If `f(x)={{:((1-cospx)/(xsinx)",",x ne0),((1)/(2)",",x=0):}` is continuous at x=0 then p is equal to

Options

A. 2
B. `-2`
C. 1,-1
D. none of these

Answer

To determine the value of \( p \) for which the function

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

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

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

### Step 1: Find \( f(0) \)

From the definition of the function, we have:

\[ f(0) = \frac{1}{2}. \]

### Step 2: Calculate \( \lim_{x \to 0} f(x) \)

We need to evaluate the limit:

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

### Step 3: Apply L'Hôpital's Rule

Since both the numerator and denominator approach \( 0 \) as \( x \to 0 \) (i.e., \( 1 - \cos(px) \to 0 \) and \( x \sin x \to 0 \)), we can apply L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{1 - \cos(px)}{x \sin x} = \lim_{x \to 0} \frac{(p \sin(px))}{\sin x + x \cos x}. \]

### Step 4: Evaluate the limit again

Now we evaluate the limit again. As \( x \to 0 \):

- \( \sin(px) \to px \)
- \( \sin x \to x \)
- \( \cos x \to 1 \)

Thus, we have:

\[ \lim_{x \to 0} \frac{p \sin(px)}{\sin x + x \cos x} = \lim_{x \to 0} \frac{p(px)}{x + x} = \lim_{x \to 0} \frac{p p x}{2x} = \frac{p^2}{2}. \]

### Step 5: Set the limit equal to \( f(0) \)

For continuity at \( x = 0 \), we set:

\[ \frac{p^2}{2} = \frac{1}{2}. \]

### Step 6: Solve for \( p \)

Multiplying both sides by \( 2 \):

\[ p^2 = 1. \]

Taking the square root of both sides gives:

\[ p = 1 \quad \text{or} \quad p = -1. \]

### Conclusion

Since the problem does not specify the sign of \( p \), we can conclude that \( p \) can be either \( 1 \) or \( -1 \). However, the context of the options suggests we should consider only the positive value. Therefore, the value of \( p \) is:

\[ \boxed{1}. \]