Find the values of 'a' and 'b' so that the following function is continuous at x = 3 and x = 5 :
`f(x)={{:(1", if "xle3),(ax+b", if "3ltxlt5),(7", if "5lex):}`.

Answer

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

\[ f(x) = \begin{cases} 1 & \text{if } x \leq 3 \\ ax + b & \text{if } 3 < x < 5 \\ 7 & \text{if } x \geq 5 \end{cases} \]

is continuous at \( x = 3 \) and \( x = 5 \), we need to ensure that the left-hand limit, right-hand limit, and the function value at those points are equal.

### Step 1: Check continuity at \( x = 3 \)

The function value at \( x = 3 \) is:

\[ f(3) = 1 \]

The left-hand limit as \( x \) approaches 3 is:

\[ f(3^-) = 1 \]

The right-hand limit as \( x \) approaches 3 is given by the expression for \( f(x) \) when \( 3 < x < 5 \):

\[ f(3^+) = a(3) + b = 3a + b \]

For continuity at \( x = 3 \), we set the left-hand limit equal to the right-hand limit:

\[ 3a + b = 1 \quad \text{(1)} \]

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

The function value at \( x = 5 \) is:

\[ f(5) = 7 \]

The left-hand limit as \( x \) approaches 5 is given by:

\[ f(5^-) = a(5) + b = 5a + b \]

The right-hand limit as \( x \) approaches 5 is:

\[ f(5^+) = 7 \]

For continuity at \( x = 5 \), we set the left-hand limit equal to the function value:

\[ 5a + b = 7 \quad \text{(2)} \]

### Step 3: Solve the system of equations

Now we have a system of two equations:

1. \( 3a + b = 1 \) (Equation 1)
2. \( 5a + b = 7 \) (Equation 2)

We can subtract Equation 1 from Equation 2 to eliminate \( b \):

\[ (5a + b) - (3a + b) = 7 - 1 \]

This simplifies to:

\[ 2a = 6 \]

Thus, we find:

\[ a = 3 \]

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

Now, substitute \( a = 3 \) back into Equation 1:

\[ 3(3) + b = 1 \]

This simplifies to:

\[ 9 + b = 1 \]

Solving for \( b \):

\[ b = 1 - 9 = -8 \]

### Final Values

Thus, the values of \( a \) and \( b \) are:

\[ a = 3, \quad b = -8 \]