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问题 解答题

已知二次函数f（x）满足f（x+1）+f（x-1）=2x2-4x；

（1）求f（x）的解析式    

（2）求当x∈[0，a]（a为大于0的常数）时f（x）的最小值．

答案

（1）设f（x）=ax2+bx+c（a≠0），则有f（x+1）+f（x-1）=2ax2+2bx+2a+2c=2x2-4x

对任意实数x恒成立

∴

2a=2
2b=-4
2a+2c=0

解之得a=1，b=-2，c=-1

∴f（x）=x2-2x-1

（2）当0＜a＜1时，f（x）的最小值为f（a）=a2-2a-1

当a≥1时，f（x）的最小值为f（1）=-2

填空题

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I don’t really know (11) are needed, but obviously a degree in modern languages or (12) would be very useful. But, a degree in itself probably wouldn’t (13) . It wouldn’t be enough to get you job. It wouldn’t (14) to make an editor decide to (15) . So it’s difficult to define the (16) . You’re going to need (17) and the ability to assimilate information (18) and then produce stories which (19) of the newspaper or radio or television station you (20) .
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单项选择题

男性，26岁。到沿海旅游，因进食过海产品，餐后4h突然发热、剧烈腹痛、腹泻、呕吐，大便10余次，呈血水样，偶有粘液血性，无里急后重住院。体检：轻度脱水征，BP12／8kPa（90／60mmHg），腹软有压痛。化验：大便光镜检WBC5～8／HP，RBC10～25／HP，最可能诊断是（）

A.霍乱

B.副霍乱

C.沙门菌食物中毒

D.副溶血弧菌食物中毒

E.细菌性痢疾
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