Question

已知不等式x²-3x+2t＜0的解集是（1,m）,m＞1,t∈R.

(1)求m,t的值；

(2)若函數(shù)f(x)=tx³+mx²-2ax在區(qū)間（1，+∞）上是單調(diào)增函數(shù)，求a的取值范圍.

Answer 1

解:（1）不等式x²-3x+2t＜0,解集是(1,m),m＞1,1、m是方程x²-3x+2t=0的根.由韋達(dá)定理.?

解得m=2,t=1.不等式x²-3x+2＜0,其解集是(1,2).                                                     ?

（2）函數(shù)f(x)=x³+2x²-2ax的導(dǎo)數(shù)f′(x)=3x²+4x-2a.                                            ?

當(dāng)f′(x)＞0,x∈(1,+∞)時(shí)f(x)是增函數(shù).?

3x²+4x-2a＞0,對(duì)x∈(1,+∞)恒成立，即2a＜3x²+4x成立.                                    ?

令g(x)=3x²+4x,g(x)在(1,+∞)上為增函數(shù)g(x)＞g(1)=7,?

∴a≤.                                                                                                          

當(dāng)a＞時(shí)，經(jīng)檢驗(yàn)g(x)在(1,+∞)上不是增函數(shù).??

綜上a的取值范圍是a∈(-∞,］.