Question

If $f(x) = \begin{cases} ax + 7 & \text{if } x<1 \\ 2x - 3 & \text{if } x = 1 \\ \frac{x+b}{b} & \text{if } x>1 \end{cases}$ is continuous at $x = 1$, then

• A. $a = 3, b = 2$
• B. $a = -8, b = -2$
• C. $a = 8, b = -2$
• D. $a = -8, b = 2$

Hint:

When dealing with blurry exam text, solve the clear parts first (here, finding $a=-8$). Use the derived information and the multiple-choice options to "reverse engineer" the unclear part of the equation.

Answer 1

The Correct Option is D

Step 1: Continuity condition
For continuity at $x=1$: \[ \text{LHL} = \text{RHL} = f(1) \]
Step 2: Find $f(1)$
Using middle function: \[ f(1) = 2(1) - 3 = -1 \]
Step 3: Left-Hand Limit (LHL)
\[ \lim_{x \to 1^-} (ax + 7) = a + 7 \] Equate with $f(1)$: \[ a + 7 = -1 \Rightarrow a = -8 \]
Step 4: Right-Hand Limit (RHL)
\[ \lim_{x \to 1^+} \frac{x - 3}{b} = \frac{1 - 3}{b} = \frac{-2}{b} \] Equate with $f(1)$: \[ \frac{-2}{b} = -1 \Rightarrow b = 2 \] Final Answer:
\[ \boxed{a = -8, b = 2} \]