`e^(|sinx|)+e^(-|sinx|)+4a=0` will have exactly four different solutions in `[0,2pi]` if. `a in R` (b) `a in [-3/4,-1/4]` `a in [(-1-e^2)/(4e),oo]` (d) none of these

Options

A. `a in R`
B. `a in [- e/4, -1/4]`
C. `a in [(-1-e^(2))/(4e), oo]`
D. none of these

Explanation

To solve the equation \( e^{|\sin x|} + e^{-|sin x|} + 4a = 0 \) and determine the conditions on \( a \) for which it has exactly four different solutions in the interval \([0, 2\pi]\), we can follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ e^{|\sin x|} + e^{-|sin x|} + 4a = 0 \]
This can be rewritten as:
\[ e^{|\sin x|} + \frac{1}{e^{|\sin x|}} + 4a = 0 \]

### Step 2: Substitute \( t \)
Let \( t = e^{|\sin x|} \). Since \( |\sin x| \) varies between 0 and 1, \( t \) will vary between \( e^0 = 1 \) and \( e^1 = e \).

### Step 3: Transform the Equation
Substituting \( t \) into the equation gives:
\[ t + \frac{1}{t} + 4a = 0 \]
This simplifies to:
\[ t + \frac{1}{t} = -4a \]

### Step 4: Analyze the Function
Define the function:
\[ f(t) = t + \frac{1}{t} \]
We need to analyze this function for \( t \) in the interval \([1, e]\).

### Step 5: Find the Range of \( f(t) \)
To find the minimum and maximum values of \( f(t) \):
- The derivative \( f'(t) = 1 - \frac{1}{t^2} \).
- Setting \( f'(t) = 0 \) gives \( t^2 = 1 \) or \( t = 1 \) (since \( t \) is positive).
- Evaluate \( f(t) \) at the endpoints:
- \( f(1) = 1 + 1 = 2 \)
- \( f(e) = e + \frac{1}{e} \)

Calculating \( f(e) \):
\[ f(e) = e + \frac{1}{e} \approx 2.718 + 0.368 = 3.086 \]
Thus, the range of \( f(t) \) for \( t \in [1, e] \) is:
\[ [2, e + \frac{1}{e}] \]

### Step 6: Set Up the Inequality
For the equation \( -4a \) to fall within the range of \( f(t) \):
\[ -4a \in [2, e + \frac{1}{e}] \]
This leads to:
\[ 2 \leq -4a \leq e + \frac{1}{e} \]

### Step 7: Solve for \( a \)
From \( -4a \geq 2 \):
\[ a \leq -\frac{1}{2} \]
From \( -4a \leq e + \frac{1}{e} \):
\[ a \geq -\frac{e + \frac{1}{e}}{4} \]

### Step 8: Final Range for \( a \)
Thus, we have:
\[ -\frac{e + \frac{1}{e}}{4} \leq a \leq -\frac{1}{2} \]

### Step 9: Check the Options
Now we check the provided options:
- (a) \( a \in \mathbb{R} \) - Incorrect.
- (b) \( a \in \left[-\frac{3}{4}, -\frac{1}{4}\right] \) - Incorrect.
- (c) \( a \in \left[-\frac{1 + e^2}{4e}, \infty\right) \) - Incorrect.
- (d) None of these - Correct.

### Conclusion
The correct answer is:
**(d) none of these.**