12、已知函数$f(x) = e^x + ax + b - 3$（$a, b \in R$）在区间$[1, 2]$上总存在零点，则$a^2 + (b - 4)^2$的最小值（ ）
$A. \frac{(e + 1)^2}{2}$ $B. \frac{4}{13}$ $C. \frac{(e^2 + 1)^2}{5}$ $D. \frac{8}{e^4}$
解：函数$f(x) = e^x + ax + b - 3$（$a, b \in R$）在区间$[1, 2]$上总存在零点
$\therefore$存在$x \in [1, 2]$使$e^x + ax + b - 3 = 0$
$\therefore ax + b - 4 = -e^x - 1$
$\therefore (ax + b - 4)^2 = (e^x + 1)^2$
$\because \left[ a^2 + (b - 4)^2 \right](x^2 + 1) \geq (ax + b - 4)^2$（柯西不等式）
$\therefore a^2 + (b - 4)^2 \geq \frac{(e^x + 1)^2}{x^2 + 1}$
设$g(x) = \frac{(e^x + 1)^2}{x^2 + 1}$，则$g'(x) = \frac{(e^{2t} + e^t)(t^2 - t + 1) + e^t - t}{(x^2 + 1)^2}$
$\because t^2 - t + 1 > 0, e^t - t > 0$
$\therefore g'(x) > 0$，$g(x)$单增
$\therefore x \in [1, 2]$时，$g(x)_{\min} = g(1) = \frac{(e + 1)^2}{2}$
$\therefore a^2 + (b - 4)^2 \geq \frac{(e + 1)^2}{2}$，选A