25. If the function $f\left(x\right)=\{\begin{array}{ll}\frac{k\cos x}{\pi -2x}& ,\; x\; \ne \; \pi /2\\ 3& ,\; x\; =\; \pi /2\end{array}$

$\begin{array}{l}\end{array}$
is continuous at $x=\frac{\pi }{2}$ , then k is equal to:

The correct option is 6 (which corresponds to the first option).

To find the value of $k$ for which the function $f\left(x\right)$ is continuous at $x=\frac{\pi }{2}$, we recall the definition of continuity.
A function $f\left(x\right)$ is continuous at a point $x=a$ if the limit of the function as $x$ approaches $a$ exists and is equal to the value of the function at $a$.
Therefore, for continuity at $x=\frac{\pi }{2}$, we must have:

$$\underset{x\to \frac{\pi }{2}}{lim}f\left(x\right)=f\left(\frac{\pi }{2}\right)$$

From the definition of the function, we are given that:
$f\left(\frac{\pi }{2}\right)=3$
So, we need to evaluate the limit:

$$\underset{x\to \frac{\pi }{2}}{lim}\frac{kcosx}{\pi -2x}=3$$

To evaluate this limit, let us introduce a substitution. Let:
$y=\frac{\pi }{2}-x$
This implies that:
$x=\frac{\pi }{2}-y$
As $x\to \frac{\pi }{2}$, we have $y\to 0$.

Now, substitute these expressions into the limit:
The numerator becomes:
$kcosx=kcos\left(\frac{\pi }{2}-y\right)=ksiny$ (since $cos\left(\frac{\pi }{2}-y\right)=siny$)
The denominator becomes:
$\pi -2x=\pi -2\left(\frac{\pi }{2}-y\right)=\pi -\pi +2y=2y$

Substituting these back into the limit expression gives:

$$\underset{y\to 0}{lim}\frac{ksiny}{2y}=3$$

We can factor out the constant terms from the limit:

$$\frac{k}{2}\left(\underset{y\to 0}{lim}\frac{siny}{y}\right)=3$$

Using the standard trigonometric limit, we know that:
$\underset{y\to 0}{lim}\frac{siny}{y}=1$
Substituting this value into the equation yields:

$$\frac{k}{2}·1=3$$

Solving for $k$:
$k=3·2=6$

Thus, the function is continuous at $x=\frac{\pi }{2}$ when $k=6$.