25~26高三上·安徽黄山一模·第14题

$\left(-\dfrac{1}{2},\dfrac{5}{8}\right]$

【题目】若$a^2 – 4b^2 = 2$，则$\dfrac{1}{a^2} – \dfrac{b}{a}$的取值范围为______.

【答案】$\left(-\dfrac{1}{2},\dfrac{5}{8}\right]$

【解析】记$x = \dfrac{b}{a}$，则$b = ax$，
由$a^2 – 4b^2 = 2$，得$a^2 – 4a^2x^2 = 2$，
即$a^2 = \dfrac{2}{1 – 4x^2}$.

因为$a^2 = \dfrac{2}{1 – 4x^2} > 0$，
所以$1 – 4x^2 > 0$，
所以$-\dfrac{1}{2} < x < \dfrac{1}{2}$.

则
$
\dfrac{1}{a^2} – \dfrac{b}{a} = \dfrac{2a^2 – b}{2a^2} – \dfrac{b}{a} $$= \dfrac{a^2 – 4b^2}{2a^2} – \dfrac{b}{a} $$= -2\left(\dfrac{b}{a}\right)^2 – \dfrac{b}{a} + \dfrac{1}{2} $$= -2x^2 – x + \dfrac{1}{2},
$
其中$-\dfrac{1}{2} < x < \dfrac{1}{2}$.

因为$y = -2x^2 – x + \dfrac{1}{2} $$= -2\left(x + \dfrac{1}{4}\right)^2 + \dfrac{5}{8}$，开口向下，其对称轴为$x = -\dfrac{1}{4}$，

所以当$-\dfrac{1}{2} < x < \dfrac{1}{2}$时，
$
y = -2\left(x + \dfrac{1}{4}\right)^2 + \dfrac{5}{8} \in \left(-\dfrac{1}{2},\dfrac{5}{8}\right].
$

所以$\dfrac{1}{a^2} – \dfrac{b}{a}$的取值范围为$\left(-\dfrac{1}{2},\dfrac{5}{8}\right]$.

故答案为：$\left(-\dfrac{1}{2},\dfrac{5}{8}\right]$.