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

Options

A. `pm 1`
B. `-1`
C. `pm (1)/(2)`
D. `(1)/(2)`

Answer

To determine the value of \( k \) (where \( k < 0 \)) for which the function

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

is continuous at \( x = 0 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches \( 0 \) equals \( f(0) = \frac{1}{2} \).

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

We need to calculate:

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

### Step 2: Use the limit properties

Using the Taylor series expansion, we know that:

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

And for \( \sin(x) \):

\[ \sin(x) \approx x \text{ as } x \to 0 \]

### Step 3: Substitute the approximations into the limit

Substituting these approximations into the limit gives:

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

### Step 4: Set the limit equal to \( f(0) \)

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

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

### Step 5: Solve for \( k^2 \)

Multiplying both sides by \( 2 \):

\[ k^2 = 1 \]

### Step 6: Find \( k \)

Taking the square root of both sides:

\[ k = \pm 1 \]

Since we are given that \( k < 0 \), we conclude that:

\[ k = -1 \]

### Conclusion

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

\[ \boxed{-1} \]