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We are given that the function \( f(x) \) is continuous at \( x = 0 \). The function is defined in three parts:

\[
f(x) =
\begin{cases}
\frac{2a}{x}(e^x - 1), & \text{if } -1 < x < 0, \\
1, & \text{if } x = 0, \\
\frac{b x \cos x}{1 - \sqrt{1 - x}}, & \text{if } 0 < x < 1.
\end{cases}
\]

To determine the values of constants \( a \) and \( b \), we will use the condition of continuity at \( x = 0 \). This requires that the left-hand limit as \( x \to 0^- \), the value of the function at \( x = 0 \), and the right-hand limit as \( x \to 0^+ \) must all be equal. That is:

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

### Step 1: Continuity at \( x = 0 \) from the left-hand side

Consider the left-hand limit as \( x \to 0^- \):
\[
f(x) = \frac{2a}{x}(e^x - 1).
\]
Using the fact that \( \lim_{x \to 0} \frac{e^x - 1}{x} = 1 \), we have:
\[
\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{2a}{x}(e^x - 1) = 2a.
\]

### Step 2: Continuity at \( x = 0 \)

We are given that \( f(0) = 1 \). Therefore, from the condition of continuity:
\[
2a = 1.
\]
Solving for \( a \):
\[
a = \frac{1}{2}.
\]

### Step 3: Continuity at \( x = 0 \) from the right-hand side

Now consider the right-hand limit as \( x \to 0^+ \):
\[
f(x) = \frac{b x \cos x}{1 - \sqrt{1 - x}}.
\]
We need to evaluate this limit as \( x \to 0^+ \). First, use a series expansion for the square root term:
\[
\sqrt{1 - x} \approx 1 - \frac{x}{2} \text{ as } x \to 0.
\]
Thus,
\[
1 - \sqrt{1 - x} \approx \frac{x}{2} \text{ as } x \to 0.
\]
Now, we substitute this into the expression for \( f(x) \) for \( x > 0 \):
\[
f(x) \approx \frac{b x \cos x}{\frac{x}{2}} = 2b \cos x.
\]
As \( x \to 0^+ \), \( \cos x \to 1 \), so:
\[
\lim_{x \to 0^+} f(x) = 2b.
\]

### Step 4: Continuity condition

Since \( f(0) = 1 \), we also have from the right-hand limit:
\[
2b = 1.
\]
Solving for \( b \):
\[
b = \frac{1}{2}.
\]

### Conclusion:

The values of the constants are:
\[
a = \frac{1}{2}, \quad b = \frac{1}{2}.
\]

Would you like further details on any steps? Here are 5 related questions to consider:

1. How do you compute limits involving series expansions for functions like \( e^x \) or \( \sqrt{1-x} \)?
2. What is the significance of a function being continuous at a point?
3. How would the solution change if the function was required to be differentiable at \( x = 0 \)?
4. What are some other common techniques for determining limits of complex functions?
5. How can you use the definition of the derivative to evaluate limits?

**Tip**: Always check if you can simplify a complex expression using series expansions or small approximations when evaluating limits.

The function f is continuous at x = 0 and is defined for -1 < x < 1 by: f(x) = { (2a/x)(e^x - 1) if -1 < x < 0, 1 if x = 0, (bx cos(x)) / (1 - sqrt(1 - x)) if 0 < x < 1 }. Determine the values of the constants a and b.

This problem involves determining constants a and b to ensure the continuity of a piecewise function at x = 0. The solution involves limits, series expansion, and the condition of continuity. Learn how to compute limits involving exponential and square root functions.

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