The function f(x) Is defined as follows :
`f(x)={{:(x^(2)+ax+b" , "0lexlt2),(3x+2" , "2lexle4),(2ax+5b" , "4ltxle8):}`.
If f(x) is continuous on [0,8], find the values of 'a' and 'b'.

Answer

To find the values of \( a \) and \( b \) such that the function \( f(x) \) is continuous on the interval \([0, 8]\), we need to ensure continuity at the points where the definition of \( f(x) \) changes, which are \( x = 2 \) and \( x = 4 \).

The function is defined as follows:
\[ f(x) = \begin{cases} x^2 + ax + b & \text{if } 0 \leq x < 2 \\ 3x + 2 & \text{if } 2 \leq x < 4 \\ 2ax + 5b & \text{if } 4 \leq x \leq 8 \end{cases} \]

### Step 1: Ensure continuity at \( x = 2 \)

1. **Calculate \( f(2^-) \)** (left-hand limit as \( x \) approaches 2):
\[ f(2^-) = 2^2 + 2a + b = 4 + 2a + b \]

2. **Calculate \( f(2^+) \)** (right-hand limit as \( x \) approaches 2):
\[ f(2^+) = 3(2) + 2 = 6 + 2 = 8 \]

3. **Set the limits equal for continuity**:
\[ 4 + 2a + b = 8 \]
Simplifying gives us:
\[ 2a + b = 4 \quad \text{(Equation 1)} \]

### Step 2: Ensure continuity at \( x = 4 \)

1. **Calculate \( f(4^-) \)** (left-hand limit as \( x \) approaches 4):
\[ f(4^-) = 3(4) + 2 = 12 + 2 = 14 \]

2. **Calculate \( f(4^+) \)** (right-hand limit as \( x \) approaches 4):
\[ f(4^+) = 2a(4) + 5b = 8a + 5b \]

3. **Set the limits equal for continuity**:
\[ 8a + 5b = 14 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations

We now have the following system of equations:
1. \( 2a + b = 4 \) (Equation 1)
2. \( 8a + 5b = 14 \) (Equation 2)

**From Equation 1**, we can express \( b \) in terms of \( a \):
\[ b = 4 - 2a \]

**Substituting \( b \) into Equation 2**:
\[ 8a + 5(4 - 2a) = 14 \]
Expanding gives:
\[ 8a + 20 - 10a = 14 \]
Combining like terms:
\[ -2a + 20 = 14 \]
Solving for \( a \):
\[ -2a = 14 - 20 \]
\[ -2a = -6 \implies a = 3 \]

**Now substituting \( a = 3 \) back into Equation 1 to find \( b \)**:
\[ 2(3) + b = 4 \]
\[ 6 + b = 4 \implies b = 4 - 6 = -2 \]

### Final Values
The values of \( a \) and \( b \) are:
\[ a = 3, \quad b = -2 \]