\(f\left( x \right) = \left\{ \begin{array}{l}a\dfrac{{{x^2} - x - 6}}{{x - 3}} + 2b\,\,\,\,\,khi\,\,x > 3\\\,\,\,\,\,\,\,\,\,\,\,\,\,8\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,khi\,\,x = 3\\a\dfrac{{x - 3}}{{\sqrt {12 - x} - 3}} - 2b\,\,khi\,\,x < 3\end{array} \right.\) tại \(x = 3\).
Tìm a, b để hàm số liên tục tại điểm đã chỉ ra:
Câu 396985: \(f\left( x \right) = \left\{ \begin{array}{l}a\dfrac{{{x^2} - x - 6}}{{x - 3}} + 2b\,\,\,\,\,khi\,\,x > 3\\\,\,\,\,\,\,\,\,\,\,\,\,\,8\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,khi\,\,x = 3\\a\dfrac{{x - 3}}{{\sqrt {12 - x} - 3}} - 2b\,\,khi\,\,x < 3\end{array} \right.\) tại \(x = 3\).
A. \(a = 16,\,\,b =- 44\).
B. \(a = - 16,\,\,b =- 44\).
C. \(a = 44,\,\,b = -16\).
D. \(a = - 16,\,\,b = 44\).
Hàm số \(y = f\left( x \right)\) liên tục tại \(x = {x_0}\) khi và chỉ khi \(\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) = f\left( {{x_0}} \right)\).
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Đáp án : D(3) bình luận (0) lời giải
Giải chi tiết:
TXĐ: \(D = \mathbb{R}\) và \(x = 0 \in D\).
Ta có:
\(\begin{array}{l}\mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} \left[ {a\dfrac{{{x^2} - x - 6}}{{x - 3}} + 2b} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {3^ + }} \left[ {a\dfrac{{\left( {x - 3} \right)\left( {x + 2} \right)}}{{x - 3}} + 2b} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {3^ + }} \left[ {a\left( {x + 2} \right) + 2b} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 5a + 2b\\\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ - }} \left( {a\dfrac{{x - 3}}{{\sqrt {12 - x} - 3}} - 2b} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {3^ - }} \left( {a\dfrac{{\left( {x - 3} \right)\left( {\sqrt {12 - x} + 3} \right)}}{{3 - x}} - 2b} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {3^ - }} \left( { - a\left( {\sqrt {12 - x} + 3} \right) - 2b} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 6a - 2b\\f\left( 3 \right) = 8\end{array}\)
Để hàm số đã cho liên tục tại \(x = 3\) thì \(\mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = f\left( 3 \right)\)\( \Leftrightarrow 5a + 2b = - 6a - 2b = 8\).
\( \Leftrightarrow \left\{ \begin{array}{l}5a + 2b = 8\\ - 6a - 2b = 8\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = - 16\\b = 44\end{array} \right.\).
Vậy \(a = - 16,\,\,b = 44\).
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