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Calantha
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Đáp án:
$a=1$
Giải thích các bước giải:
Ta có:
$f\left( x \right) = \left\{ \begin{array}{l}
2x + a,x \le 1\\
\dfrac{{{x^3} – {x^2} + 2x – 2}}{{x – 1}},x > 1
\end{array} \right.$
$\to $ Hàm $f(x)$ liên tục trên $R\backslash \left\{ 1 \right\}$
Như vậy:
Hàm số $f(x)$ liên tục trên $R$
$ \Leftrightarrow $ Hàm số $f(x)$ liên tục tại $x=1$
$ \Leftrightarrow \mathop {\lim }\limits_{x \to {1^ – }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = f\left( 1 \right)$
Ta có:
$\mathop {\lim }\limits_{x \to {1^ – }} f\left( x \right) = f\left( 1 \right) = a + 2$
Lại có:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)\\
= \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{{x^3} – {x^2} + 2x – 2}}{{x – 1}}\\
= \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{\left( {x – 1} \right)\left( {{x^2} + 2} \right)}}{{x – 1}}\\
= \mathop {\lim }\limits_{x \to {1^ + }} \left( {{x^2} + 2} \right)\\
= {1^2} + 2\\
= 3
\end{array}$
Như vậy:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to {1^ – }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)\\
\Leftrightarrow a + 2 = 3\\
\Leftrightarrow a = 1
\end{array}$
Vậy $a=1$
$\lim\limits_{x\to 1^+}f(x)$
$=\lim\limits_{x\to 1^+}\dfrac{x^3-x^2+2x-2}{x-1}$
$=\lim\limits_{x\to 1^+}\dfrac{x^2(x-1)+2(x-1)}{x-1}$
$=\lim\limits_{x\to 1^+}(x^2+2)$
$=1^2+2=3$
$\lim\limits_{x\to 1^-}f(x)$
$=\lim\limits_{x\to 1^-}(2x+a)$
$=a+2=f(1)$
$\to a+2=3$
$\to a=1$