Question

The relationship between \( a \) and \( b \) so that the function \( f(x) \) defined by

\[ f(x) = \begin{cases} ax + 1, & \text{if } x \leq 3 \\ bx + 3, & \text{if } x > 3 \end{cases} \]

is continuous at \( x = 3 \), is:

• A. \( a = b + \frac{2}{3} \)
• B. \( a - b = \frac{3}{2} \)
• C. \( a + b = \frac{2}{3} \)
• D. \( a + b = 2 \)

Hint:

For a piecewise function to be continuous at a point, the function values from both sides of the point must match.

Answer 1

The Correct Option is A

Step 1: For the function to be continuous at \( x = 3 \), the left-hand limit and the right-hand limit must be equal to the value of the function at \( x = 3 \). 
The left-hand limit is the value of \( f(x) \) for \( x \leq 3 \), i.e., \( f(3) = 3a + 1 \). 
The right-hand limit is the value of \( f(x) \) for \( x>3 \), i.e., \( f(3) = 3b + 3 \). 
Step 2: For continuity at \( x = 3 \), set the two expressions equal: \[ 3a + 1 = 3b + 3. \] 
Step 3: Solve for \( a \) in terms of \( b \): \[ 3a - 3b = 2 \quad \Rightarrow \quad a - b = \frac{2}{3}. \] Thus, the relationship is \( a = b + \frac{2}{3} \).