Question

If the function $$f(x) = \begin{array}{cc} 3ax + b, & \text{for } x < 1 \\ 11, & \text{for } x = 1 \\ 5ax - 2b, & \text{for } x > 1 \end{array}$$ is continuous at $x = 1$, then the values of $a$ and $b$ are

• A. $a = 2, b = 3$
• B. $a = 3, b = 3$
• C. $a = 2, b = 2$
• D. $a = 3, b = 2$

Hint:

To solve multiple-choice limit systems even faster, plug the options directly into the linear boundary constraints! Testing option (D) gives $3(3) + 2 = 11$ and $5(3) - 2(2) = 15 - 4 = 11$. Since both conditions match the central value of $11$ perfectly, it confirms the correct option instantly without any manual elimination steps.

Answer 1

The Correct Option is D

Step 1: State what continuity needs.
For the piecewise function to be continuous at $x=1$, the left limit, the right limit, and the value $f(1)=11$ must all agree.
Step 2: Write the left piece.
For $x<1$ the rule is $3ax+b$, so the left limit is $3a(1)+b = 3a+b$.
Step 3: Write the right piece.
For $x>1$ the rule is $5ax-2b$, so the right limit is $5a(1)-2b = 5a-2b$.
Step 4: Form two equations.
Matching each to $f(1)=11$ gives $3a+b=11$ and $5a-2b=11$.
Step 5: Eliminate $b$ by combination.
Double the first: $6a+2b=22$. Add the second: $(6a+2b)+(5a-2b)=22+11$, so $11a=33$ and $a=3$.
Step 6: Back-substitute for $b$.
Put $a=3$ into $3a+b=11$: $9+b=11$, hence $b=2$. A quick check: $5(3)-2(2)=15-4=11$, which confirms the right limit too.
\[ \boxed{a = 3,\ b = 2} \]