Find 'a' and 'b' if the function :
`f(x)={{:((sinx)/(x)" , "-2lexlt0),(a.2^(x)" , "0lexle1),(b+x", "1ltxle2):}`
is a continuous function on [-2, 2].

Answer

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

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

is continuous on the interval \([-2, 2]\), we need to ensure that the function is continuous at the points where the definition of the function changes, specifically at \(x = 0\) and \(x = 1\).

### Step 1: Continuity at \(x = 0\)

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

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

Calculating the left-hand limit:

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

Calculating the value of the function at \(x = 0\):

\[ f(0) = a \cdot 2^0 = a \]

Setting the left-hand limit equal to the function value:

\[ 1 = a \]

### Step 2: Continuity at \(x = 1\)

For the function to be continuous at \(x = 1\), we need:

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

Calculating the left-hand limit:

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

Calculating the value of the function at \(x = 1\):

\[ f(1) = b + 1 \]

Setting the left-hand limit equal to the function value:

\[ 2a = b + 1 \]

### Step 3: Substitute the value of 'a'

From Step 1, we found that \(a = 1\). Substituting this value into the equation from Step 2:

\[ 2(1) = b + 1 \]

This simplifies to:

\[ 2 = b + 1 \implies b = 1 \]

### Conclusion

The values of \(a\) and \(b\) are:

\[ a = 1, \quad b = 1 \]