Determine the constants 'a' and 'b' so that the function 'f' defined below is continuous everywhere :
`f(x)={{:(x+2", "xle2),(ax+b", "2ltxlt5),(3x-2", "xge5):}`.

Answer

To determine the constants \( a \) and \( b \) so that the function \( f \) is continuous everywhere, we need to ensure that the function is continuous at the points where the definition of the function changes, specifically at \( x = 2 \) and \( x = 5 \).

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

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

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

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

Calculating the left-hand limit as \( x \) approaches 2:
\[ \lim_{x \to 2^-} f(x) = 4 \]

Calculating the right-hand limit as \( x \) approaches 2:
\[ \lim_{x \to 2^+} f(x) = a(2) + b = 2a + b \]

Setting the limits equal for continuity:
\[ 2a + b = 4 \quad \text{(1)} \]

### Step 2: Check continuity at \( x = 5 \)

To ensure continuity at \( x = 5 \), we need:
\[ \lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x) = f(5) \]

Calculating \( f(5) \):
\[ f(5) = 3(5) - 2 = 15 - 2 = 13 \]

Calculating the left-hand limit as \( x \) approaches 5:
\[ \lim_{x \to 5^-} f(x) = a(5) + b = 5a + b \]

Calculating the right-hand limit as \( x \) approaches 5:
\[ \lim_{x \to 5^+} f(x) = 13 \]

Setting the limits equal for continuity:
\[ 5a + b = 13 \quad \text{(2)} \]

### Step 3: Solve the system of equations

Now we have a system of two equations:
1. \( 2a + b = 4 \)
2. \( 5a + b = 13 \)

Subtract equation (1) from equation (2):
\[ (5a + b) - (2a + b) = 13 - 4 \]
\[ 3a = 9 \]
\[ a = 3 \]

### Step 4: Substitute \( a \) back to find \( b \)

Substituting \( a = 3 \) into equation (1):
\[ 2(3) + b = 4 \]
\[ 6 + b = 4 \]
\[ b = 4 - 6 = -2 \]

### Conclusion

The constants are:
\[ a = 3, \quad b = -2 \]