If `f(x)` is continuous at `x=0`, where `f(x)={:{((1-cos k x)/(x^(2))", for " x!=0),(1/2 ", for " x=0):}`, then `k=`

Options

A. 1
B. -1
C. `pm1`
D. None of these

Answer

To solve the problem, we need to determine the value of \( k \) such that the function \( f(x) \) is continuous at \( x = 0 \). The function is defined as follows:

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

### Step 1: Establish Continuity Condition
For \( f(x) \) to be continuous at \( x = 0 \), we need to ensure that:

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

This means we need to find \( \lim_{x \to 0} f(x) \) when \( x \neq 0 \) and set it equal to \( \frac{1}{2} \).

### Step 2: Calculate the Limit
We need to evaluate:

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

Using the trigonometric identity \( 1 - \cos(\theta) = 2 \sin^2\left(\frac{\theta}{2}\right) \), we can rewrite the limit as:

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

### Step 3: Simplify the Limit
We can express \( \sin\left(\frac{kx}{2}\right) \) in terms of \( x \):

\[ \lim_{x \to 0} \frac{2 \left(\frac{kx}{2}\right)^2}{x^2} \cdot \frac{1}{\left(\frac{k}{2}\right)^2} = \lim_{x \to 0} \frac{2 \cdot \frac{k^2 x^2}{4}}{x^2} = \lim_{x \to 0} \frac{k^2}{2} \]

### Step 4: Set the Limit Equal to \( f(0) \)
Now we set the limit equal to \( f(0) \):

\[ \frac{k^2}{2} = \frac{1}{2} \]

### Step 5: Solve for \( k \)
To solve for \( k \), we multiply both sides by 2:

\[ k^2 = 1 \]

Taking the square root of both sides gives:

\[ k = \pm 1 \]

### Conclusion
Thus, the values of \( k \) that make \( f(x) \) continuous at \( x = 0 \) are:

\[ k = 1 \quad \text{or} \quad k = -1 \]