Determine the values of 'a' and 'b' such that the following function is continuous at x = 0 :
`f(x)={{:((x+sinx)/(sin(a+1)x)", if "-piltxlt0),(2", if "x=0),(2(e^(sinbx)-1)/(bx)", if "xgt0):}`.

Answer

To determine the values of 'a' and 'b' such that the function

\[ f(x) = \begin{cases} \frac{x + \sin x}{\sin((a + 1)x)} & \text{if } x < 0 \\ 2 & \text{if } x = 0 \\ \frac{2(e^{\sin(bx)} - 1)}{bx} & \text{if } x > 0 \end{cases} \]

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

### Step 1: Evaluate the left-hand limit as \( x \to 0^- \)

For \( x < 0 \):

\[ f(x) = \frac{x + \sin x}{\sin((a + 1)x)} \]

We can use the limit:

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

Using the fact that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), we rewrite the limit:

\[ \lim_{x \to 0^-} \frac{x + \sin x}{\sin((a + 1)x)} = \lim_{x \to 0^-} \frac{x + \sin x}{(a + 1)x} \cdot \frac{(a + 1)x}{\sin((a + 1)x)} \]

### Step 2: Simplify the limit

Now, we can separate the limit:

\[ = \lim_{x \to 0^-} \frac{x + \sin x}{(a + 1)x} \cdot \lim_{x \to 0^-} \frac{(a + 1)x}{\sin((a + 1)x)} \]

For the first part, we know that:

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \implies \lim_{x \to 0} \frac{x + \sin x}{x} = 2 \implies \lim_{x \to 0} \frac{x + \sin x}{(a + 1)x} = \frac{2}{a + 1} \]

For the second part:

\[ \lim_{x \to 0} \frac{(a + 1)x}{\sin((a + 1)x)} = a + 1 \]

Combining these limits gives:

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

### Step 3: Set the left-hand limit equal to \( f(0) \)

Since \( f(0) = 2 \):

\[ 2 = 2 \]

This condition is satisfied for any \( a \) as long as \( a + 1 \neq 0 \).

### Step 4: Evaluate the right-hand limit as \( x \to 0^+ \)

For \( x > 0 \):

\[ f(x) = \frac{2(e^{\sin(bx)} - 1)}{bx} \]

Using the limit:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{2(e^{\sin(bx)} - 1)}{bx} \]

Using the fact that \( \lim_{x \to 0} \frac{e^x - 1}{x} = 1 \):

\[ = \lim_{x \to 0^+} \frac{2\sin(bx)}{bx} \cdot \frac{e^{\sin(bx)}}{\sin(bx)} \]

### Step 5: Simplify the limit

This becomes:

\[ = 2 \cdot 1 \cdot 1 = 2 \]

### Step 6: Set the right-hand limit equal to \( f(0) \)

Since \( f(0) = 2 \):

\[ 2 = 2 \]

This condition is satisfied for any \( b \) as long as \( b \neq 0 \).

### Conclusion

From the analysis, we have:

- \( a \) can be any value except \( -1 \) (to avoid division by zero).
- \( b \) must be \( 0 \) for the limit to hold.

Thus, the values are:

\[ \boxed{(a \text{ can be any value except } -1, b = 0)} \]