Q59 / 150 · MHT CET 2026 (20 April Shift 2) · Mathematics
If 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 $x = 3$, then $(a - b) =$ ..........
- Correct answerA$\dfrac{2}{3}$
- B$\dfrac{3}{2}$
- C$2$
- D$3$
Solution
For the function $f(x)$ to be continuous at $x = 3$, the left-hand limit must be equal to the right-hand limit and the value of the function at $x = 3$.
$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)$
Evaluating the left-hand limit:
$\lim_{x \to 3^-} (ax + 1) = 3a + 1$
Evaluating the right-hand limit:
$\lim_{x \to 3^+} (bx + 3) = 3b + 3$
Equating the limits:
$3a + 1 = 3b + 3$
$3a - 3b = 2$
$3(a - b) = 2$
$a - b = \dfrac{2}{3}$
Answer: $\dfrac{2}{3}$
$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)$
Evaluating the left-hand limit:
$\lim_{x \to 3^-} (ax + 1) = 3a + 1$
Evaluating the right-hand limit:
$\lim_{x \to 3^+} (bx + 3) = 3b + 3$
Equating the limits:
$3a + 1 = 3b + 3$
$3a - 3b = 2$
$3(a - b) = 2$
$a - b = \dfrac{2}{3}$
Answer: $\dfrac{2}{3}$
Hence, the correct answer is option: 'A'.