Question

If the function \(f(x)\) defined by \(f(x) = \{\begin{array}{cc}ax+1 & \text{if }x\leq 3 \\ bx+3 & \text{if }x > 3\end{array}\) is continuous at \(x = 3\), then \((a-b) =\) ..........

• A. \(\frac{2}{3}\)
• B. \(\frac{3}{2}\)
• C. \(2\)
• D. \(3\)

Hint:

Continuity at x = 3 means the left and right values are equal.

Answer 1

The Correct Option is A

Step 1: Approach
Subtract the two expressions at the join point.

Step 2: Join condition
Continuity requires $(3a+1)-(3b+3)=0$.

Step 3: Simplify
$3a-3b-2=0$, so $3(a-b)=2$ and $a-b=\dfrac23$.

Step 4: Meaning
The right piece gives $3b+3$ at the joint and the left piece gives $3a+1$. Making these two numbers equal removes the jump, and the algebra above is just that equality rearranged. The result $\dfrac23$ is option (A).

Final Answer:
Equating the two pieces at x = 3 gives a - b = 2/3, option (A). \[ \boxed{\frac{2}{3}} \]