Let `f(x)={:{((kcosx)/(pi-2x)',xne(pi)/(2)),(3",",x=(pi)/(2).):}`
If `lim_(xto(pi)/(2))f(x)=f((pi)/(2)),` find the value of k.

Answer

To solve the problem, we need to find the value of \( k \) such that

\[ \lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right). \]

Given the function:

\[ f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x} & \text{if } x \neq \frac{\pi}{2} \\ 3 & \text{if } x = \frac{\pi}{2} \end{cases} \]

### Step 1: Find \( f\left(\frac{\pi}{2}\right) \)

Since \( f\left(\frac{\pi}{2}\right) = 3 \), we have:

\[ f\left(\frac{\pi}{2}\right) = 3. \]

### Step 2: Calculate the limit \( \lim_{x \to \frac{\pi}{2}} f(x) \)

We need to calculate the limit as \( x \) approaches \( \frac{\pi}{2} \) from both sides. We will use the expression for \( f(x) \) when \( x \neq \frac{\pi}{2} \):

\[ \lim_{x \to \frac{\pi}{2}} f(x) = \lim_{x \to \frac{\pi}{2}} \frac{k \cos x}{\pi - 2x}. \]

### Step 3: Substitute \( x = \frac{\pi}{2} + h \)

Let \( x = \frac{\pi}{2} + h \), where \( h \to 0 \). Then we have:

\[ \lim_{h \to 0} f\left(\frac{\pi}{2} + h\right) = \lim_{h \to 0} \frac{k \cos\left(\frac{\pi}{2} + h\right)}{\pi - 2\left(\frac{\pi}{2} + h\right)}. \]

### Step 4: Simplify the limit expression

Using the fact that \( \cos\left(\frac{\pi}{2} + h\right) = -\sin(h) \) and simplifying the denominator:

\[ \pi - 2\left(\frac{\pi}{2} + h\right) = \pi - \pi - 2h = -2h. \]

Thus, we have:

\[ \lim_{h \to 0} \frac{k (-\sin(h))}{-2h} = \lim_{h \to 0} \frac{k \sin(h)}{2h}. \]

### Step 5: Evaluate the limit

Using the fact that \( \lim_{h \to 0} \frac{\sin(h)}{h} = 1 \):

\[ \lim_{h \to 0} \frac{k \sin(h)}{2h} = \frac{k}{2}. \]

### Step 6: Set the limits equal

Now, we set the limit equal to \( f\left(\frac{\pi}{2}\right) \):

\[ \frac{k}{2} = 3. \]

### Step 7: Solve for \( k \)

Multiplying both sides by 2, we find:

\[ k = 6. \]

### Conclusion

Thus, the value of \( k \) is

\[ \boxed{6}. \]