Đáp án:
$\begin{array}{l}
3)\\
+ \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {{x^2} - 3x + 4} \right) = {2^2} - 3.2 + 4 = 2\\
+ f\left( 2 \right) = 5\\
+ \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {2x + 1} \right) = 2.2 + 1 = 5\\
\Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) \ne f\left( 2 \right) = \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right)
\end{array}$
=> hs ko liên tục trên R
$\begin{array}{l}
2)\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {x + 1} \right) = 1 + 1 = 2\\
\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {3 - a.{x^2}} \right) = 3 - a\\
\Rightarrow \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right)\\
\Rightarrow 2 = 3 - a\\
\Rightarrow a = 1\\
3)\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^3} - {x^2} + 2x - 2}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right)\left( {{x^2} + 2} \right)}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \left( {{x^2} + 2} \right)\\
= 3\\
\Rightarrow \mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right)\\
\Rightarrow 3 = 3 + a\\
\Rightarrow a = 0
\end{array}$
