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[转载] 两道牛津数学系入学加试题,找到原英文题目

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发表于 2008-2-1 10:27:34 | 显示全部楼层 |阅读模式

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Q1.
In this question we shall consider the function f (x) defined by
f (x) = x^2-px + 3
where p is a constant.
(i) Show that the function f (x) has one stationary value in the range 0 < x < 1 if0 < p < 1,
and no stationary values in that range otherwise.

In the remainder of the question we shall be interested in the smallest value attained by f (x) in the range 0 ≤ x ≤ 1. Of course, this value, which we shall call m, will depend
on p.
(ii) Show that if p > 1 then m = 4-2p.
(iii) What is the value of m if p ≤ 0?
(iv) Obtain a formula for m in terms of p, valid for 0 < p < 1.
(v)Using the axes opposite, sketch the graph of m as a function of p in the range
-2 ≤ p ≤ 2.

Q2
Songs of the Martian classical period had just two notes (let us call them x and y) and
were constructed according to rigorous rules:
I. the sequence consisting of no notes was deemed to be a song (perhaps the most
pleasant);
II. a sequence starting with x, followed by two repetitions of an existing song and
ending with y was also a song;
III. the sequence of notes obtained by interchanging xs and ys in a song was also a
song.
All songs were constructed using those rules.
(i) Write down four songs of length six (that is, songs with exactly six notes).
(ii) Show that if there are k songs of length m then there are 2k songs of length 2m+2.
Deduce that for each natural number there are 2^n songs of length 2^(n+1)-2.
Songs of the Martian later period were constructed using also the rule:
IV. if a song ended in y then the sequence of notes obtained by omitting that y was
also a song.
(iii)What lengths do songs of the later period have? That is, for which natural numbers n
is there a song with exactly n notes? Justify your answer.

本文转自:http://pm.yhlive.com/thread-28-1-2.html

[ 本帖最后由 282842712474 于 2008-2-1 11:16 编辑 ]
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2008-2-1 11:12:41 | 显示全部楼层
毕竟是入学考试,不难。不过不知道是本科还是研究生入学考试,怎么已经用到微积分了。
不过不懂stationary value的意义,查了下mathworld.wolfram.com才知道,
就是极值的意思。不过奇怪,应该是0<p<2时在0<x<1当中有唯一的极值点 (x0=p/2,f(x0)=3-p^2/4)
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2008-2-1 13:22:22 | 显示全部楼层

这是原版:)

http://www.admissions.ox.ac.uk/interviews/tests/Math.pdf
或直接从论坛下载: Math.pdf (182.58 KB, 下载次数: 9)
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2008-2-1 14:45:14 | 显示全部楼层
我说呢,
f (x) = x^2-px + 3

应该改为
f (x) = x^2-2px + 3

不然根本不成立
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
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