For the function `f(x) = e^(cos x)`, Rolle's theorem is applicable when

Options

A. `pi/2 le x le (3pi)/2`
B. `-pi/2 le x le pi/2`
C. `0 le x le 2pi`
D. `-pi/4 le x le pi/4`

Answer

To determine when Rolle's theorem is applicable for the function \( f(x) = e^{\cos x} \), we need to check the three conditions of Rolle's theorem:

1. The function \( f(x) \) must be continuous on the closed interval \([a, b]\).
2. The function \( f(x) \) must be differentiable on the open interval \((a, b)\).
3. The function must take the same value at the endpoints, i.e., \( f(a) = f(b) \).

### Step-by-Step Solution:

**Step 1: Identify the Function**
The given function is \( f(x) = e^{\cos x} \).

**Step 2: Check Continuity and Differentiability**
- The function \( e^{\cos x} \) is a composition of continuous functions (exponential and cosine functions). Therefore, it is continuous for all \( x \).
- Similarly, since \( e^{\cos x} \) is differentiable everywhere (as both \( e^x \) and \( \cos x \) are differentiable), it is differentiable for all \( x \).

**Step 3: Choose Intervals and Check Endpoint Values**
We need to check different intervals to see if \( f(a) = f(b) \) holds true.

**Option 1: Interval \([0, \frac{\pi}{2}]\)**
- Calculate \( f(0) \):
\[ f(0) = e^{\cos(0)} = e^1 = e \]
- Calculate \( f(\frac{\pi}{2}) \):
\[ f\left(\frac{\pi}{2}\right) = e^{\cos\left(\frac{\pi}{2}\right)} = e^0 = 1 \]
- Since \( f(0) \neq f\left(\frac{\pi}{2}\right) \), this interval does not satisfy the third condition.

**Option 2: Interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\)**
- Calculate \( f(-\frac{\pi}{2}) \):
\[ f\left(-\frac{\pi}{2}\right) = e^{\cos\left(-\frac{\pi}{2}\right)} = e^0 = 1 \]
- Calculate \( f\left(\frac{\pi}{2}\right) \):
\[ f\left(\frac{\pi}{2}\right) = e^{\cos\left(\frac{\pi}{2}\right)} = e^0 = 1 \]
- Since \( f(-\frac{\pi}{2}) = f\left(\frac{\pi}{2}\right) \), this interval satisfies all conditions.

**Option 3: Interval \([0, 2\pi]\)**
- Calculate \( f(0) \):
\[ f(0) = e^{\cos(0)} = e^1 = e \]
- Calculate \( f(2\pi) \):
\[ f(2\pi) = e^{\cos(2\pi)} = e^1 = e \]
- Since \( f(0) = f(2\pi) \), this interval satisfies all conditions.

**Option 4: Interval \([- \frac{\pi}{4}, \frac{\pi}{4}]\)**
- Calculate \( f(-\frac{\pi}{4}) \):
\[ f\left(-\frac{\pi}{4}\right) = e^{\cos\left(-\frac{\pi}{4}\right)} = e^{\frac{1}{\sqrt{2}}} \]
- Calculate \( f\left(\frac{\pi}{4}\right) \):
\[ f\left(\frac{\pi}{4}\right) = e^{\cos\left(\frac{\pi}{4}\right)} = e^{\frac{1}{\sqrt{2}}} \]
- Since \( f(-\frac{\pi}{4}) = f\left(\frac{\pi}{4}\right) \), this interval satisfies all conditions.

### Conclusion
Rolle's theorem is applicable for the intervals:
- Option 2: \([- \frac{\pi}{2}, \frac{\pi}{2}]\)
- Option 3: \([0, 2\pi]\)
- Option 4: \([- \frac{\pi}{4}, \frac{\pi}{4}]\)