没有最变态，只有更变态的验证码设计

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Highly Ranked

LittleNew @ Sep 2nd 2008 8:52PM

两道计算题都很简单
第一个是求x偏导，其实偏导符号是唬人的，就是对x求导，求导结果是28cos(7x-pi/2)，当x=0时，这个式子就是0。
第二个是求极限，也是很弱的题目arctanx*sin1/x的开方看上去很唬人，但其实sin1/x不会大于1，而arctanx在本题条件下极限是0。所以这道弱题的结果是ln2

Neutral

LittleNew @ Sep 2nd 2008 8:54PM

BTW，其实我觉得这个验证码，还是猜hot的挺难，我看来看去，勉强找到两个，实在找不到第三个。

Neutral

kirc @ Sep 2nd 2008 10:20PM

而且还有两个照片是一样的.
我估计正确答案可能有4\5个. 脱裙子的和牛仔裤都看不清..

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Reader Comments (Page 1 of 1)

LittleNew @ Sep 2nd 2008 8:52PM

LittleNew @ Sep 2nd 2008 8:54PM

kirc @ Sep 2nd 2008 10:20PM

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