If `lim_(x to (pi)/(2))(2^(-cosx)-1)/(x(x-(pi)/(2)))=(2)/(pi)logk`, `k gt 0` , the k is equal to

Options

A. `2`
B. `4`
C. `(1)/(2)`
D. none of these

Answer

To solve the limit problem given in the question, we will follow these steps:

### Step 1: Identify the limit expression
We need to evaluate the limit:
\[ \lim_{x \to \frac{\pi}{2}} \frac{2^{-\cos x} - 1}{x \left(x - \frac{\pi}{2}\right)} \]

### Step 2: Substitute \(x = \frac{\pi}{2}\)
Substituting \(x = \frac{\pi}{2}\) into the expression:
- \(\cos\left(\frac{\pi}{2}\right) = 0\)
- Thus, \(2^{-\cos\left(\frac{\pi}{2}\right)} = 2^0 = 1\)
- Therefore, the numerator becomes \(1 - 1 = 0\).
- The denominator becomes \(\frac{\pi}{2} \left(\frac{\pi}{2} - \frac{\pi}{2}\right) = \frac{\pi}{2} \cdot 0 = 0\).

This gives us a \(0/0\) indeterminate form.

### Step 3: Apply L'Hôpital's Rule
Since we have a \(0/0\) form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator separately.

#### Derivative of the numerator:
The numerator is \(2^{-\cos x} - 1\).
Using the chain rule:
\[ \frac{d}{dx}(2^{-\cos x}) = 2^{-\cos x} \cdot \ln(2) \cdot \sin x \]
Thus, the derivative of the numerator is:
\[ 2^{-\cos x} \ln(2) \sin x \]

#### Derivative of the denominator:
The denominator is \(x\left(x - \frac{\pi}{2}\right)\).
Using the product rule:
\[ \frac{d}{dx}(x(x - \frac{\pi}{2})) = (x - \frac{\pi}{2}) + x \cdot 1 = 2x - \frac{\pi}{2} \]

### Step 4: Rewrite the limit
Now, we rewrite the limit using the derivatives:
\[ \lim_{x \to \frac{\pi}{2}} \frac{2^{-\cos x} \ln(2) \sin x}{2x - \frac{\pi}{2}} \]

### Step 5: Substitute \(x = \frac{\pi}{2}\) again
Substituting \(x = \frac{\pi}{2}\):
- \(2^{-\cos\left(\frac{\pi}{2}\right)} = 1\)
- \(\sin\left(\frac{\pi}{2}\right) = 1\)
- The denominator becomes \(2\left(\frac{\pi}{2}\right) - \frac{\pi}{2} = \frac{\pi}{2}\).

Thus, the limit becomes:
\[ \frac{1 \cdot \ln(2) \cdot 1}{\frac{\pi}{2}} = \frac{2 \ln(2)}{\pi} \]

### Step 6: Set the limit equal to the given expression
According to the problem, we have:
\[ \frac{2 \ln(2)}{\pi} = \frac{2}{\pi} \log k \]

### Step 7: Solve for \(k\)
By comparing both sides:
\[ \ln(2) = \log k \]
This implies:
\[ k = 2 \]

### Final Answer
Thus, the value of \(k\) is:
\[ \boxed{2} \]