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
- A \(a = -5, b = 2\)
- B \(a = -5, b = -2\)
- C \(a = 5, b = -2\)
- D \(a = -5, b = 8\)
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: \(a = -5, b = 8\)
\(f(1) = 3(1) - 1 = 2\)
\(\lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{-}} (ax + 7) = a + 7\)
\(\lim_{x \to 1^{+}} f(x) = \lim_{x \to 1^{+}} \dfrac{b}{x + 3} = \dfrac{b}{4}\)
Equating these values:
\(a + 7 = 2 \Rightarrow a = -5\)
\(\dfrac{b}{4} = 2 \Rightarrow b = 8\)
Answer: \(a = -5, b = 8\)
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