The number of real solutions of the equation `sin(e^x)=2^x+2^(-x)` is

Options

A. 1
B. 0
C. 2
D. Infinite.

Answer

To solve the equation \( \sin(e^x) = 2^x + 2^{-x} \), we will analyze both sides of the equation step by step.

### Step 1: Analyze the right-hand side
The right-hand side of the equation is \( 2^x + 2^{-x} \). We can rewrite this expression using the properties of exponents:

\[ 2^{-x} = \frac{1}{2^x} \]

Thus, we can express the right-hand side as:

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

### Step 2: Find the minimum value of the right-hand side
To find the minimum value of \( 2^x + 2^{-x} \), we can use the AM-GM inequality:

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

This implies:

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

The equality holds when \( 2^x = 2^{-x} \), or equivalently when \( x = 0 \). Therefore, the minimum value of \( 2^x + 2^{-x} \) is 2.

### Step 3: Analyze the left-hand side
The left-hand side of the equation is \( \sin(e^x) \). The sine function oscillates between -1 and 1 for all real values of its argument. Therefore, we have:

\[ -1 \leq \sin(e^x) \leq 1 \]

### Step 4: Compare both sides
From the previous steps, we have established:

1. The minimum value of \( 2^x + 2^{-x} \) is 2.
2. The maximum value of \( \sin(e^x) \) is 1.

Since the minimum value of the right-hand side (2) is greater than the maximum value of the left-hand side (1), it follows that:

\[ \sin(e^x) < 2^x + 2^{-x} \quad \text{for all real } x \]

### Conclusion
Since there are no values of \( x \) for which \( \sin(e^x) \) can equal \( 2^x + 2^{-x} \), we conclude that the equation has no real solutions.

Thus, the final answer is:

**The number of real solutions of the equation \( \sin(e^x) = 2^x + 2^{-x} \) is 0.**

---