Step 1: Understanding the Concept:
A function f(x) is continuous at a point x = a if the limit of the function as x approaches a exists and is equal to the function's value at that point, i.e., \(\lim_{x \to a} f(x) = f(a)\).
Step 2: Key Formula or Approach:
We will apply the continuity condition at x = \(\frac{\pi}{2}\).
\[ \lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right) \]
The limit on the left side results in an indeterminate form (0/0), necessitating the use of L'Hôpital's Rule.
Step 3: Detailed Explanation:
Given that the function is continuous at x = \(\frac{\pi}{2}\), we have:
\[ \lim_{x \to \frac{\pi}{2}} \frac{k\cos x}{\pi - 2x} = f\left(\frac{\pi}{2}\right) \]
We are provided with \(f(\frac{\pi}{2}) = 3\).
Now, we evaluate the limit. As x approaches \(\frac{\pi}{2}\), \(\cos(x)\) approaches 0 and (\(\pi\) - 2x) approaches 0, resulting in the \(\frac{0}{0}\) indeterminate form.
Applying L'Hôpital's Rule, we differentiate the numerator and the denominator with respect to x:
\[ \text{Derivative of numerator:} \frac{d}{dx}(k\cos x) = -k\sin x \]
\[ \text{Derivative of denominator:} \frac{d}{dx}(\pi - 2x) = -2 \]
The limit then simplifies to:
\[ \lim_{x \to \frac{\pi}{2}} \frac{-k\sin x}{-2} = \lim_{x \to \frac{\pi}{2}} \frac{k\sin x}{2} \]
Substituting x = \(\frac{\pi}{2}\) into the simplified expression yields:
\[ \frac{k\sin(\frac{\pi}{2})}{2} = \frac{k(1)}{2} = \frac{k}{2} \]
For continuity, this limit must be equal to \(f(\frac{\pi}{2})\).
\[ \frac{k}{2} = 3 \]
\[ k = 6 \]
Step 4: Final Answer:
The value of k is 6.