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. 0

Answer

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

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

### Step 1: Apply the 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 calculate the limit as \( x \) approaches 0 of \( f(x) \) when \( x \neq 0 \).

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

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

### Step 3: Use L'Hôpital's Rule
Since direct substitution gives us the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule. We differentiate the numerator and the denominator:

- The derivative of the numerator \( 1 - \cos(kx) \) is \( k \sin(kx) \).
- The derivative of the denominator \( x^2 \) is \( 2x \).

Thus, we have:

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

### Step 4: Apply L'Hôpital's Rule Again
This limit also results in the indeterminate form \( \frac{0}{0} \). We apply L'Hôpital's Rule again:

- The derivative of the numerator \( k \sin(kx) \) is \( k^2 \cos(kx) \).
- The derivative of the denominator \( 2x \) is \( 2 \).

Now we have:

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

### Step 5: Evaluate the Limit
As \( x \to 0 \), \( \cos(kx) \) approaches \( \cos(0) = 1 \). Therefore:

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

### Step 6: Set the Limit Equal to \( f(0) \)
Since \( f(0) = \frac{1}{2} \), we set the limit equal to \( f(0) \):

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

### Step 7: Solve for \( k \)
Multiplying both sides by 2 gives:

\[ k^2 = 1 \]

Taking the square root of both sides results in:

\[ k = \pm 1 \]

### Final Answer
Thus, the values of \( k \) are:

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