Let `I_(1)=int_(0)^(alpha)(1+2cosx)/(1+e^(x))dx` and `I_(2)=int_(0)^(alpha)(1+e^(x))/(1+2cosx)dx`, where `alpha` is the root of the equation `2 cos x - e^(x)=0`. and `alpha` is positive Then,

Options

A. `I_(1)=I_(2)`
B. `I_(1) gt I_(2)`
C. `I_(1)+I_(2)=0`
D. `I_(1) ltI_(2)`

Answer

To solve the problem, we need to analyze the integrals \( I_1 \) and \( I_2 \) given by:

\[ I_1 = \int_0^{\alpha} \frac{1 + 2 \cos x}{1 + e^x} \, dx \]

\[ I_2 = \int_0^{\alpha} \frac{1 + e^x}{1 + 2 \cos x} \, dx \]

where \( \alpha \) is the positive root of the equation \( 2 \cos x - e^x = 0 \).

### Step 1: Find the value of \( \alpha \)

To find \( \alpha \), we need to solve the equation:

\[ 2 \cos x = e^x \]

This equation represents the points where the curve \( y = 2 \cos x \) intersects the curve \( y = e^x \). We can analyze the behavior of both functions to find the positive root.

- At \( x = 0 \):
\[ 2 \cos(0) = 2 \quad \text{and} \quad e^0 = 1 \]
So, \( 2 > 1 \).

- As \( x \) increases, \( e^x \) grows exponentially while \( 2 \cos x \) oscillates between -2 and 2. The first intersection after \( x = 0 \) occurs when \( 2 \cos x \) decreases below \( e^x \).

By plotting or calculating values, we can find that \( \alpha \) is approximately \( 1.0 \) (the exact value can be determined numerically).

### Step 2: Compare \( I_1 \) and \( I_2 \)

Next, we need to compare the two integrals \( I_1 \) and \( I_2 \).

From the inequality derived from the properties of the functions involved:

\[ 1 + 2 \cos x > 1 + e^x \quad \text{for} \quad x \in [0, \alpha] \]

Adding 1 to both sides gives:

\[ 1 + 2 \cos x > 1 + e^x \]

Dividing both sides by \( 1 + e^x \) (which is positive for \( x \geq 0 \)) leads to:

\[ \frac{1 + 2 \cos x}{1 + e^x} > 1 \]

This implies:

\[ \frac{1 + 2 \cos x}{1 + e^x} > \frac{1 + e^x}{1 + 2 \cos x} \]

Thus, we have:

\[ I_1 > I_2 \]

### Conclusion

From the analysis, we conclude that:

\[ I_1 > I_2 \]