Find the value of the constant k so that the function given below is continuous at `x=0` ,`f(x)={{:(,(1-cos2x)/(2x^2)(x ne 0) ,kx=0 )` .

Answer

To find the value of the constant \( k \) so that the function

\[ f(x) = \begin{cases} \frac{1 - \cos(2x)}{2x^2} & \text{if } x \neq 0 \\ k & \text{if } x = 0 \end{cases} \]

is continuous at \( x = 0 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 0 is equal to \( f(0) = k \).

### Step 1: Find the limit of \( f(x) \) as \( x \) approaches 0

We need to calculate:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{1 - \cos(2x)}{2x^2} \]

### Step 2: Use the trigonometric identity

Using the identity \( 1 - \cos(2x) = 2 \sin^2(x) \), we can rewrite the limit:

\[ \lim_{x \to 0} \frac{1 - \cos(2x)}{2x^2} = \lim_{x \to 0} \frac{2 \sin^2(x)}{2x^2} = \lim_{x \to 0} \frac{\sin^2(x)}{x^2} \]

### Step 3: Simplify the limit

The expression simplifies to:

\[ \lim_{x \to 0} \left(\frac{\sin(x)}{x}\right)^2 \]

### Step 4: Evaluate the limit

We know that:

\[ \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \]

Thus:

\[ \lim_{x \to 0} \left(\frac{\sin(x)}{x}\right)^2 = 1^2 = 1 \]

### Step 5: Set the limit equal to \( k \)

For the function to be continuous at \( x = 0 \), we need:

\[ k = \lim_{x \to 0} f(x) = 1 \]

### Final Answer

The value of the constant \( k \) is:

\[ \boxed{1} \]