The number of real solution of equation `e^(6x)+e^(4x)+2e^(3x)+12e^(2x)+e^(x)-1=0`

Options

A. 0
B. 1
C. 6
D. 8

Answer

To find the number of real solutions of the equation

\[ e^{6x} + e^{4x} + 2e^{3x} + 12e^{2x} + e^{x} - 1 = 0, \]

we can define a function \( f(x) \) as follows:

\[ f(x) = e^{6x} + e^{4x} + 2e^{3x} + 12e^{2x} + e^{x} - 1. \]

### Step 1: Differentiate the function

We will differentiate \( f(x) \) to analyze its behavior:

\[ f'(x) = \frac{d}{dx}(e^{6x}) + \frac{d}{dx}(e^{4x}) + \frac{d}{dx}(2e^{3x}) + \frac{d}{dx}(12e^{2x}) + \frac{d}{dx}(e^{x}). \]

Using the derivative of \( e^{kx} \) which is \( ke^{kx} \):

\[ f'(x) = 6e^{6x} + 4e^{4x} + 6e^{3x} + 24e^{2x} + e^{x}. \]

### Step 2: Analyze the derivative

Notice that all terms in \( f'(x) \) are positive for all \( x \):

- \( 6e^{6x} > 0 \)
- \( 4e^{4x} > 0 \)
- \( 6e^{3x} > 0 \)
- \( 24e^{2x} > 0 \)
- \( e^{x} > 0 \)

Thus, we conclude that:

\[ f'(x) > 0 \quad \text{for all } x \in \mathbb{R}. \]

### Step 3: Determine the nature of the function

Since \( f'(x) > 0 \), the function \( f(x) \) is strictly increasing. This means that \( f(x) \) can cross the x-axis at most once.

### Step 4: Evaluate the function at specific points

Now, we will evaluate \( f(x) \) at two points to determine the sign of the function:

1. **At \( x = 0 \)**:

\[ f(0) = e^{0} + e^{0} + 2e^{0} + 12e^{0} + e^{0} - 1 = 1 + 1 + 2 + 12 + 1 - 1 = 16 > 0. \]

2. **As \( x \to -\infty \)**:

As \( x \) approaches \(-\infty\), all terms involving \( e^{kx} \) (where \( k > 0 \)) approach 0:

\[ f(-\infty) = 0 + 0 + 0 + 0 + 0 - 1 = -1 < 0. \]

### Step 5: Conclusion

Since \( f(-\infty) < 0 \) and \( f(0) > 0 \), and \( f(x) \) is strictly increasing, it must cross the x-axis exactly once. Therefore, there is exactly one real solution to the equation.

Thus, the number of real solutions of the equation

\[ e^{6x} + e^{4x} + 2e^{3x} + 12e^{2x} + e^{x} - 1 = 0 \]

is

\[ \boxed{1}. \]