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

Options

A. 0
B. 1
C. 2
D. infinitely many

Explanation

To find the number of real solutions of 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 \( 2^x + 2^{-x} \)

The expression \( 2^x + 2^{-x} \) can be rewritten using the property of exponents.

**Hint:** Recall that \( 2^{-x} = \frac{1}{2^x} \).

### Step 2: Apply the AM-GM inequality

Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we can state that:

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

This implies:

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

**Hint:** Remember that the AM-GM inequality states that the average of non-negative numbers is greater than or equal to the geometric mean.

### Step 3: Analyze the left-hand side \( \sin(e^x) \)

The function \( \sin(e^x) \) oscillates between -1 and 1 for all real values of \( x \). Therefore, we have:

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

**Hint:** Recall the range of the sine function, which is between -1 and 1.

### Step 4: Compare both sides

From our analysis, we have:

- The right-hand side \( 2^x + 2^{-x} \geq 2 \)
- The left-hand side \( \sin(e^x) \leq 1 \)

Since the smallest value of \( 2^x + 2^{-x} \) is 2 and the maximum value of \( \sin(e^x) \) is 1, we can conclude that:

\[ \sin(e^x) < 2^x + 2^{-x} \text{ for all } x \in \mathbb{R} \]

**Hint:** Consider the implications of the ranges of both functions.

### Step 5: Conclusion

Since \( \sin(e^x) \) cannot equal \( 2^x + 2^{-x} \) for any real number \( x \), we conclude that the equation has no real solutions.

Thus, the number of real solutions of the equation \( \sin(e^x) = 2^x + 2^{-x} \) is:

\[ \boxed{0} \]