If the function `f(x)= {(3ax +b", " x gt 1 ),(11 ", "x=1),(5 ax-2b", " x lt 1):}`
continuous at x= 1 then ( a, b) =?

Options

A. (3,2)
B. (2,3)
C. (1,4)
D. (4,1)

Answer

To determine the values of \( a \) and \( b \) such that the function

\[ f(x) = \begin{cases} 3ax + b & \text{if } x > 1 \\ 11 & \text{if } x = 1 \\ 5ax - 2b & \text{if } x < 1 \end{cases} \]

is continuous at \( x = 1 \), we need to ensure that the left-hand limit, right-hand limit, and the function value at that point are all equal.

### Step 1: Calculate the Right-Hand Limit as \( x \) approaches 1

The right-hand limit is given by:

\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (3ax + b) \]

Substituting \( x = 1 \):

\[ \lim_{x \to 1^+} f(x) = 3a(1) + b = 3a + b \]

### Step 2: Calculate the Left-Hand Limit as \( x \) approaches 1

The left-hand limit is given by:

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (5ax - 2b) \]

Substituting \( x = 1 \):

\[ \lim_{x \to 1^-} f(x) = 5a(1) - 2b = 5a - 2b \]

### Step 3: Set Up the Continuity Condition

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

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

This gives us the equations:

1. \( 3a + b = 11 \) (from the right-hand limit)
2. \( 5a - 2b = 11 \) (from the left-hand limit)

### Step 4: Solve the System of Equations

We have the following system of equations:

1. \( 3a + b = 11 \) (Equation 1)
2. \( 5a - 2b = 11 \) (Equation 2)

From Equation 1, we can express \( b \) in terms of \( a \):

\[ b = 11 - 3a \]

Now, substitute this expression for \( b \) into Equation 2:

\[ 5a - 2(11 - 3a) = 11 \]

Expanding this gives:

\[ 5a - 22 + 6a = 11 \]

Combining like terms:

\[ 11a - 22 = 11 \]

Adding 22 to both sides:

\[ 11a = 33 \]

Dividing by 11:

\[ a = 3 \]

### Step 5: Substitute \( a \) Back to Find \( b \)

Now substitute \( a = 3 \) back into the equation for \( b \):

\[ b = 11 - 3(3) = 11 - 9 = 2 \]

### Final Answer

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

\[ (a, b) = (3, 2) \]