Question

若函數(shù)f （x）=-（a²-11a+10）x²-（a-1）x+2對(duì)一切實(shí)數(shù)x恒為正值，則實(shí)數(shù)a的取值范圍是

1. A.
   1≤a≤9
2. B.
   1＜a＜9
3. C.
   a≤1或a＞9
4. D.
   1≤a＜9

Answer 1

D
分析：對(duì)于函數(shù)f（x）=ax²+bx+c，首先對(duì)二次項(xiàng)的系數(shù)分a=0和a≠0討論，然后對(duì)a≠0再分與解出即可．
解答：①當(dāng)-（a²-11a+10）=0時(shí)，解得a=1或a=10．
當(dāng)a=10時(shí)，f（x）=-9x+2不滿足對(duì)一切實(shí)數(shù)x恒為正值，故舍去．
當(dāng)a=1時(shí)，f（x）=2滿足對(duì)一切實(shí)數(shù)x恒為正值，因此a=1適合題意．
②當(dāng)-（a²-11a+10）＞0時(shí)，解得1＜a＜10．
要使函數(shù)f （x）=-（a²-11a+10）x²-（a-1）x+2對(duì)一切實(shí)數(shù)x恒為正值，
則必有△=（a-1）²+8（a²-11a+10）＜0，又1＜a＜10，
解得1＜a＜9，滿足題意．
③當(dāng)-（a²-11a+10）＜0時(shí)，解得a＜1或a＞10．
要使函數(shù)f （x）=-（a²-11a+10）x²-（a-1）x+2對(duì)一切實(shí)數(shù)x恒為正值，
則必有△=（a-1）²+8（a²-11a+10）＜0，又a＜1或a＞10，
解得a∈∅．
綜上可知：實(shí)數(shù)a的取值范圍是1≤a＜9．
故選D．
點(diǎn)評(píng)：熟練掌握三個(gè)“二次”與判別式△的關(guān)系是解題的關(guān)鍵．