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

Options

A. `-1`
B. `-1//2`
C. `1//2`
D. 1

Answer

To determine the value of \( p \) such that 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 ensure that the function is continuous at \( x = 0 \).

### Step 1: Check continuity at \( x = 0 \)

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

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

### Step 2: Calculate \( f(0) \)

For \( x \) in the interval \( [0, 1] \):

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

### Step 3: Calculate \( \lim_{x \to 0^-} f(x) \)

For \( x \) in the interval \( [-1, 0) \):

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0} \frac{\sqrt{1+px} - \sqrt{1-px}}{x} \]

This limit is of the form \( \frac{0}{0} \) as both the numerator and denominator approach 0 when \( x \to 0 \). We can apply L'Hôpital's Rule:

### Step 4: Apply L'Hôpital's Rule

Differentiate the numerator and denominator:

1. **Numerator**:
\[ \frac{d}{dx}(\sqrt{1+px} - \sqrt{1-px}) = \frac{p}{2\sqrt{1+px}} + \frac{p}{2\sqrt{1-px}} \]

2. **Denominator**:
\[ \frac{d}{dx}(x) = 1 \]

Thus, we have:

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

### Step 5: Evaluate the limit as \( x \to 0 \)

Substituting \( x = 0 \):

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

### Step 6: Set the limits equal for continuity

Now, set the limits equal to ensure continuity at \( x = 0 \):

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

### Conclusion

Thus, the value of \( p \) that makes the function continuous on the interval \([-1, 1]\) is

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