题目

已知f(x)=ax2+bx+c(a、b、c∈R,且a≠0).证明方程f(x)=0有两个不相等的实数解的充要条件是:存在x0∈R,使af(x0)<0. 答案：证明:(1)充分性:若存在x0∈R,使af(x0)<0,则b2-4ac=b2-4a［f(x0)-ax02-bx0］=b2+4abx0+4a2x02-4af(x0)=(b+2ax0)2-4af(x0)>0.∴方程f(x)=0有两个不等实根.(2)必要性:若方程f(x)=0有两个不等实根,则b2-4ac>0.设x0=-,a·f(x0)=a［a(-)2+b(-)+c］=-+ac=<0.听一段材料，回答下面几个小题。【1】What is the conversation about?A.A film.B.A book.C.A story.【2】What does the girl think of the film?A.She thinks it is terrible.B.She thinks it is good.C.She thinks it is wonderful.【3】According to the girl,what did the film director do to the story?A.He followed the story.B.He improved the story.C.He changed the story.