ExamBro
ExamBro
KCET · Maths · Continuity and Differentiability

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

  1. A \(a = -5, b = 2\)
  2. B \(a = -5, b = -2\)
  3. C \(a = 5, b = -2\)
  4. D \(a = -5, b = 8\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(a = -5, b = 8\)

Step-by-step Solution

Detailed explanation

For the function to be continuous at \(x = 1\), the left-hand limit, right-hand limit, and the value of the function at \(x = 1\) must be equal.

\(f(1) = 3(1) - 1 = 2\)

\(\lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{-}} (ax + 7) = a + 7\)

\(\lim_{x \to 1^{+}} f(x) = \lim_{x \to 1^{+}} \dfrac{b}{x + 3} = \dfrac{b}{4}\)

Equating these values:

\(a + 7 = 2 \Rightarrow a = -5\)

\(\dfrac{b}{4} = 2 \Rightarrow b = 8\)

Answer: \(a = -5, b = 8\)