问题
选择题
已知函数y=
|
答案
函数定义域是R，则mx2+mx+1＞0恒成立
①m=0，则1＞0恒成立
②m≠0，mx2+mx+1＞0恒成立应满足：m＞0且△=m2-4m＜0，解得0＜m＜4
综上所述实数m的取值范围是[0，4），
故选D．
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已知函数y=
|
函数定义域是R，则mx2+mx+1＞0恒成立
①m=0，则1＞0恒成立
②m≠0，mx2+mx+1＞0恒成立应满足：m＞0且△=m2-4m＜0，解得0＜m＜4
综上所述实数m的取值范围是[0，4），
故选D．
| 完形填空。 | |||
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