$\text{If}f\left(x\right)=\{\begin{array}{cc}\frac{tan(\frac{\pi }{4}-x)}{cot2x}& x\ne \frac{\pi }{4}\\ k& x=\frac{\pi }{4}\end{array}\text{is continuous at}\mathrm{ x}=\frac{\pi }{4},\text{then the value of }k\text{ will be equal to}$

The correct option is 1/2.

We are given the function:

$$f\left(x\right)=\{\begin{array}{l}\frac{tan(\frac{\pi }{4}-x)}{cot2x},x\ne \frac{\pi }{4}\\ k,x=\frac{\pi }{4}\end{array}$$

For the function $f\left(x\right)$ to be continuous at $x=\frac{\pi }{4}$, the limit of $f\left(x\right)$ as $x$ approaches $\frac{\pi }{4}$ must exist and be equal to the value of the function at $x=\frac{\pi }{4}$. That is:

$$\underset{x\to \frac{\pi }{4}}{lim}f\left(x\right)=f\left(\frac{\pi }{4}\right)=k$$

Let us compute this limit:

$$L=\underset{x\to \frac{\pi }{4}}{lim}\frac{tan(\frac{\pi }{4}-x)}{cot2x}$$

To evaluate this limit, we can perform a substitution. Let $y=\frac{\pi }{4}-x$. As $x\to \frac{\pi }{4}$, we have $y\to 0$.
From this substitution, we can express $x$ in terms of $y$:

$$x=\frac{\pi }{4}-y$$

Substituting these variables into the denominator term, we get:

$$cot2x=cot\left(2\right(\frac{\pi }{4}-y\left)\right)=cot(\frac{\pi }{2}-2y)$$

Using the trigonometric co-function identity $cot(\frac{\pi }{2}-\theta )=tan\theta$, we obtain:

$$cot2x=tan2y$$

Now, we rewrite the limit in terms of $y$:

$$L=\underset{y\to 0}{lim}\frac{tany}{tan2y}$$

We can solve this limit using standard limit properties. We know that $\underset{\theta \to 0}{lim}\frac{tan\theta }{\theta }=1$. We manipulate the expression accordingly:

$$L=\underset{y\to 0}{lim}\left(\frac{tany}{y}\right)\cdot \left(\frac{2y}{tan2y}\right)\cdot \frac{y}{2y}$$

Simplifying the variables, we have:

$$L=\frac{1}{2}\cdot \left[\underset{y\to 0}{lim}\frac{tany}{y}\right]\cdot \left[\underset{y\to 0}{lim}\frac{2y}{tan2y}\right]$$

Applying the standard limits:

$$L=\frac{1}{2}\cdot 1\cdot 1=\frac{1}{2}$$

Since the function is continuous, we set the function value equal to the limit value:

$$k=L=\frac{1}{2}$$

Thus, the value of $k$ must be equal to 1/2.