The complete set of values of x satisfying the equation `x^2. 2^(x+1)+2^(|x-3|+2)=x^2. 2^(|x-3|+4)+2^(x-1)` is

Options

A. `[3,oo)`
B. `[-(1)/(2),(1)/(2)]uu(3,oo)`
C. `(-oo-(1)/(2))`
D. `{-(1)/(2),(1)/(2)}uu[3,oo)`

Answer

To solve the equation \( x^2 \cdot 2^{(x+1)} + 2^{|x-3| + 2} = x^2 \cdot 2^{|x-3| + 4} + 2^{(x-1)} \), we will analyze two cases based on the value of \( x \) in relation to 3, due to the absolute value function \( |x-3| \).

### Step 1: Case 1 - \( x \geq 3 \)

In this case, \( |x-3| = x-3 \).

Substituting this into the equation:

\[ x^2 \cdot 2^{(x+1)} + 2^{(x-3) + 2} = x^2 \cdot 2^{(x-3) + 4} + 2^{(x-1)} \]

This simplifies to:

\[ x^2 \cdot 2^{(x+1)} + 2^{(x-1)} = x^2 \cdot 2^{(x+1)} + 2^{(x-1)} \]

Both sides are equal, which means the equation holds true for all \( x \geq 3 \).

**Hint for Step 1:** Check how the absolute value affects the equation based on the value of \( x \).

### Step 2: Case 2 - \( x < 3 \)

In this case, \( |x-3| = -(x-3) = 3-x \).

Substituting this into the equation:

\[ x^2 \cdot 2^{(x+1)} + 2^{(3-x) + 2} = x^2 \cdot 2^{(3-x) + 4} + 2^{(x-1)} \]

This simplifies to:

\[ x^2 \cdot 2^{(x+1)} + 2^{(5-x)} = x^2 \cdot 2^{(5-x)} + 2^{(x-1)} \]

Rearranging gives:

\[ x^2 \cdot 2^{(x+1)} - x^2 \cdot 2^{(5-x)} = 2^{(x-1)} - 2^{(5-x)} \]

Factoring out \( x^2 \):

\[ x^2 (2^{(x+1)} - 2^{(5-x)}) = 2^{(x-1)} - 2^{(5-x)} \]

### Step 3: Solve for \( x \)

To solve the equation, we can analyze the left and right sides separately.

1. Set \( y = 2^x \), then rewrite the equation in terms of \( y \).
2. Solve for \( x \) by substituting back \( y = 2^x \).

After simplification, we find:

\[ x^2 = \frac{1}{4} \implies x = \pm \frac{1}{2} \]

Since we are in the case where \( x < 3 \), both \( x = \frac{1}{2} \) and \( x = -\frac{1}{2} \) are valid solutions.

### Step 4: Combine Solutions

From both cases, we have:

- From Case 1: \( x \in [3, \infty) \)
- From Case 2: \( x \in \left[-\frac{1}{2}, \frac{1}{2}\right) \)

Thus, the complete set of values of \( x \) satisfying the equation is:

\[ x \in \left[-\frac{1}{2}, \frac{1}{2}\right) \cup [3, \infty) \]

**Final Answer:** \( x \in \left[-\frac{1}{2}, \frac{1}{2}\right) \cup [3, \infty) \)