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高中数学必修 第一册(648题)


已知函数\(f\left( x \right) = \left( {x - 1} \right)\left( {ax + 1} \right)\),其中\(a\in \text{R}\).


求解关于\(x\)的不等式\(f\left( x \right) < 0\).



知识点:第二章 一元二次函数、方程和不等式


参考答案:综上所述,当\(a < - 1\)时,原不等式的解集为\(\left\{ {x\left| {x < - \frac{1}{a}} \right.} \right.\)或\(\left. {x > 1} \right\}\);当\(a = - 1\)时,原不等式的解集为\(\left\{ {x\left| {x \ne 1} \right.} \right\}\);当\( - 1 < a < 0\)时,原不等式的解集为\(\left\{ {x\left| {x < 1} \right.} \right.\)或\(\left. {x > - \frac{1}{a}} \right\}\);当\(a = 0\)时,原不等式的解集为\(\left\{ {x\left| {x < 1} \right.} \right\}\);当\(a > 0\)时,原不等式的解集为\(\left\{ {x\left| { - \frac{1}{a} < x < 1} \right.} \right\}\).

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