The value of k for which `f(x)= {((1-cos 2x )/(x^2 )", " x ne 0 ),(k ", "x=0):}`
continuous at x=0, is :

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the value of \( k \) for which the function

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

is continuous at \( x = 0 \), we need to ensure that

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

### Step 1: Calculate the limit as \( x \) approaches 0.

We need to find

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

### Step 2: Use the trigonometric identity.

We can use the identity \( 1 - \cos(2x) = 2\sin^2(x) \):

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

### Step 3: Simplify the limit.

We can rewrite the limit as:

\[ \lim_{x \to 0} 2 \cdot \frac{\sin^2(x)}{x^2}. \]

### Step 4: Apply the limit property.

We know that

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

Thus,

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

### Step 5: Substitute back into the limit.

Now we can substitute this back into our limit:

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

### Step 6: Set the limit equal to \( k \).

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

\[ k = \lim_{x \to 0} f(x) = 2. \]

### Conclusion:

Thus, the value of \( k \) for which the function is continuous at \( x = 0 \) is

\[ \boxed{2}. \]

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