1975 AMC 12 真题(P版)
中英双语:每题先中文,后附英文原题;解答在同页下方「查看答案解析与下载」里。
1
下列表达式 \[ \frac{1}{2-\dfrac{1}{2-\dfrac{1}{2-\frac12}}} \] 的值是
A\(\frac34\) B\(\frac45\) C\(\frac56\) D\(\frac67\) E\(\frac65\)
English
The value of \[ \frac{1}{2-\dfrac{1}{2-\dfrac{1}{2-\frac12}}} \] is
(A) \(\frac34\)(B) \(\frac45\)(C) \(\frac56\)(D) \(\frac67\)(E) \(\frac65\)
2
对于哪些实数 \(m\),联立方程 \[ \begin{aligned} y&=mx+3,\\ y&=(2m-1)x+4 \end{aligned} \] 至少有一组实数解 \((x,y)\)?
A所有 \(m\) B所有 \(m\ne0\) C所有 \(m\ne\frac12\) D所有 \(m\ne1\) E不存在这样的 \(m\)
English
For which real values of \(m\) are the simultaneous equations \[ \begin{aligned} y&=mx+3,\\ y&=(2m-1)x+4 \end{aligned} \] satisfied by at least one pair of real numbers \((x,y)?\)
(A) all \(m\)(B) all \(m\ne0\)(C) all \(m\ne\frac12\)(D) all \(m\ne1\)(E) no values of \(m\)
3
对于所有实数 \(a\)、\(b\)、\(c\)、\(x\)、\(y\)、\(z\),若它们满足 \(x\lt a\)、\(y\lt b\)、\(z\lt c\),则下列哪些不等式恒成立?
I. \(xy+yz+zx\lt ab+bc+ca\)
II. \(x^2+y^2+z^2\lt a^2+b^2+c^2\)
III. \(xyz\lt abc\)
A没有一个恒成立。 B只有 \(\mathrm{I}\) C只有 \(\mathrm{II}\) D只有 \(\mathrm{III}\) E全部恒成立。
English
Which of the following inequalities are satisfied for all real numbers \(a,\) \(b,\) \(c,\) \(x,\) \(y,\) \(z\) which satisfy \(x\lt a,\) \(y\lt b,\) and \(z\lt c?\)
I. \(xy+yz+zx\lt ab+bc+ca\)
II. \(x^2+y^2+z^2\lt a^2+b^2+c^2\)
III. \(xyz\lt abc\)
(A) None are satisfied.(B) \(\mathrm{I}\) only(C) \(\mathrm{II}\) only(D) \(\mathrm{III}\) only(E) All are satisfied.
4
若第一个正方形的一条边等于第二个正方形的一条对角线,则第一个正方形与第二个正方形的面积之比是多少?
A\(2\) B\(\sqrt2\) C\(\frac12\) D\(2\sqrt2\) E\(4\)
English
If the side of one square is the diagonal of a second square, what is the ratio of the area of the first square to the area of the second?
(A) \(2\)(B) \(\sqrt2\)(C) \(\frac12\)(D) \(2\sqrt2\)(E) \(4\)
5
将多项式 \((x+y)^9\) 按 \(x\) 的降幂展开。在 \(x=p\)、\(y=q\) 处,展开式的第二项与第三项取值相等,其中 \(p\)、\(q\) 均为正数且和为一。\(p\) 的值是多少?
A\(\frac15\) B\(\frac45\) C\(\frac14\) D\(\frac34\) E\(\frac89\)
English
The polynomial \((x+y)^9\) is expanded in decreasing powers of \(x.\) The second and third terms have equal values when evaluated at \(x=p\) and \(y=q,\) where \(p\) and \(q\) are positive numbers whose sum is one. What is the value of \(p?\)
(A) \(\frac15\)(B) \(\frac45\)(C) \(\frac14\)(D) \(\frac34\)(E) \(\frac89\)
6
前八十个正偶数之和减去前八十个正奇数之和等于
A\(0\) B\(20\) C\(40\) D\(60\) E\(80\)
English
The sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is
(A) \(0\)(B) \(20\)(C) \(40\)(D) \(60\)(E) \(80\)
7
对于哪些非零实数 \(x\),\(\frac{|x-|x||}{x}\) 是正整数?
A仅当 \(x\) 为负数时 B仅当 \(x\) 为正数时 C仅当 \(x\) 为偶整数时 D对所有非零实数 \(x\) E不存在这样的非零实数 \(x\)
English
For which nonzero real numbers \(x\) is \(\frac{|x-|x||}{x}\) a positive integer?
(A) for negative \(x\) only(B) for positive \(x\) only(C) only for \(x\) an even integer(D) for all nonzero real numbers \(x\)(E) for no nonzero real numbers \(x\)
8
若命题“这家商店的所有衬衫都在打折”为假,则下列哪些命题一定为真?
I. 这家商店的所有衬衫都按原价出售。
II. 这家商店里至少有一件衬衫不打折。
III. 这家商店里没有衬衫在打折。
IV. 这家商店的衬衫并非全都在打折。
注:命题 I 原文为:这家商店的所有衬衫都不打折。
A只有 \(\mathrm{II}\) B只有 \(\mathrm{IV}\) C只有 \(\mathrm{I}\) 和 \(\mathrm{III}\) D只有 \(\mathrm{II}\) 和 \(\mathrm{IV}\) E只有 \(\mathrm{I}\)、\(\mathrm{II}\) 和 \(\mathrm{IV}\)
English
If the statement “All shirts in this store are on sale.” is false, then which of the following statements must be true?
I. All shirts in this store are at non-sale prices.
II. There is some shirt in this store not on sale.
III. No shirt in this store is on sale.
IV. Not all shirts in this store are on sale.
Note: Originally, statement I read: All shirts in this store are not on sale.
(A) \(\mathrm{II}\) only(B) \(\mathrm{IV}\) only(C) \(\mathrm{I}\) and \(\mathrm{III}\) only(D) \(\mathrm{II}\) and \(\mathrm{IV}\) only(E) \(\mathrm{I},\) \(\mathrm{II}\) and \(\mathrm{IV}\) only
9
已知 \(a_1\)、\(a_2\)、\(\ldots\) 与 \(b_1\)、\(b_2\)、\(\ldots\) 都是等差数列,且 \(a_1=25\)、\(b_1=75\)、\(a_{100}+b_{100}=100\)。求数列 \(a_1+b_1\)、\(a_2+b_2\)、\(\ldots\) 的前一百项之和。
A\(0\) B\(100\) C\(10{,}000\) D\(505{,}000\) E所给信息不足以求解
English
Let \(a_1,\) \(a_2,\) \(\ldots\) and \(b_1,\) \(b_2,\) \(\ldots\) be arithmetic progressions such that \(a_1=25,\) \(b_1=75,\) and \(a_{100}+b_{100}=100.\) Find the sum of the first one hundred terms of the progression \(a_1+b_1,\) \(a_2+b_2,\) \(\ldots.\)
(A) \(0\)(B) \(100\)(C) \(10{,}000\)(D) \(505{,}000\)(E) not enough information given to solve the problem
10
设 \(n\) 是正整数,则 \(\left(10^{4n^2+8}+1\right)^2\) 的十进制各位数字之和为
A\(4\) B\(4n\) C\(2+2n\) D\(4n^2\) E\(n^2+n+2\)
English
The sum of the digits in base ten of \(\left(10^{4n^2+8}+1\right)^2,\) where \(n\) is a positive integer, is
(A) \(4\)(B) \(4n\)(C) \(2+2n\)(D) \(4n^2\)(E) \(n^2+n+2\)
11
设 \(P\) 是圆 \(K\) 内异于圆 \(K\) 圆心的一点。作圆 \(K\) 中所有经过 \(P\) 的弦,并取这些弦的中点。这些中点的轨迹是
A去掉一点的圆 B若 \(P\) 到圆 \(K\) 圆心的距离小于圆 \(K\) 半径的一半,则为一个圆;否则为小于 \(360^\circ\) 的一段圆弧 C去掉一点的半圆 D一个半圆 E一个圆
English
Let \(P\) be an interior point of circle \(K\) other than the center of \(K.\) Form all chords of \(K\) which pass through \(P,\) and determine their midpoints. The locus of these midpoints is
(A) a circle with one point deleted(B) a circle if the distance from \(P\) to the center of \(K\) is less than one half the radius of \(K;\) otherwise a circular arc of less than \(360^\circ\)(C) a semicircle with one point deleted(D) a semicircle(E) a circle
12
若 \(a\ne b\)、\(a^3-b^3=19x^3\)、\(a-b=x\),则下列哪个结论正确?
A\(a=3x\) B\(a=3x\) 或 \(a=-2x\) C\(a=-3x\) 或 \(a=2x\) D\(a=3x\) 或 \(a=2x\) E\(a=2x\)
English
If \(a\ne b,\) \(a^3-b^3=19x^3,\) and \(a-b=x,\) which of the following conclusions is correct?
(A) \(a=3x\)(B) \(a=3x\) or \(a=-2x\)(C) \(a=-3x\) or \(a=2x\)(D) \(a=3x\) or \(a=2x\)(E) \(a=2x\)
13
方程 \(x^6-3x^5-6x^3-x+8=0\) 有
A没有实根 B恰有两个不同的负根 C恰有一个负根 D没有负根,但至少有一个正根 E以上都不对
English
The equation \(x^6-3x^5-6x^3-x+8=0\) has
(A) no real roots(B) exactly two distinct negative roots(C) exactly one negative root(D) no negative roots, but at least one positive root(E) none of these
14
若当“乙量”等于“丙量”且“丁量加丁量”等于“丙量”\(\cdot\)“丁量”时,“甲量”等于“丁量”,那么当“乙量”等于“丁量”、“丁量加丁量”等于“丁量”\(\cdot\)“丁量”且“丙量”等于二时,“乙量”\(\cdot\)“甲量”等于什么(甲量、乙量、丙量和丁量都是取正值的变量)?
A乙量 \(\cdot\) 丙量 \(\cdot\) 丁量 B乙量 C丙量 D丁量 E丁量加丁量
English
If the whatsis is so when the whosis is is and the so and so is is \(\cdot\) so, what is the whosis \(\cdot\) whatsis when the whosis is so, the so and so is so \(\cdot\) so, and the is is two (whatsis, whosis, is and so are variables taking positive values)?
(A) whosis \(\cdot\) is \(\cdot\) so(B) whosis(C) is(D) so(E) so and so
15
数列 \(1\)、\(3\)、\(2\)、\(\ldots\) 中,从第三项起,每一项都等于紧邻它的前一项减去再前一项。该数列前一百项之和为
A\(5\) B\(4\) C\(2\) D\(1\) E\(-1\)
English
In the sequence of numbers \(1,\) \(3,\) \(2,\) \(\ldots,\) each term after the first two is equal to the term preceding it minus the term preceding that. The sum of the first one hundred terms of the sequence is
(A) \(5\)(B) \(4\)(C) \(2\)(D) \(1\)(E) \(-1\)
16
一个无穷等比级数的首项为正整数,公比为正整数的倒数,且级数之和为 \(3\),则其前两项之和为
A\(\frac13\) B\(\frac23\) C\(\frac83\) D\(2\) E\(\frac92\)
English
If the first term of an infinite geometric series is a positive integer, the common ratio is the reciprocal of a positive integer, and the sum of the series is \(3,\) then the sum of the first two terms of the series is
(A) \(\frac13\)(B) \(\frac23\)(C) \(\frac83\)(D) \(2\)(E) \(\frac92\)
17
一名男子上下班可以乘火车或公交车。如果早晨乘火车上班,他下午就乘公交车回家;如果下午乘火车回家,他早晨就乘公交车上班。在总共 \(x\) 个工作日中,他早晨乘公交车上班 \(8\) 次,下午乘公交车回家 \(15\) 次,并且共乘火车通勤(早晨或下午)\(9\) 次。求 \(x\)。
A\(19\) B\(18\) C\(17\) D\(16\) E所给信息不足以求解
English
A man can commute either by train or by bus. If he goes to work on the train in the morning, he comes home on the bus in the afternoon; and if he comes home in the afternoon on the train, he took the bus in the morning. During a total of \(x\) working days, the man took the bus to work in the morning \(8\) times, came home by bus in the afternoon \(15\) times, and commuted by train (either morning or afternoon) \(9\) times. Find \(x.\)
(A) \(19\)(B) \(18\)(C) \(17\)(D) \(16\)(E) not enough information given to solve the problem
18
从所有十进制三位正整数中等可能地随机选取一个正整数 \(N\)。\(\log_2N\) 为整数的概率是
A\(0\) B\(\frac3{899}\) C\(\frac1{225}\) D\(\frac1{300}\) E\(\frac1{450}\)
English
A positive integer \(N\) with three digits in its base ten representation is chosen at random, with each three-digit number having an equal chance of being chosen. The probability that \(\log_2N\) is an integer is
(A) \(0\)(B) \(\frac3{899}\)(C) \(\frac1{225}\)(D) \(\frac1{300}\)(E) \(\frac1{450}\)
19
哪些正数 \(x\) 满足方程 \((\log_3x)(\log_x5)=\log_35\)?
A只有 \(3\) 和 \(5\) B只有 \(3\)、\(5\) 和 \(15\) C只有形如 \(5^n\cdot3^m\) 的数,其中 \(n\) 和 \(m\) 为正整数 D所有满足 \(x\ne1\) 的正数 E以上都不对
English
Which positive numbers \(x\) satisfy the equation \((\log_3x)(\log_x5)=\log_35?\)
(A) \(3\) and \(5\) only(B) \(3,\) \(5\) and \(15\) only(C) only numbers of the form \(5^n\cdot3^m,\) where \(n\) and \(m\) are positive integers(D) all positive \(x\ne1\)(E) none of these
20
如图,在三角形 \(ABC\) 中,\(AB=4\)、\(AC=8\)。若 \(M\) 是 \(BC\) 的中点且 \(AM=3\),则 \(BC\) 的长度是多少?
A\(2\sqrt{26}\) B\(2\sqrt{31}\) C\(9\) D\(4+2\sqrt{13}\) E所给信息不足以求解
English
In the adjoining figure triangle \(ABC\) is such that \(AB=4\) and \(AC=8.\) If \(M\) is the midpoint of \(BC\) and \(AM=3,\) what is the length of \(BC?\)
(A) \(2\sqrt{26}\)(B) \(2\sqrt{31}\)(C) \(9\)(D) \(4+2\sqrt{13}\)(E) not enough information given to solve the problem
21
设 \(f(x)\) 对所有实数 \(x\) 都有定义;\(f(x)\gt0\) 对所有 \(x\) 成立;且 \(f(a)f(b)=f(a+b)\) 对所有 \(a\) 和 \(b\) 成立。下列哪些命题为真?
I. \(f(0)=1\)
II. \(f(-a)=\frac{1}{f(a)}\) 对所有 \(a\) 成立
III. \(f(a)=\sqrt[3]{f(3a)}\) 对所有 \(a\) 成立
IV. \(f(b)\gt f(a)\),只要 \(b\gt a\)
A只有 \(\mathrm{III}\) 和 \(\mathrm{IV}\) B只有 \(\mathrm{I}\)、\(\mathrm{III}\) 和 \(\mathrm{IV}\) C只有 \(\mathrm{I}\)、\(\mathrm{II}\) 和 \(\mathrm{IV}\) D只有 \(\mathrm{I}\)、\(\mathrm{II}\) 和 \(\mathrm{III}\) E全部为真。
English
Suppose \(f(x)\) is defined for all real numbers \(x;\) \(f(x)\gt0\) for all \(x;\) and \(f(a)f(b)=f(a+b)\) for all \(a\) and \(b.\) Which of the following statements are true?
I. \(f(0)=1\)
II. \(f(-a)=\frac{1}{f(a)}\) for all \(a\)
III. \(f(a)=\sqrt[3]{f(3a)}\) for all \(a\)
IV. \(f(b)\gt f(a)\) if \(b\gt a\)
(A) \(\mathrm{III}\) and \(\mathrm{IV}\) only(B) \(\mathrm{I},\) \(\mathrm{III}\) and \(\mathrm{IV}\) only(C) \(\mathrm{I},\) \(\mathrm{II}\) and \(\mathrm{IV}\) only(D) \(\mathrm{I},\) \(\mathrm{II}\) and \(\mathrm{III}\) only(E) All are true.
22
若 \(p\) 和 \(q\) 都是质数,且 \(x^2-px+q=0\) 有两个不同的正整数根,则下列哪些命题为真?
I. 两根之差为奇数。
II. 至少有一个根是质数。
III. \(p^2-q\) 是质数。
IV. \(p+q\) 是质数。
A只有 \(\mathrm{I}\) B只有 \(\mathrm{II}\) C只有 \(\mathrm{II}\) 和 \(\mathrm{III}\) D只有 \(\mathrm{I}\)、\(\mathrm{II}\) 和 \(\mathrm{IV}\) E全部为真。
English
If \(p\) and \(q\) are primes and \(x^2-px+q=0\) has distinct positive integral roots, then which of the following statements are true?
I. The difference of the roots is odd.
II. At least one root is prime.
III. \(p^2-q\) is prime.
IV. \(p+q\) is prime.
(A) \(\mathrm{I}\) only(B) \(\mathrm{II}\) only(C) \(\mathrm{II}\) and \(\mathrm{III}\) only(D) \(\mathrm{I},\) \(\mathrm{II}\) and \(\mathrm{IV}\) only(E) All are true.
23
如图,\(AB\) 和 \(BC\) 是正方形 \(ABCD\) 的相邻两边;\(M\) 是 \(AB\) 的中点;\(N\) 是 \(BC\) 的中点;\(AN\) 与 \(CM\) 交于 \(O\)。四边形 \(AOCD\) 与正方形 \(ABCD\) 的面积之比为
A\(\frac56\) B\(\frac34\) C\(\frac23\) D\(\frac{\sqrt3}{2}\) E\(\frac{\sqrt3-1}{2}\)
English
In the adjoining figure \(AB\) and \(BC\) are adjacent sides of square \(ABCD;\) \(M\) is the midpoint of \(AB;\) \(N\) is the midpoint of \(BC;\) and \(AN\) and \(CM\) intersect at \(O.\) The ratio of the area of \(AOCD\) to the area of \(ABCD\) is
(A) \(\frac56\)(B) \(\frac34\)(C) \(\frac23\)(D) \(\frac{\sqrt3}{2}\)(E) \(\frac{\sqrt3-1}{2}\)
24
在三角形 \(ABC\) 中,\(\angle C=\theta\)、\(\angle B=2\theta\),其中 \(0^\circ\lt\theta\lt60^\circ\)。以 \(A\) 为圆心、\(AB\) 为半径的圆与 \(AC\) 交于 \(D\),并与 \(BC\)(必要时延长)交于 \(B\) 和 \(E\)(\(E\) 可能与 \(B\) 重合)。则等式 \(EC=AD\)
A对任何 \(\theta\) 都不成立 B仅当 \(\theta=45^\circ\) 时成立 C仅当 \(0^\circ\lt\theta\le45^\circ\) 时成立 D仅当 \(45^\circ\le\theta\lt60^\circ\) 时成立 E对所有 \(\theta\) 都成立,其中 \(0^\circ\lt\theta\lt60^\circ\)
English
In triangle \(ABC,\) \(\angle C=\theta\) and \(\angle B=2\theta,\) where \(0^\circ\lt\theta\lt60^\circ.\) The circle with center \(A\) and radius \(AB\) intersects \(AC\) at \(D\) and intersects \(BC,\) extended if necessary, at \(B\) and at \(E\) (\(E\) may coincide with \(B\)). Then \(EC=AD\)
(A) for no values of \(\theta\)(B) only if \(\theta=45^\circ\)(C) only if \(0^\circ\lt\theta\le45^\circ\)(D) only if \(45^\circ\le\theta\lt60^\circ\)(E) for all \(\theta\) such that \(0^\circ\lt\theta\lt60^\circ\)
25
一名女子、她的兄弟、她的儿子和她的女儿都是国际象棋棋手(所有亲属关系均为血缘关系)。水平最差者的双胞胎手足也是这四人之一,且与水平最佳者性别相反。水平最差者与水平最佳者年龄相同。谁的水平最差?
A这名女子 B她的儿子 C她的兄弟 D她的女儿 E不存在符合所给信息的答案。
English
A woman, her brother, her son and her daughter are chess players (all relations by birth). The worst player’s twin (who is one of the four players) and the best player are of opposite sex. The worst player and the best player are the same age. Who is the worst player?
(A) the woman(B) her son(C) her brother(D) her daughter(E) No solution is consistent with the given information.
26
在锐角三角形 \(ABC\) 中,\(\angle A\) 的角平分线与边 \(BC\) 交于 \(D\)。以 \(B\) 为圆心、\(BD\) 为半径的圆与边 \(AB\) 交于 \(M\);以 \(C\) 为圆心、\(CD\) 为半径的圆与边 \(AC\) 交于 \(N\)。则恒有
A\(\begin{aligned}\angle CND+\angle BMD\\{}-\angle DAC=120^\circ\end{aligned}\) B\(AMDN\) 是梯形 C\(BC\) 平行于 \(MN\) D\(AM-AN=\frac{3(DB-DC)}2\) E\(AB-AC=\frac{3(DB-DC)}2\)
English
In acute triangle \(ABC\) the bisector of \(\angle A\) meets side \(BC\) at \(D.\) The circle with center \(B\) and radius \(BD\) intersects side \(AB\) at \(M;\) and the circle with center \(C\) and radius \(CD\) intersects side \(AC\) at \(N.\) Then it is always true that
(A) \(\begin{aligned}\angle CND+\angle BMD\\{}-\angle DAC=120^\circ\end{aligned}\)(B) \(AMDN\) is a trapezoid(C) \(BC\) is parallel to \(MN\)(D) \(AM-AN=\frac{3(DB-DC)}2\)(E) \(AB-AC=\frac{3(DB-DC)}2\)
27
若 \(p\)、\(q\)、\(r\) 是方程 \(x^3-x^2+x-2=0\) 的三个不同根,则 \(p^3+q^3+r^3\) 等于
A\(-1\) B\(1\) C\(3\) D\(5\) E以上都不对
English
If \(p,\) \(q,\) and \(r\) are distinct roots of \(x^3-x^2+x-2=0,\) then \(p^3+q^3+r^3\) equals
(A) \(-1\)(B) \(1\)(C) \(3\)(D) \(5\)(E) none of these
28
如图,在三角形 \(ABC\) 中,\(M\) 是边 \(BC\) 的中点,\(AB=12\)、\(AC=16\)。点 \(E\) 和 \(F\) 分别取在 \(AC\) 和 \(AB\) 上,直线 \(EF\) 与 \(AM\) 交于 \(G\)。若 \(AE=2AF\),则 \(\frac{EG}{GF}\) 等于
A\(\frac32\) B\(\frac43\) C\(\frac54\) D\(\frac65\) E所给信息不足以求解
English
In triangle \(ABC\) shown in the adjoining figure, \(M\) is the midpoint of side \(BC,\) \(AB=12,\) and \(AC=16.\) Points \(E\) and \(F\) are taken on \(AC\) and \(AB,\) respectively, and lines \(EF\) and \(AM\) intersect at \(G.\) If \(AE=2AF,\) then \(\frac{EG}{GF}\) equals
(A) \(\frac32\)(B) \(\frac43\)(C) \(\frac54\)(D) \(\frac65\)(E) not enough information given to solve the problem
29
大于 \((\sqrt3+\sqrt2)^6\) 的最小整数是多少?
A\(972\) B\(971\) C\(970\) D\(969\) E\(968\)
English
What is the smallest integer larger than \((\sqrt3+\sqrt2)^6?\)
(A) \(972\)(B) \(971\)(C) \(970\)(D) \(969\)(E) \(968\)
30
令 \(x=\cos36^\circ-\cos72^\circ\)。则 \(x\) 等于
A\(\frac13\) B\(\frac12\) C\(3-\sqrt6\) D\(2\sqrt3-3\) E以上都不对
English
Let \(x=\cos36^\circ-\cos72^\circ.\) Then \(x\) equals
(A) \(\frac13\)(B) \(\frac12\)(C) \(3-\sqrt6\)(D) \(2\sqrt3-3\)(E) none of these
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| 题号 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 答案 | B | D | A | A | B | E | E | D | C |
| 题号 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
| 答案 | A | E | B | D | E | A | C | D | D |
| 题号 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 |
| 答案 | D | B | D | E | C | E | B | C | E |
| 题号 | 28 | 29 | 30 |
| 答案 | A | C | B |
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逐题解析
1 答案 B
由内向外计算,\[ \begin{aligned} 2-\frac12&=\frac32,\\ 2-\frac1{\frac{3}{2}}&=\frac43,\\ 2-\frac1{\frac{4}{3}}&=\frac54\text{。} \end{aligned} \] 因此所求值为 \(\frac{1}{\frac{5}{4}}=\frac{4}{5}\)。
因此,正确答案是 B。
English
Working outward, \[ \begin{aligned} 2-\frac12&=\frac32,\\ 2-\frac1{\frac{3}{2}}&=\frac43,\\ 2-\frac1{\frac{4}{3}}&=\frac54. \end{aligned} \] The given value is therefore \(\frac{1}{\frac{5}{4}}=\frac{4}{5}.\)
Therefore, the correct answer is B.
2 答案 D
两直线只有在 \(m=2m-1\) 时才可能不相交,由此得 \(m=1\)。它们在 \(y\) 轴上的截距分别为 \(3\) 和 \(4\),所以此时是两条不重合的平行直线。其余每个 \(m\) 都对应一个交点。
因此,正确答案是 D。
English
The lines fail to intersect only if \(m=2m-1,\) which gives \(m=1.\) Their \(y\)-intercepts are \(3\) and \(4,\) so at that value they are distinct parallel lines. Every other \(m\) gives one intersection.
Therefore, the correct answer is D.
3 答案 A
取 \(a=b=1,\ c=-1\) 和 \(x=y=0,\ z=-10\)。题设的三个不等式均满足。但是 I 变为 \(0\lt-1\),II 变为 \(100\lt3\),III 变为 \(0\lt-1\),全都不成立。因此没有一个不等式恒成立。
因此,正确答案是 A。
English
Take \(a=b=1,\ c=-1,\) and \(x=y=0,\ z=-10.\) The required three coordinate inequalities hold. But I becomes \(0\lt-1,\) II becomes \(100\lt3,\) and III becomes \(0\lt-1,\) all false. Thus none is universally true.
Therefore, the correct answer is A.
4 答案 A
设第一个正方形的边长为 \(s\)。第二个正方形的对角线长为 \(s\),所以其边长为 \(\frac{s}{\sqrt2}\),面积为 \(\frac{s^2}{2}\)。所求面积比为 \(\frac{s^2}{\frac{s^2}{2}}=2\)。
因此,正确答案是 A。
English
Let the first square have side \(s.\) The second has diagonal \(s,\) so its side is \(\frac{s}{\sqrt2}\) and its area is \(\frac{s^2}{2}.\) The ratio is \(\frac{s^2}{\frac{s^2}{2}}=2.\)
Therefore, the correct answer is A.
5 答案 B
这两项分别为 \(9p^8q\) 和 \(36p^7q^2\)。由两者相等且 \(p,q\gt0\),得 \(p=4q\)。又因 \(p+q=1\),所以 \(5q=1\),从而 \(p=\frac{4}{5}\)。
因此,正确答案是 B。
English
The terms are \(9p^8q\) and \(36p^7q^2.\) Their equality, with \(p,q\gt0,\) gives \(p=4q.\) Since \(p+q=1,\) we get \(5q=1\) and \(p=\frac{4}{5}.\)
Therefore, the correct answer is B.
6 答案 E
所求差为 \[ \begin{aligned} &(2-1)+(4-3)+\cdots\\ &\qquad +(160-159)\text{。} \end{aligned} \] 这是 \(80\) 个一的和,因此等于 \(80\)。
因此,正确答案是 E。
English
The desired difference is \[ \begin{aligned} &(2-1)+(4-3)+\cdots\\ &\qquad +(160-159). \end{aligned} \] It is the sum of \(80\) ones, hence \(80.\)
Therefore, the correct answer is E.
7 答案 E
若 \(x\gt0\),该商为 \(0\)。若 \(x\lt0\),则 \(x-|x|=2x\lt0\),所以 \(|x-|x||=-2x\),该商为 \(-2\)。这两个值都不是正整数。
因此,正确答案是 E。
English
If \(x\gt0,\) the quotient is \(0.\) If \(x\lt0,\) then \(x-|x|=2x\lt0,\) so \(|x-|x||=-2x\) and the quotient is \(-2.\) Neither value is a positive integer.
Therefore, the correct answer is E.
8 答案 D
“每件衬衫都在打折”的否定是“有些衬衫不打折”,即命题 II,也与命题 IV 等价。命题 I 和 III 都作出了更强的断言,即没有衬衫打折,而这并不能由题设推出。
因此,正确答案是 D。
English
The negation of “Every shirt is on sale” is “Some shirt is not on sale,” which is II and is equivalently phrased by IV. Statements I and III make the stronger claim that no shirt is on sale, which need not follow.
Therefore, the correct answer is D.
9 答案 C
令 \(c_i=a_i+b_i\)。这是一个等差数列,且 \(c_1=25+75=100\)、\(c_{100}=100\)。首末两项相等,故公差为 \(0\),所以全部 \(100\) 项都等于 \(100\),其和为 \(10{,}000\)。
因此,正确答案是 C。
English
Let \(c_i=a_i+b_i.\) This is an arithmetic progression with \(c_1=25+75=100\) and \(c_{100}=100.\) Equal endpoint terms force its common difference to be \(0,\) so all \(100\) terms equal \(100.\) Their sum is \(10{,}000.\)
Therefore, the correct answer is C.
10 答案 A
令 \(k=4n^2+8\),则 \[ (10^k+1)^2=10^{2k}+2\cdot10^k+1\text{。} \] 它仅有的三个非零数字为 \(1,2,1\),所以各位数字之和为 \(4\)。
因此,正确答案是 A。
English
For \(k=4n^2+8,\) \[ (10^k+1)^2=10^{2k}+2\cdot10^k+1. \] Its only nonzero digits are \(1,2,1,\) so their sum is \(4.\)
Therefore, the correct answer is A.
11 答案 E
若 \(M\) 是弦的中点、\(O\) 是圆心,则 \(OM\) 垂直于该弦,所以 \(\angle OMP=90^\circ\),但 \(M=O\) 时除外。因此 \(M\) 位于以 \(OP\) 为直径的圆上。反过来,对该圆上不是端点的每个 \(M\),都有 \(PM\perp OM\),所以直线 \(PM\) 截得圆 \(K\) 的一条以 \(M\) 为中点的弦。端点 \(M=P\) 对应经过 \(P\) 且垂直于 \(OP\) 的弦,而 \(M=O\) 对应经过 \(O\) 和 \(P\) 的直径。
因此,正确答案是 E。
English
If \(M\) is a chord midpoint and \(O\) is the center, then \(OM\) is perpendicular to the chord, so \(\angle OMP=90^\circ\) unless \(M=O.\) Thus \(M\) lies on the circle with diameter \(OP.\) Conversely, each nonendpoint \(M\) on that circle gives \(PM\perp OM,\) so the line \(PM\) cuts a chord of \(K\) whose midpoint is \(M.\) The endpoint \(M=P\) gives the chord through \(P\) perpendicular to \(OP,\) while \(M=O\) gives the diameter through \(O\) and \(P.\)
Therefore, the correct answer is E.
12 答案 B
因式分解并除以 \(x\ne0\),得 \(a^2+ab+b^2=19x^2\)。代入 \(b=a-x\),\[ 3a^2-3ax+x^2=19x^2\text{,} \] 因而 \[ \begin{aligned} a^2-ax-6x^2&=0,\\ (a-3x)(a+2x)&=0\text{。} \end{aligned} \] 所以 \(a=3x\) 或 \(a=-2x\)。
因此,正确答案是 B。
English
Factoring and dividing by \(x\ne0\) gives \(a^2+ab+b^2=19x^2.\) With \(b=a-x,\) \[ 3a^2-3ax+x^2=19x^2, \] so \[ \begin{aligned} a^2-ax-6x^2&=0,\\ (a-3x)(a+2x)&=0. \end{aligned} \] Hence \(a=3x\) or \(a=-2x.\)
Therefore, the correct answer is B.
13 答案 D
当 \(x\lt0\) 时,\(x^6,-3x^5,-6x^3,-x\) 和 \(8\) 均为正数,因此没有负根。该多项式在 \(x=0\) 时的值为 \(8\),在 \(x=1\) 时的值为 \(-1\)。由连续性可知,在 \(0\) 与 \(1\) 之间至少有一个正根。
因此,正确答案是 D。
English
For \(x\lt0,\) each of \(x^6,-3x^5,-6x^3,-x,\) and \(8\) is positive, so there is no negative root. At \(x=0\) the polynomial is \(8,\) while at \(x=1\) it is \(-1.\) Continuity therefore gives a positive root between \(0\) and \(1.\)
Therefore, the correct answer is D.
14 答案 E
第一个条件说明,当 \(H=I\) 且 \(2S=IS\) 时有 \(W=S\)。由于 \(S\gt0\),后一个等式给出 \(I=2\)。在所求情形中,\(H=S,\ 2S=S^2\) 且 \(I=2\),由各量为正得 \(H=I=S=2\)。因此 \(W=S=2\),且 \(HW=4=S+S\)。
因此,正确答案是 E。
English
The first condition says that \(H=I\) and \(2S=IS\) imply \(W=S.\) Since \(S\gt0,\) the latter equation gives \(I=2.\) In the requested case, \(H=S,\ 2S=S^2,\) and \(I=2,\) so positivity gives \(H=I=S=2.\) Hence \(W=S=2\) and \(HW=4=S+S.\)
Therefore, the correct answer is E.
15 答案 A
该数列开头为 \[ 1,3,2,-1,-3,-2,1,3,\ldots \] 每 \(6\) 项重复一次,且一个周期内各项之和为 \(0\)。前 \(96\) 项之和为 \(0\),最后四项之和为 \(1+3+2-1=5\)。
因此,正确答案是 A。
English
The sequence begins \[ 1,3,2,-1,-3,-2,1,3,\ldots \] and repeats every \(6\) terms, with period sum \(0.\) The first \(96\) terms sum to \(0,\) and the last four sum to \(1+3+2-1=5.\)
Therefore, the correct answer is A.
16 答案 C
设首项为 \(a\),公比为 \(\frac{1}{n}\)。由级数收敛知 \(n\gt1\),且 \[ \frac{a}{1-\frac{1}{n}}=3,\qquad a=3-\frac3n\text{。} \] 因而 \(n\) 整除 \(3\),所以 \(n=3\)、\(a=2\)。前两项之和为 \(2+\frac{2}{3}=\frac{8}{3}\)。
因此,正确答案是 C。
English
Let the first term be \(a\) and the ratio \(\frac{1}{n}.\) Convergence gives \(n\gt1,\) and \[ \frac{a}{1-\frac{1}{n}}=3,\qquad a=3-\frac3n. \] Thus \(n\) divides \(3,\) so \(n=3\) and \(a=2.\) The first two terms sum to \(2+\frac{2}{3}=\frac{8}{3}.\)
Therefore, the correct answer is C.
17 答案 D
公交车行程共有 \(8+15=23\) 次,火车行程共有 \(9\) 次,因此单程通勤一共 \(32\) 次。每个工作日对应两次行程,所以 \(2x=32\),从而 \(x=16\)。题中的条件相互一致,但这次计数并不需要用到它们。
因此,正确答案是 D。
English
There were \(8+15=23\) bus trips and \(9\) train trips, hence \(32\) one-way trips in all. Since each working day contributes two trips, \(2x=32\) and \(x=16.\) The conditional statements are consistent but not needed for this count.
Therefore, the correct answer is D.
18 答案 D
三位整数共有 \(999-100+1=900\) 个。其中 \(2\) 的幂有三个,分别是 \(128,256\) 和 \(512\)。因此所求概率为 \(\frac{3}{900}=\frac{1}{300}\)。
因此,正确答案是 D。
English
There are \(999-100+1=900\) three-digit integers. The three powers of \(2\) among them are \(128,256,\) and \(512.\) Thus the probability is \(\frac{3}{900}=\frac{1}{300}.\)
Therefore, the correct answer is D.
19 答案 D
当 \(x\gt0,\ x\ne1\) 时,换底公式将左边化为 \[ \frac{\log x}{\log3}\cdot\frac{\log5}{\log x}\text{。} \] 约去 \(\log x\) 后得到 \(\frac{\log5}{\log3}=\log_35\)。当 \(x=1\) 时原式无定义。
因此,正确答案是 D。
English
For \(x\gt0,\ x\ne1,\) change of base makes the left side \[ \frac{\log x}{\log3}\cdot\frac{\log5}{\log x}. \] Canceling \(\log x\) leaves \(\frac{\log5}{\log3}=\log_35.\) The expression is undefined at \(x=1.\)
Therefore, the correct answer is D.
20 答案 B
令 \(BM=CM=x\),则 \(BC=2x\)。由阿波罗尼奥斯定理,\[ 4^2+8^2=2(3^2+x^2)\text{。} \] 因此 \(80=18+2x^2,\ x^2=31\),从而 \(BC=2\sqrt{31}\)。
因此,正确答案是 B。
English
Let \(BM=CM=x,\) so \(BC=2x.\) By Apollonius’ theorem, \[ 4^2+8^2=2(3^2+x^2). \] Hence \(80=18+2x^2,\ x^2=31,\) and \(BC=2\sqrt{31}.\)
Therefore, the correct answer is B.
21 答案 D
取 \(a=0\),结合函数值为正可得 \(f(0)=1\)。取 \(b=-a\),得 \(f(a)f(-a)=1\)。另外,\(f(3a)=f(a)^3\),所以由函数值为正,可以在 III 中取正的立方根。但是 \(f(x)=1\) 满足该函数方程却不是严格递增函数,因此 IV 不一定成立。
因此,正确答案是 D。
English
Taking \(a=0\) and using positivity gives \(f(0)=1.\) Taking \(b=-a\) gives \(f(a)f(-a)=1.\) Also \(f(3a)=f(a)^3,\) so positivity permits the positive cube root in III. But \(f(x)=1\) satisfies the functional equation and is not strictly increasing, so IV need not hold.
Therefore, the correct answer is D.
22 答案 E
两个正整数根的乘积为 \(q\),所以两根是 \(1\) 和 \(q\)。它们的和为 \(p=q+1\),只有当 \(q=2\) 时这个和才是质数,此时 \(p=3\),两根为 \(1,2\)。两根之差为 \(1\),其中一个根是质数,\(p^2-q=7\),且 \(p+q=5\);因此四个命题全部成立。
因此,正确答案是 E。
English
The positive integer roots have product \(q,\) so they are \(1\) and \(q.\) Their sum is \(p=q+1,\) which is prime only when \(q=2,\) giving \(p=3\) and roots \(1,2.\) Their difference is \(1,\) one root is prime, \(p^2-q=7,\) and \(p+q=5;\) therefore all four statements hold.
Therefore, the correct answer is E.
23 答案 C
取 \(A=(0,0),B=(1,0)\)、\(C=(1,1),D=(0,1)\)。因为 \(O\) 是三角形 \(ABC\) 的重心,所以 \(O=(\frac{2}{3},\frac{1}{3})\)。三角形 \(AOB\) 和 \(COB\) 的面积都为 \(\frac{1}{6}\)。从单位正方形中去掉这两个三角形,得到 \[ [AOCD]=1-\frac16-\frac16=\frac23\text{。} \]
因此,正确答案是 C。
English
Take \(A=(0,0),B=(1,0),\) \(C=(1,1),D=(0,1).\) Since \(O\) is the centroid of triangle \(ABC,\) \(O=(\frac{2}{3},\frac{1}{3}).\) Triangles \(AOB\) and \(COB\) each have area \(\frac{1}{6}.\) Removing them from the unit square leaves \[ [AOCD]=1-\frac16-\frac16=\frac23. \]
Therefore, the correct answer is C.
24 答案 E
由于 \(AB=AE\),三角形 \(ABE\) 是等腰三角形。当 \(0^\circ\lt\theta\lt45^\circ\) 时,\(E\) 位于 \(B,C\) 之间;三角形 \(AEC\) 的外角定理给出 \(2\theta=\angle EAC+\theta\),所以 \(\angle EAC=\theta\)。当 \(45^\circ\lt\theta\lt60^\circ\) 时,\(E\) 位于 \(B\) 的外侧;此时 \(\angle AEC=180^\circ-2\theta\),于是 \[ \begin{aligned} \angle EAC &=180^\circ\\ &\quad-(180^\circ-2\theta)-\theta\\ &=\theta\text{。} \end{aligned} \] 当 \(\theta=45^\circ,\ E=B\) 时,结论显然成立。因此三角形 \(AEC\) 始终是等腰三角形,所以 \(EC=EA=AD\)。
因此,正确答案是 E。
English
Since \(AB=AE,\) triangle \(ABE\) is isosceles. For \(0^\circ\lt\theta\lt45^\circ,\) \(E\) lies between \(B,C\); the exterior-angle theorem in triangle \(AEC\) gives \(2\theta=\angle EAC+\theta,\) so \(\angle EAC=\theta.\) For \(45^\circ\lt\theta\lt60^\circ,\) \(E\) lies beyond \(B\); then \(\angle AEC=180^\circ-2\theta,\) hence \[ \begin{aligned} \angle EAC &=180^\circ\\ &\quad-(180^\circ-2\theta)-\theta\\ &=\theta. \end{aligned} \] At \(\theta=45^\circ,\ E=B\) and the conclusion is immediate. Thus triangle \(AEC\) is always isosceles, so \(EC=EA=AD.\)
Therefore, the correct answer is E.
25 答案 B
若儿子水平最差,则女儿可以是他的双胞胎手足,女子的兄弟可以是水平最佳者;儿子与这位兄弟可能同岁,因此所有条件都能满足。若女子水平最差,则她的兄弟是其双胞胎,女儿必须水平最佳,但母女不可能同岁。若兄弟水平最差,则女子是其双胞胎,儿子必须水平最佳,年龄条件仍不可能满足。若女儿水平最差,则儿子是其双胞胎,女子必须水平最佳,同样不可能同岁。因此只有儿子符合条件。
因此,正确答案是 B。
English
If the son is worst, the daughter can be his twin and the brother can be best; the son and brother may have the same age, so all conditions can hold. If the woman is worst, the brother is her twin and the daughter must be best, but mother and daughter cannot be the same age. If the brother is worst, the woman is his twin and the son must be best, again impossible in age. If the daughter is worst, her twin is the son and the woman must be best, also impossible in age. Thus only the son works.
Therefore, the correct answer is B.
26 答案 C
由角平分线定理,\[ \frac{BD}{CD}=\frac{AB}{AC}\text{。} \] 因为 \(BM=BD\) 且 \(CN=CD\),所以 \(\frac{BM}{CN}=\frac{AB}{AC}\)。因此,\(M\) 与 \(N\) 在 \(AB\) 与 \(AC\) 上分别从 \(B\) 与 \(C\) 两端按相同比例分割,由三角形一边平行线判定定理可得 \(MN\parallel BC\)。
因此,正确答案是 C。
English
The angle bisector theorem gives \[ \frac{BD}{CD}=\frac{AB}{AC}. \] Since \(BM=BD\) and \(CN=CD,\) we have \(\frac{BM}{CN}=\frac{AB}{AC}.\) Therefore \(M\) and \(N\) divide \(AB\) and \(AC\) proportionally from \(B\) and \(C,\) so the converse of the side-splitter theorem gives \(MN\parallel BC.\)
Therefore, the correct answer is C.
27 答案 E
由韦达定理得 \(p+q+r=1\) 且 \(pq+pr+qr=1\)。因此 \[ \begin{aligned} p^2+q^2+r^2 &=(p+q+r)^2\\ &\quad-2(pq+pr+qr)\\ &=-1\text{。} \end{aligned} \] 每个根 \(t\) 都满足 \(t^3=t^2-t+2\)。将三个根对应的等式相加,得到 \[ p^3+q^3+r^3=-1-1+6=4\text{。} \] 这个数不在选项 A 到 D 中。
因此,正确答案是 E。
English
Vieta gives \(p+q+r=1\) and \(pq+pr+qr=1.\) Thus \[ \begin{aligned} p^2+q^2+r^2 &=(p+q+r)^2\\ &\quad-2(pq+pr+qr)\\ &=-1. \end{aligned} \] Each root \(t\) satisfies \(t^3=t^2-t+2.\) Summing over the three roots yields \[ p^3+q^3+r^3=-1-1+6=4. \] This is not among choices A–D.
Therefore, the correct answer is E.
28 答案 A
将 \(A\) 置于原点,并将 \(B,C\) 的位置向量分别记为 \(\mathbf b,\mathbf c\)。若 \(F=t\mathbf b\),则 \(AF=12t\)、\(AE=24t\),所以 \(E=(\frac{3t}{2})\mathbf c\)。令 \(G=(1-u)F+uE\)。因为 \(G\) 也位于中线 \(AM\) 上,所以它在 \(\mathbf b\) 与 \(\mathbf c\) 方向的系数相等:\[ (1-u)t=\frac32ut\text{。} \] 因此 \(u=\frac{2}{5}\)。沿着 \(FE,\frac{FG}{FE}=\frac{2}{5}\),并有 \(\frac{EG}{FE}=\frac{3}{5}\),所以 \(\frac{EG}{GF}=\frac{3}{2}\)。
因此,正确答案是 A。
English
Put \(A\) at the origin and write the position vectors of \(B,C\) as \(\mathbf b,\mathbf c.\) If \(F=t\mathbf b,\) then \(AF=12t\) and \(AE=24t,\) so \(E=(\frac{3t}{2})\mathbf c.\) Write \(G=(1-u)F+uE.\) Because \(G\) also lies on the median \(AM,\) its \(\mathbf b\) and \(\mathbf c\) coefficients are equal: \[ (1-u)t=\frac32ut. \] Hence \(u=\frac{2}{5}.\) Along \(FE,\frac{FG}{FE}=\frac{2}{5}\) and \(\frac{EG}{FE}=\frac{3}{5},\) so \(\frac{EG}{GF}=\frac{3}{2}.\)
Therefore, the correct answer is A.
29 答案 C
令 \(\alpha=\sqrt3+\sqrt2\)、\(\beta=\sqrt3-\sqrt2\)。将它们的六次方相加,所有含奇次根式的项都会抵消:\[ \begin{aligned} \alpha^6+\beta^6 &=2\bigl(3^3+15(3^2)(2)\\ &\qquad+15(3)(2^2)+2^3\bigr)\\ &=970\text{。} \end{aligned} \] 因为 \(0\lt\beta\lt1\),所以 \(969\lt\alpha^6=970-\beta^6\lt970\)。因此,大于它的最小整数为 \(970\)。
因此,正确答案是 C。
English
Let \(\alpha=\sqrt3+\sqrt2\) and \(\beta=\sqrt3-\sqrt2.\) Adding their sixth powers cancels all odd radical terms: \[ \begin{aligned} \alpha^6+\beta^6 &=2\bigl(3^3+15(3^2)(2)\\ &\qquad+15(3)(2^2)+2^3\bigr)\\ &=970. \end{aligned} \] Since \(0\lt\beta\lt1,\) we have \(969\lt\alpha^6=970-\beta^6\lt970.\) Thus the smallest larger integer is \(970.\)
Therefore, the correct answer is C.
30 答案 B
令 \(w=\cos36^\circ\)、\(y=\cos72^\circ\)。由二倍角公式,\[ y=2w^2-1,\qquad w=1-2y^2\text{。} \] 两式相加得 \[ \begin{aligned} w+y&=2(w^2-y^2)\\ &=2(w-y)(w+y)\text{。} \end{aligned} \] 因为 \(w+y\ne0\),两边相除得 \(w-y=\frac{1}{2}\)。因此 \(x=\frac{1}{2}\)。
因此,正确答案是 B。
English
Set \(w=\cos36^\circ\) and \(y=\cos72^\circ.\) The double-angle identities give \[ y=2w^2-1,\qquad w=1-2y^2. \] Adding yields \[ \begin{aligned} w+y&=2(w^2-y^2)\\ &=2(w-y)(w+y). \end{aligned} \] Since \(w+y\ne0,\) division gives \(w-y=\frac{1}{2}.\) Thus \(x=\frac{1}{2}.\)
Therefore, the correct answer is B.