InterviewSolution
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InterviewSolution
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Find the value of 'a' and 'b' if \(\lim\limits_{x \to 2}f(x)\) and \(\lim\limits_{x \to 4}f(x)\)exists where.\(f(x) = \begin{cases} x^2+as+b & \quad \text, 0\leq{x} <{2}\\ 3x+2, & \quad \text, 2\leq{x}\leq{4}\\2ax+5b & \quad \text,4<x\leq{8} \end{cases} \) |
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Answer» Given, \(f(x) = \begin{cases} x^2+as+b & \quad \text, 0\leq{x} <{2}\\ 3x+2, & \quad \text, 2\leq{x}\leq{4}\\2ax+5b & \quad \text,4<x\leq{8} \end{cases} \) To find \(\lim\limits_{x \to 2}f(x)\) L.H.L \(=\lim\limits_{x \to 2^-}f(x)\) \(=\lim\limits_{x \to 2^-}(x^2+ax+b)\) = 22 + a∙2 + b = 2a + b + 4 And, R.H.L \(=\lim\limits_{x \to 2^+}f(x)\) \(=\lim\limits_{x \to 2^+}(3x+2)\) = 3 ∙ 2 + 2 = 8 Since \(\lim\limits_{x \to 2}f(x) \) exists, ∴ \(\lim\limits_{x \to 2^-}f(x)\) \(=\lim\limits_{x \to 2^+}f(x)\) ⇒ 2a + b + 4 = 8 ⇒ 2a + b = 4 … (1) To find \(\lim\limits_{x \to 2}f(x)\). L.H.L \(=\lim\limits_{x \to 4^-}f(x)\) \(=\lim\limits_{x \to 4^-}(3x+2)\) = 3 ∙ 4 + 2 = 14 And R.H.L \(=\lim\limits_{x \to 4^+}f(x)\) \(=\lim\limits_{x \to 4^+}(2ax+5b)\) = 2a. 4 + 5b = 8a + 5b Since \(\lim\limits_{x \to 4}f(x) \) exists. ∴ \(\lim\limits_{x \to 4^-}f(x) \) = \(\lim\limits_{x \to 4^+}f(x) \) ⇒ 8a + 5b = 14 … (2) From (1) and (2), a = 3 and b = -2. |
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