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 find the values of \( a \) and \( b \) such that the function \( f(x) \) is continuous at \( x = 1 \), we need to ensure that the left-hand limit, the right-hand limit, and the function value at \( x = 1 \) are all equal.

The function is defined as follows:
- \( f(x) = 5ax - 2b \) for \( x < 1 \)
- \( f(1) = 11 \)
- \( f(x) = 3ax + b \) for \( x > 1 \)

### Step 1: Find the left-hand limit as \( x \) approaches 1
The left-hand limit as \( x \) approaches 1 is given by:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (5ax - 2b) = 5a(1) - 2b = 5a - 2b \]

### Step 2: Find the right-hand limit as \( x \) approaches 1
The right-hand limit as \( x \) approaches 1 is given by:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (3ax + b) = 3a(1) + b = 3a + b \]

### Step 3: Set the limits equal to each other and to the function value at \( x = 1 \)
For the function to be continuous at \( x = 1 \), we need:
\[ \lim_{x \to 1^-} f(x) = f(1) = \lim_{x \to 1^+} f(x) \]
This gives us two equations:
1. \( 5a - 2b = 11 \) (from the left-hand limit)
2. \( 3a + b = 11 \) (from the right-hand limit)

### Step 4: Solve the system of equations
We have the following system of equations:
1. \( 5a - 2b = 11 \) (Equation 1)
2. \( 3a + b = 11 \) (Equation 2)

From Equation 2, we can express \( b \) in terms of \( a \):
\[ b = 11 - 3a \]

Now, substitute \( b \) in Equation 1:
\[ 5a - 2(11 - 3a) = 11 \]
Expanding this gives:
\[ 5a - 22 + 6a = 11 \]
Combining like terms:
\[ 11a - 22 = 11 \]
Adding 22 to both sides:
\[ 11a = 33 \]
Dividing by 11:
\[ a = 3 \]

Now substitute \( a = 3 \) back into the expression for \( b \):
\[ b = 11 - 3(3) = 11 - 9 = 2 \]

### Final Answer
Thus, the values of \( a \) and \( b \) are:
\[ (a, b) = (3, 2) \]