Number of roots ofequation `3^|x|-|2-|x||=1` is

Options

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

Answer

To solve the equation \( 3^{|x|} - |2 - |x|| = 1 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 3^{|x|} - |2 - |x|| = 1 \]
This can be rearranged to:
\[ 3^{|x|} = 1 + |2 - |x|| \]

### Step 2: Analyze the right-hand side
The expression \( |2 - |x|| \) can be broken down into two cases based on the value of \( |x| \):
1. When \( |x| \leq 2 \), \( |2 - |x|| = 2 - |x| \)
2. When \( |x| > 2 \), \( |2 - |x|| = |x| - 2 \)

### Step 3: Solve for each case

#### Case 1: \( |x| \leq 2 \)
In this case, we have:
\[ 3^{|x|} = 1 + (2 - |x|) \implies 3^{|x|} = 3 - |x| \]
Rearranging gives:
\[ 3^{|x|} + |x| - 3 = 0 \]

Let \( f(|x|) = 3^{|x|} + |x| - 3 \). We need to find the roots of this function for \( |x| \in [0, 2] \).

- At \( |x| = 0 \):
\[ f(0) = 3^0 + 0 - 3 = 1 - 3 = -2 \]
- At \( |x| = 2 \):
\[ f(2) = 3^2 + 2 - 3 = 9 + 2 - 3 = 8 \]

Since \( f(0) < 0 \) and \( f(2) > 0 \), by the Intermediate Value Theorem, there is at least one root in the interval \( (0, 2) \).

#### Case 2: \( |x| > 2 \)
In this case, we have:
\[ 3^{|x|} = 1 + (|x| - 2) \implies 3^{|x|} = |x| - 1 \]
Rearranging gives:
\[ 3^{|x|} - |x| + 1 = 0 \]

Let \( g(|x|) = 3^{|x|} - |x| + 1 \). We need to find the roots of this function for \( |x| > 2 \).

- At \( |x| = 2 \):
\[ g(2) = 3^2 - 2 + 1 = 9 - 2 + 1 = 8 \]
- As \( |x| \to \infty \), \( 3^{|x|} \) grows much faster than \( |x| \), so \( g(|x|) \to \infty \).

Since \( g(2) > 0 \) and \( g(|x|) \to \infty \) as \( |x| \) increases, we need to check if there is a root in the interval \( (2, \infty) \).

### Step 4: Conclusion
From the analysis:
- We found one root in the interval \( (0, 2) \).
- We found one root in the interval \( (2, \infty) \).

Thus, the total number of roots for the equation \( 3^{|x|} - |2 - |x|| = 1 \) is **2**.

### Final Answer:
The number of roots of the equation is **2**.