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一道2015年北大博雅自主招生试题

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青青子衿 Posted at 2015-7-25 16:00:09 |Read mode
两校均取消了中学或校长推荐,改为学生自荐,并要求考生成绩要在高中拔尖。
清华领军和北大博雅计划的初试时间也都与普通自招错开,考生可参加两种自招考试。
6、已知点集\(M=\{{(x,y)|\sqrt{(1 - x^2) (1 - y^2)}\geqslant xy}\}\),则平面直角坐标系中区域M的面积为_______.

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kuing Posted at 2015-7-25 16:32:57
这个怎么发来高数区?明明是初等而且非常简单的题啊
首先二、四象限里的是正方形,一、三象限里的两边平方就变成 $x^2+y^2\leqslant1$,所以就是两块四分之一圆……

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 Author| 青青子衿 Posted at 2015-7-25 16:36:51
回复 2# kuing
刚才只想用二重积分算一算~

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longzaifei Posted at 2015-12-20 12:31:50
题目是不是有问题,区域在二、四象限好像不是一个封闭区域!

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kuing Posted at 2015-12-20 14:28:58
回复 4# longzaifei

谢谢提醒,确实!!!$x\le -1, y\ge1$ 和 $x\ge 1, y\le-1$ 都符合

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kuing Posted at 2015-12-20 16:31:16
看来得改成 $\sqrt{1-x^2}\sqrt{1-y^2}\ge xy$ 才行

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血狼王 Posted at 2015-12-28 21:45:43
自招也有出错题的时候……

看来学会出题和学会解题一样重要啊……

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 Author| 青青子衿 Posted at 2021-7-4 18:58:23
Last edited by 青青子衿 at 2021-7-8 14:39:00回复 7# 血狼王
自招也有出错题的时候……
血狼王 发表于 2015-12-28 21:45
翻了一下,是贴吧回忆版记错了,原题目
确实和k版写的一样
$\sqrt{1-x^2}\sqrt{1-y^2}\ge xy$

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isee Posted at 2021-7-8 13:40:01
回复 8# 青青子衿


哈哈哈这几年了都,还记得

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2025-4-21 14:41 GMT+8

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