Đáp án:
$a = \pm 2\sqrt 2 ;b = - 1$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
Miny = - 2\\
\Leftrightarrow \dfrac{{ax + b}}{{{x^2} + 1}} \ge - 2,\forall x\\
\Leftrightarrow ax + b \ge - 2{x^2} - 2,\forall x\\
\Leftrightarrow 2{x^2} + ax + b + 2 \ge 0,\forall x\\
\Leftrightarrow \Delta = 0\\
\Leftrightarrow {a^2} - 4.2.\left( {b + 2} \right) = 0\\
\Leftrightarrow {a^2} - 8b - 16 = 0\left( 1 \right)
\end{array}$
Lại có:
$\begin{array}{l}
Maxy = 1\\
\Leftrightarrow \dfrac{{ax + b}}{{{x^2} + 1}} \le 1,\forall x\\
\Leftrightarrow {x^2} - ax + 1 - b \ge 0,\forall x\\
\Leftrightarrow \Delta = 0\\
\Leftrightarrow {\left( { - a} \right)^2} - 4.1.\left( {1 - b} \right) \le 0\\
\Leftrightarrow {a^2} + 4b - 4 = 0\left( 2 \right)
\end{array}$
Từ $\left( 1 \right),\left( 2 \right)$ ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
{a^2} - 8b - 16 = 0\\
{a^2} + 4b - 4 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{a^2} - 8b - 16 = 0\\
{a^2} - 8b - 16 + 2\left( {{a^2} + 4b - 4} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{a^2} - 8b - 16 = 0\\
3{a^2} - 24 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = \dfrac{{{a^2} - 16}}{8}\\
{a^2} = 8
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
b = - 1\\
\left[ \begin{array}{l}
a = 2\sqrt 2 \\
a = - 2\sqrt 2
\end{array} \right.
\end{array} \right.
\end{array}$
Vậy $a = \pm 2\sqrt 2 ;b = - 1$
