1950 AMC 12 真题（P版）

1 答案 C

这个比共有 \(2+4+6=12\) 个等份。因此，最小的一部分为 \[ 64\left(\frac{2}{12}\right)=\frac{32}{3}=10\frac{2}{3}\text{。}\]

因此，正确答案是 C。

English

The ratio contains \(2+4+6=12\) equal shares. The smallest part is therefore \[ 64\left(\frac{2}{12}\right)=\frac{32}{3}=10\frac{2}{3}. \]

Thus, the correct answer is C.

2 答案 D

代入 \(S=8\) 和 \(R=16\)，得到 \(16=8g-4\)，所以 \(g=\tfrac52\)。因此，当 \(S=10\) 时， \[ R=\frac52(10)-4=21\text{。}\]

因此，正确答案是 D。

English

Substituting \(S=8\) and \(R=16\) gives \(16=8g-4,\) so \(g=\tfrac52.\) Hence, when \(S=10,\) \[ R=\frac52(10)-4=21. \]

Thus, the correct answer is D.

3 答案 E

方程的标准形式为 \(4x^2-8x+5=0\)。由韦达定理，两根之和为 \[ -\frac{-8}{4}=2\text{。}\]这个值不在选项 A 至 D 中。

因此，正确答案是 E。

English

In standard form the equation is \(4x^2-8x+5=0.\) By Vieta’s formulas, the sum of its roots is \[ -\frac{-8}{4}=2. \] This value is not among choices A through D.

Thus, the correct answer is E.

4 答案 A

将各个分子和分母因式分解，得到 \[ \begin{aligned} &\frac{a^2-b^2}{ab} -\frac{ab-b^2}{ab-a^2}\\ &=\left(\frac{a}{b}-\frac{b}{a}\right)\\ &\quad-\frac{b(a-b)}{a(b-a)}\\ &=\frac{a}{b}-\frac{b}{a} +\frac{b}{a}\\ &=\frac{a}{b} \end{aligned}\text{。}\]

因此，正确答案是 A。

English

Factoring the numerators and denominator gives \[ \begin{aligned} &\frac{a^2-b^2}{ab} -\frac{ab-b^2}{ab-a^2}\\ &=\left(\frac{a}{b}-\frac{b}{a}\right)\\ &\quad-\frac{b(a-b)}{a(b-a)}\\ &=\frac{a}{b}-\frac{b}{a} +\frac{b}{a}\\ &=\frac{a}{b}. \end{aligned} \]

Thus, the correct answer is A.

5 答案 A

从第一项 \(8\) 到第七项 \(5832\) 需要连续乘六次公比。因此， \[ q^6=\frac{5832}{8}=729=3^6\text{，}\]所以正公比为 \(q=3\)。第五项是 \[ 8q^4=8\cdot3^4=648\text{。}\]

因此，正确答案是 A。

English

There are six common-ratio steps from the first term \(8\) to the seventh term \(5832.\) Thus \[ q^6=\frac{5832}{8}=729=3^6, \] so the positive common ratio is \(q=3.\) The fifth term is \[ 8q^4=8\cdot3^4=648. \]

Thus, the correct answer is A.

6 答案 C

由第二个方程可得 \(x=-\tfrac{y+3}{2}\)。将其代入第一个方程并乘以 \(2\)，得到 \[ \begin{aligned} &(y+3)^2-6(y+3)\\ &\quad+10y+2=0,\\ &y^2+10y-7=0 \end{aligned}\text{。}\]

因此，正确答案是 C。

English

From the second equation, \(x=-\tfrac{y+3}{2}.\) Substituting into the first equation and multiplying by \(2\) gives \[ \begin{aligned} &(y+3)^2-6(y+3)\\ &\quad+10y+2=0,\\ &y^2+10y-7=0. \end{aligned} \]

Thus, the correct answer is C.

7 答案 B

这个两位数是 \(10t+u\)。在它后面添上 \(1\)，得到 \[ \begin{aligned} &10(10t+u)+1\\ &=100t+10u+1 \end{aligned}\text{。}\]

因此，正确答案是 B。

English

The two digit number is \(10t+u.\) Placing \(1\) after it produces \[ \begin{aligned} &10(10t+u)+1\\ &=100t+10u+1. \end{aligned} \]

Thus, the correct answer is B.

8 答案 C

将 \(r\) 增加 \(100\%\) 后，半径变为 \(2r\)。面积从 \(\pi r^2\) 变为 \(\pi(2r)^2=4\pi r^2\)，增加了 \(3\pi r^2\)，即原面积的 \(300\%\)。

因此，正确答案是 C。

English

Increasing \(r\) by \(100\%\) changes it to \(2r.\) The area changes from \(\pi r^2\) to \(\pi(2r)^2=4\pi r^2,\) an increase of \(3\pi r^2.\) This is \(300\%\) of the original area.

Thus, the correct answer is C.

9 答案 A

半圆内三角形的底边至多为直径 \(2r\)，高至多为 \(r\)。因此，其面积至多为 \[ \frac12(2r)(r)=r^2\text{。}\]取直径为底边、半圆的最高点为第三个顶点时，可以达到这个上界。

因此，正确答案是 A。

English

A triangle in the semicircle has base at most the diameter \(2r\) and altitude at most \(r.\) Therefore its area is at most \[ \frac12(2r)(r)=r^2. \] This bound is attained by using the diameter as the base and the topmost point of the semicircle as the third vertex.

Thus, the correct answer is A.

10 答案 D

分子、分母同乘原分子的共轭式，得到 \[ \begin{aligned} &\frac{\sqrt3-\sqrt2}{\sqrt3} \cdot\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}\\ &=\frac{1}{\sqrt3(\sqrt3+\sqrt2)}\\ &=\frac{1}{3+\sqrt6} \end{aligned}\text{。}\]

因此，正确答案是 D。

English

Multiplying by the conjugate of the numerator gives \[ \begin{aligned} &\frac{\sqrt3-\sqrt2}{\sqrt3} \cdot\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}\\ &=\frac{1}{\sqrt3(\sqrt3+\sqrt2)}\\ &=\frac{1}{3+\sqrt6}. \end{aligned} \]

Thus, the correct answer is D.

11 答案 B

将公式改写为 \[ C=\frac{e}{\frac{R}{n}+r}\text{。}\]当正数 \(n\) 增大时，\(\frac{R}{n}\) 减小，所以正分母减小，而分子保持不变。因此，\(C\) 增大。

因此，正确答案是 B。

English

Rewrite the formula as \[ C=\frac{e}{\frac{R}{n}+r}. \] As positive \(n\) increases, \(\frac{R}{n}\) decreases, so the positive denominator decreases while the numerator remains fixed. Therefore \(C\) increases.

Thus, the correct answer is B.

12 答案 C

沿多边形走一周，并在每个顶点转过相应的外角，恰好完成一整圈。因此，无论边数多少，外角和始终为 \(360^\circ\)。

因此，正确答案是 C。

English

Traversing the polygon and turning through each exterior angle makes one complete turn. Hence the sum is always \(360^\circ,\) independent of the number of sides.

Thus, the correct answer is C.

13 答案 D

由于 \(x^2-3x+2=(x-1)(x-2)\)，原方程化为 \[ (x-1)(x-2)x(x-4)=0\text{。}\]它的根为 \(0\)、\(1\)、\(2\) 和 \(4\)。

因此，正确答案是 D。

English

Since \(x^2-3x+2=(x-1)(x-2),\) the equation becomes \[ (x-1)(x-2)x(x-4)=0. \] Its roots are \(0,\) \(1,\) \(2,\) and \(4.\)

Thus, the correct answer is D.

14 答案 D

第二个方程的左边是 \(-2(2x-3y)\)。若第一个方程成立，这个式子必须等于 \(-2\cdot8=-16\)，但第二个方程却规定它等于 \(9\)。这个矛盾说明不存在同时满足两个方程的有序数对。

因此，正确答案是 D。

English

The left side of the second equation is \(-2(2x-3y).\) If the first equation holds, that expression must equal \(-2\cdot8=-16,\) but the second equation says it equals \(9.\) This contradiction means that no ordered pair satisfies both equations.

Thus, the correct answer is D.

15 答案 E

若有实数一次因式，就会有实根。然而，\(x^2+4=0\) 可推出 \(x^2=-4\)，它没有实数解。因此，这个多项式没有实数一次因式。

因此，正确答案是 E。

English

A real linear factor would give a real root. But \(x^2+4=0\) implies \(x^2=-4,\) which has no real solution. Therefore the polynomial has no real linear factors.

Thus, the correct answer is E.

16 答案 B

利用平方差公式， \[ \begin{aligned} &\big[(a+3b)^2(a-3b)^2\big]^2\\ &=\big[(a^2-9b^2)^2\big]^2\\ &=(a^2-9b^2)^4 \end{aligned}\text{。}\]其二项式展开中，指数的每个取值 \(0,1,2,3,4\) 都对应一个非零项，因此共有五项。

因此，正确答案是 B。

English

Using the difference of squares, \[ \begin{aligned} &\big[(a+3b)^2(a-3b)^2\big]^2\\ &=\big[(a^2-9b^2)^2\big]^2\\ &=(a^2-9b^2)^4. \end{aligned} \] Its binomial expansion has one nonzero term for each exponent choice \(0,1,2,3,4,\) so it has five terms.

Thus, the correct answer is B.

17 答案 C

当 \(x=1\) 时，选项 A 和 C 均得到 \(90\)，而 B 得到 \(95\)，D 得到 \(18\)。再令 \(x=2\) 检验剩下的两个选项：A 得到 \(80\)，而 C 得到 \[ 100-5(2)-5(2^2)=70\text{，}\]与表格相符。

因此，正确答案是 C。

English

At \(x=1,\) choices A and C both give \(90,\) while B gives \(95\) and D gives \(18.\) Testing the two survivors at \(x=2,\) choice A gives \(80,\) whereas choice C gives \[ 100-5(2)-5(2^2)=70, \] matching the table.

Thus, the correct answer is C.

18 答案 E

等式 \(1\) 是分配律。等式 \(2\) 错误，因为 \(a^{x-y}=\frac{a^x}{a^y}\)。等式 \(3\) 错误，因为 \(\log x-\log y=\log(\frac{x}{y})\)。等式 \(4\) 把换底公式中的商误写成了差；等式 \(5\) 的右边应化为 \(a^2xy\)，而不是 \(axy\)。

因此，只有等式 \(1\) 正确，所以正确答案是 E。

English

Statement \(1\) is the distributive property. Statement \(2\) is false because \(a^{x-y}=\frac{a^x}{a^y}.\) Statement \(3\) is false because \(\log x-\log y=\log(\frac{x}{y}).\) Statement \(4\) confuses the change-of-base quotient with a difference, and statement \(5\) has right side \(a^2xy,\) not \(axy.\)

Thus, only statement \(1\) is true, so the correct answer is E.

19 答案 C

这项工作需要 \(m\cdot d=md\) 人日。若 \(m+r\) 名工人的个人效率相同，则所需天数为 \[ \frac{md}{m+r}\text{。}\]

因此，正确答案是 C。

English

The job requires \(m\cdot d=md\) man-days. With \(m+r\) men working at the same individual rate, the required number of days is \[ \frac{md}{m+r}. \]

Thus, the correct answer is C.

20 答案 D

根据余式定理，除以 \(x-1\) 所得的余数等于多项式在 \(x=1\) 处的值，即 \[ 1^{13}+1=2\text{。}\]

因此，正确答案是 D。

English

By the Remainder Theorem, the remainder upon division by \(x-1\) is the value of the polynomial at \(x=1.\) This value is \[ 1^{13}+1=2. \]

Thus, the correct answer is D.

21 答案 B

若三条棱长分别为 \(x\)、\(y\) 和 \(z\)，则三个面的面积分别为 \(xy\)、\(yz\) 和 \(xz\)。因此， \[ \begin{aligned} (xyz)^2&=(xy)(yz)(xz)\\ &=12\cdot8\cdot6=576 \end{aligned}\text{。}\]由于体积为正，所以 \(xyz=\sqrt{576}=24\) 立方英寸。

因此，正确答案是 B。

English

If the edge lengths are \(x,\) \(y,\) and \(z,\) then the three face areas are \(xy,\) \(yz,\) and \(xz.\) Therefore \[ \begin{aligned} (xyz)^2&=(xy)(yz)(xz)\\ &=12\cdot8\cdot6=576. \end{aligned} \] Since the volume is positive, \(xyz=\sqrt{576}=24\) cubic inches.

Thus, the correct answer is B.

22 答案 D

从价格 \(100\) 开始，第一次折扣后剩下 \(90\)。第二次折扣后剩下 \(0.8(90)=72\)。因此，总共减少了 \(100-72=28\)，即 \(28\%\)。

因此，正确答案是 D。

English

Starting from a price of \(100,\) the first discount leaves \(90.\) The second discount leaves \(0.8(90)=72.\) The total reduction is therefore \(100-72=28,\) or \(28\%.\)

Thus, the correct answer is D.

23 答案 B

设每月租金为 \(r\) 美元。全年租金为 \(12r\)，留出维护费后剩下其中的 \(1-0.125=\tfrac78\)。期望的年收益为 \(0.055(10,000)=550\) 美元，所以 \[ \frac78(12r)-325=550\text{。}\]于是 \(10.5r=875\)，得到 \(r=83.\overline3\)，精确到美分为 \(\$83.33\)。

因此，正确答案是 B。

English

Let the monthly rent be \(r\) dollars. The yearly rent is \(12r,\) of which \(1-0.125=\tfrac78\) remains after setting aside the upkeep money. The desired yearly return is \(0.055(10,000)=550\) dollars, so \[ \frac78(12r)-325=550. \] Hence \(10.5r=875,\) giving \(r=83.\overline3,\) which is \(\$83.33\) to the nearest cent.

Thus, the correct answer is B.

24 答案 E

将根式单独置于一边并平方，得到 \[ \begin{aligned} \sqrt{x-2}&=4-x,\\ x-2&=(4-x)^2,\\ x^2-9x+18&=0 \end{aligned}\text{。}\]所得二次方程的根为 \(3\) 和 \(6\)。其中 \(3\) 满足原方程，但将后者代回原方程得 \(6+\sqrt4=8\)，所以 \(6\) 是增根。

因此，正确答案是 E。

English

Isolating and squaring gives \[ \begin{aligned} \sqrt{x-2}&=4-x,\\ x-2&=(4-x)^2,\\ x^2-9x+18&=0. \end{aligned} \] The quadratic roots are \(3\) and \(6.\) The value \(3\) satisfies the original equation, but \(6\) gives \(6+\sqrt4=8,\) so it is extraneous.

Thus, the correct answer is E.

25 答案 D

由于 \(125=5^3\)、\(625=5^4\) 且 \(25=5^2\)， \[ \begin{aligned} &\log_5\frac{(125)(625)}{25}\\ &=\log_5\left(5^{3+4-2}\right)\\ &=\log_5(5^5)=5 \end{aligned}\text{。}\]

因此，正确答案是 D。

English

Since \(125=5^3,\) \(625=5^4,\) and \(25=5^2,\) \[ \begin{aligned} &\log_5\frac{(125)(625)}{25}\\ &=\log_5\left(5^{3+4-2}\right)\\ &=\log_5(5^5)=5. \end{aligned} \]

Thus, the correct answer is D.

26 答案 E

移项并使用对数的乘法法则，得到 \[ \begin{aligned} &\log_{10}m+\log_{10}n\\ &=\log_{10}(mn)=b \end{aligned}\text{。}\]因此 \(mn=10^b\)，所以 \(m=\dfrac{10^b}{n}\)。

因此，正确答案是 E。

English

Rearranging and using the product rule for logarithms gives \[ \begin{aligned} &\log_{10}m+\log_{10}n\\ &=\log_{10}(mn)=b. \end{aligned} \] Therefore \(mn=10^b,\) so \(m=\dfrac{10^b}{n}.\)

Thus, the correct answer is E.

27 答案 B

总路程为 \(240\) 英里。两段路程分别用时 \(4\) 小时和 \(3\) 小时，所以平均速度（单位为英里/小时）为 \[ \frac{240}{4+3}=\frac{240}{7}\approx34.29\text{。}\]这最接近每小时 \(34\) 英里。

因此，正确答案是 B。

English

The total distance is \(240\) miles. The travel times are \(4\) hours and \(3\) hours, so the average speed, in miles per hour, is \[ \frac{240}{4+3}=\frac{240}{7}\approx34.29 \] which is closest to \(34\) mph.

Thus, the correct answer is B.

28 答案 B

相遇时，\(A\) 已行驶 \(60-12=48\) 英里，\(B\) 已行驶 \(60+12=72\) 英里。由于两人行驶的时间相同，速度之比为 \(48:72=2:3\)。若 \(A\) 的速度为 \(v\)，则 \(B\) 的速度为 \(v+4\)，所以 \[ \frac{v}{v+4}=\frac23\text{。}\]于是 \(3v=2v+8\)，得到 \(v=8\)，即每小时八英里。

因此，正确答案是 B。

English

At the meeting point, \(A\) has traveled \(60-12=48\) miles and \(B\) has traveled \(60+12=72\) miles. Since their travel times are equal, their speeds are in the ratio \(48:72=2:3.\) If \(A\)’s speed is \(v,\) then \(B\)’s is \(v+4,\) so \[ \frac{v}{v+4}=\frac23. \] Thus \(3v=2v+8,\) and \(v=8\) mph.

Thus, the correct answer is B.

29 答案 B

第一台机器每分钟完成一批 \(500\) 个信封的 \(\tfrac18\)，第二台每分钟完成一批的 \(\tfrac1x\)。两台一起工作时，每分钟必须完成一批的 \(\tfrac12\)。因此，所需方程为 \[ \frac18+\frac1x=\frac12\text{。}\]

因此，正确答案是 B。

English

The first machine completes \(\tfrac18\) of a \(500\)-envelope batch per minute, and the second completes \(\tfrac1x\) of a batch per minute. Together they must complete \(\tfrac12\) of a batch per minute. Therefore the required equation is \[ \frac18+\frac1x=\frac12. \]

Thus, the correct answer is B.

30 答案 A

设最初有 \(G\) 名女孩和 \(B\) 名男孩。两个条件给出 \[ \begin{aligned} B&=2(G-15),\\ G-15&=5(B-45) \end{aligned}\text{。}\]将第一个方程代入第二个方程，得到 \[ \begin{aligned} G-15 &=5\bigl(2(G-15)-45\bigr)\\ &=10G-375 \end{aligned}\text{。}\]因此 \(9G=360\)，所以 \(G=40\)。

因此，正确答案是 A。

English

Let the original counts be \(G\) girls and \(B\) boys. The two conditions give \[ \begin{aligned} B&=2(G-15),\\ G-15&=5(B-45). \end{aligned} \] Substituting the first into the second yields \[ \begin{aligned} G-15 &=5\bigl(2(G-15)-45\bigr)\\ &=10G-375. \end{aligned} \] Hence \(9G=360,\) so \(G=40.\)

Thus, the correct answer is A.

31 答案 C

设每双蓝袜价格为 \(p\)，则每双黑袜价格为 \(2p\)；再设原订单中蓝袜有 \(n\) 双。原账单为 \((8+n)p\)。调换两种袜子的双数后，账单为 \((2n+4)p\)。后者比前者多 \(50\%\)，所以 \[ 2n+4=\frac32(n+8)\text{。}\]因此 \(n=16\)，原订单中黑袜与蓝袜的双数之比为 \(4:16=1:4\)。

因此，正确答案是 C。

English

Let a blue pair cost \(p,\) so a black pair costs \(2p,\) and let \(n\) be the original number of blue pairs. The original bill is \((8+n)p.\) After the quantities are interchanged, the bill is \((2n+4)p.\) The latter is \(50\%\) greater, so \[ 2n+4=\frac32(n+8). \] Thus \(n=16,\) and the original black-to-blue ratio is \(4:16=1:4.\)

Thus, the correct answer is C.

32 答案 D

梯顶原来的高度（单位为英尺）为 \[ \sqrt{25^2-7^2}=\sqrt{576}=24\text{。}\]梯顶下滑后，高度为 \(20\) 英尺，所以新的水平距离为 \(\sqrt{25^2-20^2}=15\) 英尺。因此，梯脚滑动了 \(15-7=8\) 英尺。

因此，正确答案是 D。

English

The initial height, in feet, is \[ \sqrt{25^2-7^2}=\sqrt{576}=24. \] After the top slips, the height is \(20\) feet, so the new horizontal distance is \(\sqrt{25^2-20^2}=15\) feet. The foot therefore slides \(15-7=8\) feet.

Thus, the correct answer is D.

33 答案 D

两种圆管的直径之比为 \(6:1\)，所以横截面积之比为 \[ 6^2:1^2=36:1\text{。}\]因此，\(36\) 根小圆管的横截面积之和等于大圆管的横截面积。

因此，正确答案是 D。

English

The ratio of the diameters is \(6:1,\) so the ratio of cross-sectional areas is \[ 6^2:1^2=36:1. \] Thus \(36\) of the smaller pipes have the same total cross-sectional area as the larger pipe.

Thus, the correct answer is D.

34 答案 D

因为 \(C=2\pi r\)，所以变化量满足 \[ 25-20=2\pi\mathopen{}\left(r_{\mathrm{new}}-r_{\mathrm{old}}\right)\mathclose{}\text{。}\]因此，半径增加了 \(\dfrac{5}{2\pi}\) 英寸。

因此，正确答案是 D。

English

Because \(C=2\pi r,\) the changes satisfy \[ 25-20=2\pi\mathopen{}\left(r_{\mathrm{new}}-r_{\mathrm{old}}\right)\mathclose{}. \] Hence the radius increases by \(\dfrac{5}{2\pi}\) inches.

Thus, the correct answer is D.

35 答案 B

由于 \(10^2+24^2=26^2\)，这个三角形是直角三角形。其面积为 \(\tfrac12(10)(24)=120\)，半周长为 \(\tfrac{10+24+26}{2}=30\)。利用 \(K=rs\)，内切圆半径（单位为英寸）为 \[ r=\frac{K}{s}=\frac{120}{30}=4\text{。}\]

因此，正确答案是 B。

English

Since \(10^2+24^2=26^2,\) the triangle is right. Its area is \(\tfrac12(10)(24)=120,\) and its semiperimeter is \(\tfrac{10+24+26}{2}=30.\) Using \(K=rs,\) the inradius, in inches, is \[ r=\frac{K}{s}=\frac{120}{30}=4. \]

Thus, the correct answer is B.

36 答案 A

设目录价和标价分别为 \(L\) 和 \(M\)。商人的成本为 \(0.75L\)，打折后的售价为 \(0.8M\)。利润等于售价的 \(25\%\)，意味着成本是售价的其余 \(75\%\)。因此， \[ 0.75L=0.75(0.8M)=0.6M\text{。}\]所以 \(M=1.25L\)，即目录价的 \(125\%\)。

因此，正确答案是 A。

English

Let the list and marked prices be \(L\) and \(M.\) The merchant’s cost is \(0.75L,\) while the selling price after the discount is \(0.8M.\) A profit equal to \(25\%\) of the selling price means that the cost is the remaining \(75\%\) of that price. Therefore \[ 0.75L=0.75(0.8M)=0.6M. \] Thus \(M=1.25L,\) or \(125\%\) of the list price.

Thus, the correct answer is A.

37 答案 E

我们有 \(\log_a1=0\) 和 \(\log_aa=1\)。实对数在 \(-1\) 处没有定义，但其复数值不是实数。当 \(a>1\) 时，在 \(0<x<1\) 上有 \(\log_a x<0\)，并且当 \(x\) 从右侧趋近于 \(0\) 时，它趋于 \(-\infty\)。因此，A 至 D 都正确，“只有一部分正确”才是不正确的说法。

因此，正确答案是 E。

English

We have \(\log_a1=0\) and \(\log_aa=1.\) A real logarithm is not defined at \(-1,\) though its complex values are nonreal. For \(a>1,\) \(\log_a x<0\) on \(0<x<1,\) and it tends to \(-\infty\) as \(x\) approaches \(0\) from the right. Thus statements A through D are all correct, making the claim that only some are correct the incorrect statement.

Thus, the correct answer is E.

38 答案 B

根据给定规则，方程化为 \[ (2x)(x)-(1)(x)=3\text{，}\]即 \(2x^2-x-3=0\)。因式分解得到 \((2x-3)(x+1)=0\)，其两个不同的解为 \(x=\tfrac32\) 和 \(x=-1\)。

因此，正确答案是 B。

English

The given rule turns the equation into \[ (2x)(x)-(1)(x)=3, \] or \(2x^2-x-3=0.\) Factoring gives \((2x-3)(x+1)=0,\) whose two distinct solutions are \(x=\tfrac32\) and \(x=-1.\)

Thus, the correct answer is B.

39 答案 E

部分和递增并趋近于 \[ \frac{2}{1-\frac12}=4\text{，}\]所以它们既不无限增大，也不无限减小。因此，说法 \(1\) 和 \(2\) 错误，而说法 \(4\) 和 \(5\) 正确。

在说法 \(3\) 中，“任意一项”指已经选定的一项，它与零的距离固定，不能再变小；正确的说法应是可以选取足够靠后的项，使其小于任意给定的正数。因此，按原文措辞，说法 \(3\) 错误。

因此，只有说法 \(4\) 和 \(5\) 正确，所以正确答案是 E。

English

The partial sums increase toward \[ \frac{2}{1-\frac12}=4, \] so they neither increase nor decrease without limit. This makes statements \(1\) and \(2\) false, while statements \(4\) and \(5\) are true.

In statement \(3,\) “any term” refers to an already selected term, whose distance from zero is fixed and cannot be made smaller; one can instead choose a sufficiently late term below any prescribed positive bound. Thus statement \(3\), as worded, is false.

Therefore only statements \(4\) and \(5\) are correct, so the correct answer is E.

40 答案 D

当 \(x\ne1\) 时， \[ \begin{aligned} \frac{x^2-1}{x-1} &=\frac{(x-1)(x+1)}{x-1}\\ &=x+1 \end{aligned}\text{。}\]因此，当 \(x\) 趋近于 \(1\) 时，极限为 \(1+1=2\)。

因此，正确答案是 D。

English

For \(x\ne1,\) \[ \begin{aligned} \frac{x^2-1}{x-1} &=\frac{(x-1)(x+1)}{x-1}\\ &=x+1. \end{aligned} \] Therefore the limit as \(x\) approaches \(1\) is \(1+1=2.\)

Thus, the correct answer is D.

41 答案 D

配方得到 \[ \begin{aligned} ax^2+bx+c &=a\left(x+\frac{b}{2a}\right)^2\\ &\quad+c-\frac{b^2}{4a} \end{aligned}\text{。}\]因为 \(a>0\)，平方项的最小值为 \(0\)。因此，函数的最小值为 \[ c-\frac{b^2}{4a}=\frac{4ac-b^2}{4a}\text{。}\]

因此，正确答案是 D。

English

Completing the square gives \[ \begin{aligned} ax^2+bx+c &=a\left(x+\frac{b}{2a}\right)^2\\ &\quad+c-\frac{b^2}{4a}. \end{aligned} \] Because \(a>0,\) the squared term has minimum \(0.\) The least value is therefore \[ c-\frac{b^2}{4a}=\frac{4ac-b^2}{4a}. \]

Thus, the correct answer is D.

42 答案 D

设无限幂塔的值为 \(T\)。去掉最底层的 \(x\) 后，指数部分仍是同一个幂塔，所以 \(T=x^T\)。由于 \(T=2\)，得到 \[ 2=x^2\text{。}\]因此，正底数为 \(x=\sqrt2\)，此时幂塔收敛。

因此，正确答案是 D。

English

Let the value of the infinite tower be \(T.\) Removing its bottom \(x\) leaves the same tower as the exponent, so \(T=x^T.\) Since \(T=2,\) we obtain \[ 2=x^2. \] The positive base is therefore \(x=\sqrt2,\) for which the tower is convergent.

Thus, the correct answer is D.

43 答案 E

将相邻两项分组后，得到一个等比级数，其首项为 \[ \frac17+\frac2{49}=\frac9{49}\text{，}\]公比为 \(\tfrac1{49}\)。因此，级数之和为 \[ \frac{\frac9{49}}{1-\frac1{49}} =\frac9{48} =\frac3{16}\text{。}\]这个值不在选项 A 至 D 中。

因此，正确答案是 E。

English

Grouping consecutive terms gives a geometric series whose first grouped term is \[ \frac17+\frac2{49}=\frac9{49} \] and whose ratio is \(\tfrac1{49}.\) Hence the sum is \[ \frac{\frac9{49}}{1-\frac1{49}} =\frac9{48} =\frac3{16}. \] This is not among choices A through D.

Thus, the correct answer is E.

44 答案 C

由于 \(\log1=0\)，图像经过 \((1,0)\)，因此与 \(x\) 轴相交。又因为 \(x=0\) 不在定义域内，所以图像不可能与 \(y\) 轴相交。

因此，正确答案是 C。

English

Since \(\log1=0,\) the graph passes through \((1,0)\) and therefore cuts the \(x\)-axis. It cannot cut the \(y\)-axis because \(x=0\) is outside its domain.

Thus, the correct answer is C.

45 答案 A

连接各对顶点共有 \(\binom{100}{2}\) 条线段，其中恰有 \(100\) 条是多边形的边，所以对角线条数为 \[ \begin{aligned} \binom{100}{2}-100 &=4950-100\\ &=4850 \end{aligned}\text{。}\]

因此，正确答案是 A。

English

There are \(\binom{100}{2}\) segments joining pairs of vertices. Exactly \(100\) of these are sides, so the number of diagonals is \[ \begin{aligned} \binom{100}{2}-100 &=4950-100\\ &=4850. \end{aligned} \]

Thus, the correct answer is A.

46 答案 E

三条新边的长度为 \(24\)、\(14\) 和 \(10\)。由于 \[ 14+10=24\text{，}\]它们构成退化三角形，三个顶点共线。因此，其面积为 \(0\)。

因此，正确答案是 E。

English

The new side lengths are \(24,\) \(14,\) and \(10.\) Because \[ 14+10=24, \] they form a degenerate triangle: all three vertices are collinear. Its area is therefore \(0.\)

Thus, the correct answer is E.

47 答案 C

由相似三角形可知，在距底边高度 \(x\) 处，三角形的宽度为 \(b(1-\tfrac{x}{h})\)，这就是内接矩形的底边。由于矩形的高是底边的一半，其底边长为 \(2x\)。因此， \[ 2x=b\left(1-\frac{x}{h}\right)\text{。}\]两边乘以 \(h\) 并求解，得到 \(x(2h+b)=bh\)，所以 \(x=\dfrac{bh}{2h+b}\)。

因此，正确答案是 C。

English

By similarity, the width of the triangle at height \(x\) above its base is \(b(1-\tfrac{x}{h}).\) This is the base of the inscribed rectangle. Since the rectangle’s altitude is half its base, its base is \(2x.\) Hence \[ 2x=b\left(1-\frac{x}{h}\right). \] Multiplying by \(h\) and solving gives \(x(2h+b)=bh,\) so \(x=\dfrac{bh}{2h+b}.\)

Thus, the correct answer is C.

48 答案 C

设等边三角形的边长为 \(s\)，高为 \(h\)，三个垂直距离为 \(d_1,d_2,d_3\)。用内部点将大三角形分成三个小三角形，得到 \[ \frac12s(d_1+d_2+d_3)=\frac12sh\text{。}\]因此 \(d_1+d_2+d_3=h\)，与所选点的位置无关。

因此，正确答案是 C。

English

Let the equilateral triangle have side length \(s\) and altitude \(h,\) and let the three perpendicular distances be \(d_1,d_2,d_3.\) Splitting the triangle at the interior point gives \[ \frac12s(d_1+d_2+d_3)=\frac12sh. \] Therefore \(d_1+d_2+d_3=h,\) independent of the selected point.

Thus, the correct answer is C.

49 答案 D

将 \(A\) 置于原点，并把 \(B\)、\(C\) 以及 \(BC\) 的中点 \(M\) 看作向量。由于 \(AM=\tfrac32\)，点 \(M\) 在以 \(A\) 为圆心、半径为 \(\tfrac32\) 的圆上运动。中点关系给出 \[ C=2M-B\text{。}\]因此，\(C\) 的轨迹是先将该圆放大 \(2\) 倍，再平移 \(-B\) 所得的图形，即以 \(-B\) 为圆心、半径为 \(3\) 的圆。

因为 \(AB=2\)，点 \(-B\) 位于射线 \(BA\) 上，且距 \(B\) 为 \(4\) 英寸。因此，正确答案是 D。

English

Put \(A\) at the origin and regard \(B,\) \(C,\) and the midpoint \(M\) of \(BC\) as vectors. Since \(AM=\tfrac32,\) the point \(M\) moves on a circle of radius \(\tfrac32\) centered at \(A.\) The midpoint relation gives \[ C=2M-B. \] Thus the locus of \(C\) is the image of that circle under a dilation by \(2\) followed by translation by \(-B.\) It is a circle of radius \(3\) centered at \(-B.\)

Because \(AB=2,\) the point \(-B\) is \(4\) inches from \(B\) along the ray \(BA.\) Hence the correct answer is D.

50 答案 E

最初两小时内，私掠船追近的英里数为 \[ 2(11-8)=6\text{，}\]所以下午 \(1{:}45\) 时，两船距离从 \(10\) 英里缩短为 \(4\) 英里。

上桅帆被刮走后，两船速度之比为 \(17:15\)。由于商船仍以每小时 \(8\) 英里航行，私掠船的新速度为每小时 \(8\cdot\tfrac{17}{15}=\tfrac{136}{15}\) 英里。两船接近的相对速度为每小时 \(\tfrac{136}{15}-8=\tfrac{16}{15}\) 英里，所以追完剩余距离所需的小时数为 \[ \frac{4}{\frac{16}{15}}=\frac{15}{4}=3\frac34\text{。}\]在下午 \(1{:}45\) 后再过 \(3\) 小时 \(45\) 分钟，即为下午 \(5{:}30\)。

因此，正确答案是 E。

English

During the first two hours, the number of miles the privateer gains is \[ 2(11-8)=6, \] reducing the gap from \(10\) miles to \(4\) miles at \(1{:}45\) p.m.

After the damage, the ships’ speeds are in the ratio \(17:15.\) Since the merchantman still travels at \(8\) mph, the privateer’s new speed is \(8\cdot\tfrac{17}{15}=\tfrac{136}{15}\) mph. The closing speed is therefore \(\tfrac{136}{15}-8=\tfrac{16}{15}\) mph, so the number of hours needed to close the remaining gap is \[ \frac{4}{\frac{16}{15}}=\frac{15}{4}=3\frac34. \] Adding \(3\) hours \(45\) minutes to \(1{:}45\) p.m. gives \(5{:}30\) p.m.

Thus, the correct answer is E.