. 1

Answer

To solve the equation \( f_1(x) = f_2(x) \), we first need to define the functions \( f_1(x) \) and \( f_2(x) \) based on the given function \( f(x) \).

### Step 1: Define the function \( f(x) \)
The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} x^2 - 1 & \text{for } -1 \leq x \leq 1 \\ \ln x & \text{for } 1 < x \leq e \end{cases} \]

### Step 2: Define \( f_1(x) \)
The function \( f_1(x) = f(|x|) \) means we evaluate \( f \) at the absolute value of \( x \). Thus, we have:
\[ f_1(x) = \begin{cases} x^2 - 1 & \text{for } 0 \leq |x| \leq 1 \\ \ln |x| & \text{for } 1 < |x| \leq e \end{cases} \]

### Step 3: Define \( f_2(x) \)
The function \( f_2(x) = |f(|x|)| \) means we take the absolute value of \( f \) evaluated at the absolute value of \( x \). Thus, we have:
\[ f_2(x) = \begin{cases} |x^2 - 1| & \text{for } 0 \leq |x| \leq 1 \\ |\ln |x|| & \text{for } 1 < |x| \leq e \end{cases} \]

### Step 4: Analyze \( f_1(x) \) and \( f_2(x) \)
1. **For \( 0 \leq |x| \leq 1 \)**:
- \( f_1(x) = x^2 - 1 \)
- \( f_2(x) = |x^2 - 1| \)
- Since \( x^2 - 1 \) is negative in this interval, \( f_2(x) = 1 - x^2 \).

Therefore, in this interval:
\[ f_1(x) = x^2 - 1 \quad \text{and} \quad f_2(x) = 1 - x^2 \]
Setting them equal gives:
\[ x^2 - 1 = 1 - x^2 \implies 2x^2 = 2 \implies x^2 = 1 \implies x = \pm 1 \]

2. **For \( 1 < |x| \leq e \)**:
- \( f_1(x) = \ln |x| \)
- \( f_2(x) = |\ln |x|| = \ln |x| \) (since \(|x|\) is positive in this range).

Therefore, in this interval:
\[ f_1(x) = \ln |x| \quad \text{and} \quad f_2(x) = \ln |x| \]
Setting them equal gives:
\[ \ln |x| = \ln |x| \quad \text{(always true)} \]
This means any \( x \) in the interval \( (1, e] \) satisfies this equation.

### Step 5: Count the integral solutions
- From the interval \( 0 \leq |x| \leq 1 \), we found the solutions \( x = 1 \) and \( x = -1 \).
- From the interval \( 1 < |x| \leq e \), the integral solutions are \( x = 2, 3, 4, 5 \) (as \( e \approx 2.718 \), we consider integers 2).

Thus, the total integral solutions are:
- From \( 0 \leq |x| \leq 1 \): \( 2 \) solutions (\( x = 1, -1 \))
- From \( 1 < |x| \leq e \): \( 1 \) solution (\( x = 2 \))

### Final Count
Total integral solutions = \( 2 + 1 = 3 \).

### Conclusion
The number of integral solutions of the equation \( f_1(x) = f_2(x) \) is \( 3 \).