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Answer» \(f(x) =\begin{cases}ax^2 & \quad \text{0<} x \text{<1}\\x+3 & \quad \text{1<}x< \text{2}\\4& \quad \text{x} = \text{1}\end{cases}\) f(x) is not continuous at x = 1 Left hand limit of f(x) at x = 1 is f(1-) = \(\lim\limits_{x \to 1^-}\) f(x) = \(\lim\limits_{x \to 1}\) ax2 = a Right hand limit of f(x) at x = 1 is f(1+) = \(\lim\limits_{x \to 1^+}\) f(x) = \(\lim\limits_{x \to 1}\) (x + 3) = 1 + 3 = 4 f(1) = 4 ∵ f(1+) = f(1) and f is not continuous at x = 1. Therefore f(1-) ≠ f(1+) ⇒a ≠ 4 (∵ f (1-)= a) Hence, a ∈ R-{4}, then f(x) is not continuous at x = 1. |
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