题目

设函数f(x)=tx2+2t2x+t-1(x∈R,t>0). (I)求f (x)的最小值h(t); (II)若h(t)<-2t+m对t∈(0,2)恒成立，求实数m的取值范围. 答案：本题主要考查函数的单调性、极值以及函数导数的应用，考查运用数学知识分析问题解决问题的能力 解：（I）∵（）， ∴当x=-t时，f(x)取最小值f(-t)=-t3+t-1, 即h(t)=-t3+t-1. (II)令g(t)=h(t)-(-2t+m)=-t3+3t-1-m, 由g’(t)=-3t2+3=0得t=1,t=-1（不合题意，舍去）. 当t变化时g’(t)、g(t)的变化情况如下表： t(0,1)1(1,2) g’(t)+0- 单词拼写（10分）【小题1】Beijing is an a_______ (古老的)city, but Dalian is young.【小题2】It is a _______(社会的)problem.【小题3】Sun Haiping is Liu Xiang’s _______(教练).【小题4】I bought a _______(数码)camera last week.【小题5】My mother often look _______(浏览)magazines before going to bed.【小题6】We must_______(遵守)the rulers of school.【小题7】Reading in sun does_______(有害的)to our eyes.【小题8】There are many kinds of_______(鲨鱼).【小题9】Australia is famous for_______(绵羊).【小题10】Tom is enjoying the _______(阳光)on the beach.