Đáp án:
\[a + b = 4\]
Giải thích các bước giải:
\(\begin{array}{l}
y = f\left( x \right) = \left\{ \begin{array}{l}
a - 4x,\,\,\,\,\,\,x < - 1\\
{x^2} + 2,\,\,\,\,\,\,\, - 1 \le x \le 1\\
2bx - 7,\,\,\,\,\,\,x > 1
\end{array} \right.\\
\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \left( {a - 4x} \right) = a - 4.\left( { - 1} \right) = a + 4\\
\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \left( {{x^2} + 2} \right) = {\left( { - 1} \right)^2} + 2 = 3\\
\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {{x^2} + 2} \right) = {1^2} + 2 = 3\\
\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {2bx - 7} \right) = 2.b.1 - 7 = 2b - 7
\end{array}\)
Hàm số đã cho liên tục trên R khi và chỉ khi:
\(\begin{array}{l}
\left\{ \begin{array}{l}
\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} f\left( x \right)\\
\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a + 4 = 3\\
2b - 7 = 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a = - 1\\
b = 5
\end{array} \right.\\
\Rightarrow a + b = \left( { - 1} \right) + 5 = 4
\end{array}\)
