If `f(x)` is continuous on [0, 8] , where `f(x)={:{(x^(2)+ax+6", for " 0 le x lt 2),(3x+2", for " 2 le x le 4),(2ax+5b", for " 4 lt x le 8):}`, then

Options

A. `a=-1, b=(22)/(5)`
B. `a=-1, b=(-8)/(5)`
C. `a=-1, b=(-22)/(5)`
D. `a=1, b=8/5`

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at the points where the definition of the function changes, specifically at \( x = 2 \) and \( x = 4 \).

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

The function is defined as:
- \( f(x) = x^2 + ax + 6 \) for \( 0 \leq x < 2 \)
- \( f(x) = 3x + 2 \) for \( 2 \leq x \leq 4 \)

To ensure continuity at \( x = 2 \), we need to check that:
\[ \lim_{x \to 2^-} f(x) = f(2) \]

Calculating \( \lim_{x \to 2^-} f(x) \):
\[ \lim_{x \to 2^-} f(x) = 2^2 + a(2) + 6 = 4 + 2a + 6 = 10 + 2a \]

Calculating \( f(2) \):
\[ f(2) = 3(2) + 2 = 6 + 2 = 8 \]

Setting these equal for continuity:
\[ 10 + 2a = 8 \]

### Step 2: Solve for \( a \)

Rearranging the equation:
\[ 2a = 8 - 10 \]
\[ 2a = -2 \]
\[ a = -1 \]

### Step 3: Check continuity at \( x = 4 \)

Now we check continuity at \( x = 4 \):
- \( f(x) = 3x + 2 \) for \( 2 \leq x \leq 4 \)
- \( f(x) = 2ax + 5b \) for \( 4 < x \leq 8 \)

We need:
\[ \lim_{x \to 4^-} f(x) = f(4) = \lim_{x \to 4^+} f(x) \]

Calculating \( \lim_{x \to 4^-} f(x) \):
\[ \lim_{x \to 4^-} f(x) = 3(4) + 2 = 12 + 2 = 14 \]

Calculating \( f(4) \):
\[ f(4) = 3(4) + 2 = 14 \]

Calculating \( \lim_{x \to 4^+} f(x) \):
\[ \lim_{x \to 4^+} f(x) = 2a(4) + 5b = 8a + 5b \]

Setting these equal for continuity:
\[ 14 = 8a + 5b \]

### Step 4: Substitute \( a \) and solve for \( b \)

Substituting \( a = -1 \):
\[ 14 = 8(-1) + 5b \]
\[ 14 = -8 + 5b \]
\[ 5b = 14 + 8 \]
\[ 5b = 22 \]
\[ b = \frac{22}{5} \]

### Final Result
The values we found are:
- \( a = -1 \)
- \( b = \frac{22}{5} \)