`f(x) = {{:((sqrt(1+kx)-sqrt(1-kx))/(x),if -1 le x lt 0),((2x+1)/(x-1),if 0 le x le 1):}` at `x = 0`. f(x) is continuous then find k

Explanation

To determine the value of \( k \) such that the function \( f(x) \) is continuous at \( x = 0 \), we need to ensure that the left-hand limit and right-hand limit at \( x = 0 \) are equal, and that they are equal to the value of the function at \( x = 0 \).

The function is defined as:
\[ f(x) = \begin{cases} \frac{\sqrt{1+kx} - \sqrt{1-kx}}{x} & \text{if } -1 \leq x < 0 \\ \frac{2x + 1}{x - 1} & \text{if } 0 \leq x \leq 1 \end{cases} \]

### Step 1: Find the Left-Hand Limit as \( x \to 0^- \)

We need to compute:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sqrt{1+kx} - \sqrt{1-kx}}{x} \]

Substituting \( x = 0 \) directly gives us:
\[ \frac{\sqrt{1+0} - \sqrt{1-0}}{0} = \frac{1 - 1}{0} = \frac{0}{0} \]
This is an indeterminate form, so we will rationalize the numerator.

### Step 2: Rationalizing the Numerator

Multiply the numerator and denominator by the conjugate:
\[ \frac{\sqrt{1+kx} - \sqrt{1-kx}}{x} \cdot \frac{\sqrt{1+kx} + \sqrt{1-kx}}{\sqrt{1+kx} + \sqrt{1-kx}} = \frac{(1+kx) - (1-kx)}{x(\sqrt{1+kx} + \sqrt{1-kx})} \]

This simplifies to:
\[ \frac{2kx}{x(\sqrt{1+kx} + \sqrt{1-kx})} \]

### Step 3: Simplifying the Expression

Now we can cancel \( x \) from the numerator and denominator:
\[ \lim_{x \to 0^-} \frac{2k}{\sqrt{1+kx} + \sqrt{1-kx}} \]

As \( x \to 0 \):
\[ \sqrt{1+kx} \to 1 \quad \text{and} \quad \sqrt{1-kx} \to 1 \]
Thus, we have:
\[ \lim_{x \to 0^-} \frac{2k}{1 + 1} = \frac{2k}{2} = k \]

### Step 4: Find the Right-Hand Limit as \( x \to 0^+ \)

Now we compute:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{2x + 1}{x - 1} \]

Substituting \( x = 0 \):
\[ \frac{2(0) + 1}{0 - 1} = \frac{1}{-1} = -1 \]

### Step 5: Value of the Function at \( x = 0 \)

From the definition of \( f(x) \) for \( 0 \leq x \leq 1 \):
\[ f(0) = \frac{2(0) + 1}{0 - 1} = -1 \]

### Step 6: Setting the Limits Equal for Continuity

For \( f(x) \) to be continuous at \( x = 0 \):
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \]
This gives us:
\[ k = -1 \]

### Conclusion

Thus, the value of \( k \) for which \( f(x) \) is continuous at \( x = 0 \) is:
\[ \boxed{-1} \]