The solution of `2^(x)+2^(|x|)ge 2 sqrt2` is

Options

A. `(-oo,log_(2)(sqrt2+1)`
B. `(0,oo)`
C. `((1)/(2),log_(2)(sqrt2-1))`
D. `(-oolog_(2)(sqrt2-1)]cup[(1)/(2),oo)`

Answer

To solve the inequality \(2^x + 2^{|x|} \geq 2\sqrt{2}\), we will consider two cases based on the definition of the absolute value function.

### Step 1: Case 1 - \(x \geq 0\)

In this case, \(|x| = x\). Therefore, the inequality becomes:

\[ 2^x + 2^x \geq 2\sqrt{2} \]

This simplifies to:

\[ 2 \cdot 2^x \geq 2\sqrt{2} \]

Dividing both sides by 2:

\[ 2^x \geq \sqrt{2} \]

We can express \(\sqrt{2}\) as \(2^{1/2}\):

\[ 2^x \geq 2^{1/2} \]

Taking logarithm base 2 of both sides:

\[ x \geq \frac{1}{2} \]

### Step 2: Case 2 - \(x < 0\)

In this case, \(|x| = -x\). Therefore, the inequality becomes:

\[ 2^x + 2^{-x} \geq 2\sqrt{2} \]

We can rewrite \(2^{-x}\) as \(\frac{1}{2^x}\). Thus, the inequality becomes:

\[ 2^x + \frac{1}{2^x} \geq 2\sqrt{2} \]

Let \(t = 2^x\) (note that \(t > 0\) since \(x < 0\)). The inequality can now be expressed as:

\[ t + \frac{1}{t} \geq 2\sqrt{2} \]

Multiplying through by \(t\) (which is positive):

\[ t^2 + 1 \geq 2\sqrt{2}t \]

Rearranging gives us a quadratic inequality:

\[ t^2 - 2\sqrt{2}t + 1 \geq 0 \]

### Step 3: Solve the Quadratic Inequality

To solve \(t^2 - 2\sqrt{2}t + 1 \geq 0\), we first find the roots using the quadratic formula:

\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2\sqrt{2} \pm \sqrt{(2\sqrt{2})^2 - 4 \cdot 1}}{2} \]

Calculating the discriminant:

\[ (2\sqrt{2})^2 - 4 = 8 - 4 = 4 \]

Thus, the roots are:

\[ t = \frac{2\sqrt{2} \pm 2}{2} = \sqrt{2} + 1 \quad \text{and} \quad \sqrt{2} - 1 \]

### Step 4: Analyze the Quadratic

The quadratic \(t^2 - 2\sqrt{2}t + 1\) opens upwards (since the coefficient of \(t^2\) is positive). Therefore, the inequality \(t^2 - 2\sqrt{2}t + 1 \geq 0\) holds outside the roots:

\[ t \leq \sqrt{2} - 1 \quad \text{or} \quad t \geq \sqrt{2} + 1 \]

### Step 5: Convert Back to \(x\)

Since \(t = 2^x\), we have:

1. \(2^x \leq \sqrt{2} - 1\) (this case is not valid since \(2^x\) is always positive and \(\sqrt{2} - 1 < 0\))
2. \(2^x \geq \sqrt{2} + 1\)

Taking logarithm base 2:

\[ x \geq \log_2(\sqrt{2} + 1) \]

### Step 6: Combine Results

From Case 1, we have \(x \geq \frac{1}{2}\). From Case 2, we have \(x \geq \log_2(\sqrt{2} + 1)\).

Now we need to compare \(\frac{1}{2}\) and \(\log_2(\sqrt{2} + 1)\):

Calculating \(\log_2(\sqrt{2} + 1)\):

\[ \sqrt{2} + 1 \approx 2.414 \quad \Rightarrow \quad \log_2(2.414) \approx 1.27 \]

Thus, \(\frac{1}{2} < \log_2(\sqrt{2} + 1)\).

### Final Solution

Combining both cases, the solution set is:

\[ x \in \left[\frac{1}{2}, +\infty\right) \cup \left[\log_2(\sqrt{2} + 1), +\infty\right) \]