`f(x)={((sqrt(1+px)-sqrt(1-px))/(x),,,-1 le x lt 0),((2x+1)/(x-2),,,0le x le 1):}`
is continuous in the interval [-1,1], then p equals :

Options

A. `-1`
B. `-(1)/(2)`
C. `(1)/(2)`
D. 4

Explanation

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

\[ f(x) = \begin{cases} \frac{\sqrt{1 + px} - \sqrt{1 - px}}{x} & \text{for } -1 \leq x < 0 \\ \frac{2x + 1}{x - 2} & \text{for } 0 \leq x \leq 1 \end{cases} \]

is continuous on the interval \([-1, 1]\), we need to check the continuity at the point where the two pieces of the function meet, which is at \( x = 0 \).

### Step 1: Find the left-hand limit as \( x \) approaches 0

The left-hand limit (LHL) as \( x \to 0^- \) is given by the first piece of the function:

\[ \text{LHL} = \lim_{x \to 0^-} \frac{\sqrt{1 + px} - \sqrt{1 - px}}{x} \]

### Step 2: Apply L'Hôpital's Rule

Since substituting \( x = 0 \) directly gives us the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that if the limit results in \( \frac{0}{0} \), we can differentiate the numerator and denominator:

\[ \text{LHL} = \lim_{x \to 0^-} \frac{\frac{d}{dx}(\sqrt{1 + px} - \sqrt{1 - px})}{\frac{d}{dx}(x)} \]

Calculating the derivatives:

- The derivative of the numerator:
\[ \frac{d}{dx}(\sqrt{1 + px}) = \frac{p}{2\sqrt{1 + px}}, \quad \frac{d}{dx}(\sqrt{1 - px}) = \frac{-p}{2\sqrt{1 - px}} \]
Thus,
\[ \frac{d}{dx}(\sqrt{1 + px} - \sqrt{1 - px}) = \frac{p}{2\sqrt{1 + px}} + \frac{p}{2\sqrt{1 - px}} = \frac{p(\sqrt{1 - px} + \sqrt{1 + px})}{2\sqrt{(1 + px)(1 - px)}} \]

- The derivative of the denominator is simply \( 1 \).

So we have:

\[ \text{LHL} = \lim_{x \to 0^-} \frac{p(\sqrt{1 - px} + \sqrt{1 + px})}{2\sqrt{(1 + px)(1 - px)}} \]

### Step 3: Evaluate the limit as \( x \to 0 \)

Substituting \( x = 0 \):

\[ \text{LHL} = \frac{p(\sqrt{1} + \sqrt{1})}{2\sqrt{1 \cdot 1}} = \frac{p(1 + 1)}{2} = \frac{p \cdot 2}{2} = p \]

### Step 4: Find the right-hand limit as \( x \) approaches 0

The right-hand limit (RHL) as \( x \to 0^+ \) is given by the second piece of the function:

\[ \text{RHL} = \lim_{x \to 0^+} \frac{2x + 1}{x - 2} \]

Substituting \( x = 0 \):

\[ \text{RHL} = \frac{2(0) + 1}{0 - 2} = \frac{1}{-2} = -\frac{1}{2} \]

### Step 5: Set the left-hand limit equal to the right-hand limit

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

\[ \text{LHL} = \text{RHL} \]

Thus,

\[ p = -\frac{1}{2} \]

### Conclusion

The value of \( p \) for which the function is continuous on the interval \([-1, 1]\) is:

\[ \boxed{-\frac{1}{2}} \]