1958 AMC 12 真题（P版）

1 答案 C

由于 \(2-3=-1\)，其倒数仍为 \(-1\)。因此 \[ \left[2-3(-1)\right]^{-1} =5^{-1}=\frac15\text{。}\]

所以正确答案为 C。

English

Since \(2-3=-1,\) its reciprocal is also \(-1.\) Therefore \[ \left[2-3(-1)\right]^{-1} =5^{-1}=\frac15. \]

Thus, the correct answer is C.

2 答案 D

通分可得 \[ \frac1z=\frac1x-\frac1y =\frac{y-x}{xy}\text{。}\]两边取倒数，得到 \[ z=\frac{xy}{y-x}\text{。}\]

所以正确答案为 D。

English

Combining the fractions, \[ \frac1z=\frac1x-\frac1y =\frac{y-x}{xy}. \] Taking reciprocals gives \[ z=\frac{xy}{y-x}. \]

Therefore, the correct answer is D.

3 答案 B

有 \[ a^{-3}-b^{-3} =\frac{b^3-a^3}{a^3b^3}\text{。}\]因此 \[ \frac{\frac{1}{ab}}{\frac{b^3-a^3}{a^3b^3}} =\frac{a^2b^2}{b^3-a^3}\text{。}\]

所以正确答案为 B。

English

We have \[ a^{-3}-b^{-3} =\frac{b^3-a^3}{a^3b^3}. \] Hence \[ \frac{\frac{1}{ab}}{\frac{b^3-a^3}{a^3b^3}} =\frac{a^2b^2}{b^3-a^3}. \]

Thus, the correct answer is B.

4 答案 E

令 \(f(x)=\frac{x+1}{x-1}\)。则 \[ \begin{aligned} f(f(x)) &=\frac{\frac{x+1}{x-1}+1} {\frac{x+1}{x-1}-1}\\ &=\frac{\frac{2x}{x-1}}{\frac{2}{x-1}} =x \end{aligned}\text{。}\]当 \(x=\frac{1}{2}\) 时，其值为 \(\frac{1}{2}\)，不在选项中。

所以正确答案为 E。

English

Let \(f(x)=\frac{x+1}{x-1}.\) Then \[ \begin{aligned} f(f(x)) &=\frac{\frac{x+1}{x-1}+1} {\frac{x+1}{x-1}-1}\\ &=\frac{\frac{2x}{x-1}}{\frac{2}{x-1}} =x. \end{aligned} \] At \(x=\frac{1}{2},\) the value is \(\frac{1}{2},\) which is not listed.

Therefore, the correct answer is E.

5 答案 A

分母有理化后得到 \[ \begin{aligned} \frac1{2+\sqrt2}&=\frac{2-\sqrt2}{2},\\ \frac1{\sqrt2-2}&=-\frac{2+\sqrt2}{2} \end{aligned}\text{。}\]两者之和为 \(-\sqrt2\)。因此整个表达式为 \[ 2+\sqrt2-\sqrt2=2\text{。}\]

所以正确答案为 A。

English

Rationalizing gives \[ \begin{aligned} \frac1{2+\sqrt2}&=\frac{2-\sqrt2}{2},\\ \frac1{\sqrt2-2}&=-\frac{2+\sqrt2}{2}. \end{aligned} \] Their sum is \(-\sqrt2.\) Therefore the whole expression is \[ 2+\sqrt2-\sqrt2=2. \]

Thus, the correct answer is A.

6 答案 B

算术平均数为 \[ \begin{aligned} &\frac12\left(\frac{x+a}{x}+\frac{x-a}{x}\right)\\ &\qquad=\frac12\left(\frac{2x}{x}\right)=1 \end{aligned}\text{。}\]

所以正确答案为 B。

English

The arithmetic mean is \[ \begin{aligned} &\frac12\left(\frac{x+a}{x}+\frac{x-a}{x}\right)\\ &\qquad=\frac12\left(\frac{2x}{x}\right)=1. \end{aligned} \]

Therefore, the correct answer is B.

7 答案 A

斜率为 \[ \frac{9-1}{3-(-1)}=2\text{。}\]利用点 \((-1,1)\)，直线方程为 \(y=2x+3\)。令 \(y=0\)，得到 \[ x=-\frac32\text{。}\]

所以正确答案为 A。

English

The slope is \[ \frac{9-1}{3-(-1)}=2. \] Using \((-1,1),\) the line is \(y=2x+3.\) Setting \(y=0\) gives \[ x=-\frac32. \]

Thus, the correct answer is A.

8 答案 D

第一个数是 \(\pi\)，为无理数。\(\frac{4}{5}\) 的立方根也是无理数。此外 \[ \sqrt[4]{0.00016} =\sqrt[4]{16\cdot10^{-5}} \] 是无理数。第四个数为 \[ (-1)\sqrt{\frac{100}{9}}=-\frac{10}{3}\text{，}\]它是有理数。

因此只有第四个数是有理数，正确答案为 D。

English

The first number is \(\pi,\) which is irrational. The cube root of \(\frac{4}{5}\) is irrational. Also \[ \sqrt[4]{0.00016} =\sqrt[4]{16\cdot10^{-5}} \] is irrational. The fourth number is \[ (-1)\sqrt{\frac{100}{9}}=-\frac{10}{3}, \] which is rational.

Therefore, only the fourth is rational, and the correct answer is D.

9 答案 C

展开右边，得到 \[ x^2+b^2=a^2-2ax+x^2\text{。}\]消去 \(x^2\) 并求解，得到 \[ \begin{aligned} 2ax&=a^2-b^2,\\ x&=\frac{a^2-b^2}{2a} \end{aligned}\text{。}\]

所以正确答案为 C。

English

Expanding the right side, \[ x^2+b^2=a^2-2ax+x^2. \] Cancelling \(x^2\) and solving gives \[ \begin{aligned} 2ax&=a^2-b^2,\\ x&=\frac{a^2-b^2}{2a}. \end{aligned} \]

Therefore, the correct answer is C.

10 答案 E

判别式为 \[ k^2-4k^2=-3k^2\text{。}\]对每个非零实数 \(k\)，它都为负，因此该二次方程没有实根。

所以正确答案为 E。

English

The discriminant is \[ k^2-4k^2=-3k^2. \] For every nonzero real \(k,\) this is negative, so the quadratic has no real roots.

Thus, the correct answer is E.

11 答案 C

因式分解得 \[ (1-x)\sqrt{5-x}=0\text{。}\]因此 \(x=1\)，或由 \(5-x=0\) 得 \(x=5\)。两个值都在定义域 \(x\le5\) 内，并且满足原方程。

所以共有两个根，正确答案为 C。

English

Factoring gives \[ (1-x)\sqrt{5-x}=0. \] Thus either \(x=1\) or \(5-x=0,\) giving \(x=5.\) Both values lie in the domain \(x\le5\) and satisfy the original equation.

Therefore, there are two roots, and the correct answer is C.

12 答案 A

取对数得 \[ \log P=\log s-n\log(1+k)\text{。}\]因此 \[ n=\frac{\log s-\log P}{\log(1+k)} =\frac{\log(\frac{s}{P})}{\log(1+k)}\text{。}\]

所以正确答案为 A。

English

Taking logarithms, \[ \log P=\log s-n\log(1+k). \] Therefore \[ n=\frac{\log s-\log P}{\log(1+k)} =\frac{\log(\frac{s}{P})}{\log(1+k)}. \]

Thus, the correct answer is A.

13 答案 B

若两数为 \(x\) 和 \(y\)，则 \[ \frac1x+\frac1y =\frac{x+y}{xy} =\frac{10}{20} =\frac12\text{。}\]

所以正确答案为 B。

English

If the numbers are \(x\) and \(y,\) then \[ \frac1x+\frac1y =\frac{x+y}{xy} =\frac{10}{20} =\frac12. \]

Therefore, the correct answer is B.

14 答案 C

数列 \(5,6,7,\ldots\) 的 \(b\) 项中，末项为 \[ 5+(b-1)=b+4\text{。}\]这个末项等于女孩总数 \(g\)。因此 \(g=b+4\)，即 \(b=g-4\)。

所以正确答案为 C。

English

The last of the \(b\) terms in the sequence \(5,6,7,\ldots\) is \[ 5+(b-1)=b+4. \] This last number equals the total number \(g\) of girls. Hence \(g=b+4,\) or \(b=g-4.\)

Thus, the correct answer is C.

15 答案 D

设各边所截的四段劣弧度数为 \(\alpha_1,\ldots,\alpha_4\)，其和为 \(360^\circ\)。与弧 \(\alpha_i\) 对应的外侧弓形内的圆周角所对的是另一段弧，因此其度数为 \[ \frac{360^\circ-\alpha_i}{2}\text{。}\]所以四个角的度数之和为 \[ \frac{4(360^\circ)-360^\circ}{2}=540^\circ\text{。}\]

所以正确答案为 D。

English

Let the four minor arcs cut off by the sides have measures \(\alpha_1,\ldots,\alpha_4,\) whose sum is \(360^\circ.\) The angle in the outside segment corresponding to arc \(\alpha_i\) subtends the other arc, so its measure is \[ \frac{360^\circ-\alpha_i}{2}. \] The sum of the four angles is therefore \[ \frac{4(360^\circ)-360^\circ}{2}=540^\circ. \]

Thus, the correct answer is D.

16 答案 D

圆的半径为 \(r=10\)，也是六边形的边心距。因此边长为 \[ 2r\tan30^\circ=\frac{20}{\sqrt3}\text{。}\]利用面积等于周长与边心距乘积的一半，六边形的面积为 \[ \frac12\left(6\cdot\frac{20}{\sqrt3}\right)(10) =200\sqrt3\text{。}\]

所以正确答案为 D。

English

The circle has radius \(r=10,\) which is the hexagon’s apothem. Thus the side length is \[ 2r\tan30^\circ=\frac{20}{\sqrt3}. \] Using one-half the product of perimeter and apothem, the hexagon’s area is \[ \frac12\left(6\cdot\frac{20}{\sqrt3}\right)(10) =200\sqrt3. \]

Therefore, the correct answer is D.

17 答案 E

由不等式得 \[ \frac12\log x\ge\log2\text{，}\]所以 \(\log x\ge2\log2=\log4\)。因此 \(x\ge4\)，故 \(x\) 的最小可能值为 \(4\)。

所以正确答案为 E。

English

The inequality gives \[ \frac12\log x\ge\log2, \] so \(\log x\ge2\log2=\log4.\) Hence \(x\ge4,\) and the minimum possible value of \(x\) is \(4.\)

Therefore, the correct answer is E.

18 答案 A

面积加倍的条件为 \[ \pi(r+n)^2=2\pi r^2\text{。}\]由于半径为正，\(r+n=\sqrt2\,r\)。因此 \[ r=\frac{n}{\sqrt2-1} =n(\sqrt2+1)\text{。}\]

所以正确答案为 A。

English

The doubled-area condition is \[ \pi(r+n)^2=2\pi r^2. \] Since the radii are positive, \(r+n=\sqrt2\,r.\) Thus \[ r=\frac{n}{\sqrt2-1} =n(\sqrt2+1). \]

Thus, the correct answer is A.

19 答案 B

由斜边高所形成的相似三角形可得 \[ a^2=cr,\qquad b^2=cs\text{。}\]因此 \[ \frac rs=\frac{a^2}{b^2} =\left(\frac13\right)^2=\frac19\text{。}\]

所以正确答案为 B。

English

Similarity from the altitude-to-hypotenuse construction gives \[ a^2=cr,\qquad b^2=cs. \] Therefore \[ \frac rs=\frac{a^2}{b^2} =\left(\frac13\right)^2=\frac19. \]

Thus, the correct answer is B.

20 答案 C

有 \[ 4^x-4^{x-1} =\frac34\,4^x=24\text{，}\]所以 \(4^x=32\)。于是 \(2^{2x}=2^5\)，得到 \(x=\frac{5}{2}\)。因此 \[ (2x)^x=5^{\frac{5}{2}}=25\sqrt5\text{。}\]

所以正确答案为 C。

English

We have \[ 4^x-4^{x-1} =\frac34\,4^x=24, \] so \(4^x=32.\) Thus \(2^{2x}=2^5,\) giving \(x=\frac{5}{2}.\) Therefore \[ (2x)^x=5^{\frac{5}{2}}=25\sqrt5. \]

Therefore, the correct answer is C.

21 答案 E

设圆的半径为 \(r\)。在图示位置中，\(\overline{CD}\) 是直径，\(E\) 是与它垂直的半径的端点，因此 \[ [CED]=\frac12(2r)(r)=r^2\text{。}\]由于弧 \(AB\) 是四分之一圆周，\(\angle AOB=90^\circ\)，且 \[ [AOB]=\frac12r^2\text{。}\]所求比为 \(2:1\)。

所以正确答案为 E。

English

Let the circle have radius \(r.\) In the pictured configuration, \(\overline{CD}\) is a diameter and \(E\) is the endpoint of the perpendicular radius, so \[ [CED]=\frac12(2r)(r)=r^2. \] Since arc \(AB\) is a quarter-circle, \(\angle AOB=90^\circ,\) and \[ [AOB]=\frac12r^2. \] The requested ratio is therefore \(2:1.\)

Thus, the correct answer is E.

22 答案 C

当 \(y=6\) 时，\[ x^2-x-12=0\text{，}\]所以 \(x=4\) 或 \(-3\)。当 \(y=-6\) 时，\(x^2-x=0\)，所以 \(x=0\) 或 \(1\)。离 \(x=4\) 最近的下方点为 \(x=1\)，离 \(x=-3\) 最近的下方点为 \(x=0\)。两种情况下的水平距离均为 \(3\)。

所以正确答案为 C。

English

For \(y=6,\) \[ x^2-x-12=0, \] so \(x=4\) or \(-3.\) For \(y=-6,\) \(x^2-x=0,\) so \(x=0\) or \(1.\) The nearest lower point to \(x=4\) is \(x=1,\) and the nearest to \(x=-3\) is \(x=0.\) Either horizontal distance is \(3.\)

Therefore, the correct answer is C.

23 答案 A

增加时，\[ \begin{aligned} &(x+a)^2-3-(x^2-3)\\ &\qquad=2ax+a^2 \end{aligned}\text{。}\]减少时，\[ \begin{aligned} &(x-a)^2-3-(x^2-3)\\ &\qquad=-2ax+a^2 \end{aligned}\text{。}\]因而这两种变化量合写为 \(\pm2ax+a^2\)。

所以正确答案为 A。

English

For an increase, \[ \begin{aligned} &(x+a)^2-3-(x^2-3)\\ &\qquad=2ax+a^2. \end{aligned} \] For a decrease, \[ \begin{aligned} &(x-a)^2-3-(x^2-3)\\ &\qquad=-2ax+a^2. \end{aligned} \] Together these changes are \(\pm2ax+a^2.\)

Thus, the correct answer is A.

24 答案 B

北行速度为 \(30\) 英里每小时，南行速度为 \(120\) 英里每小时。由于往返距离相等，全程平均速度为 \[ \frac{2(30)(120)}{30+120}=48\text{ 英里/小时}\text{。}\]距离 \(m\) 会约去。

所以正确答案为 B。

English

The northbound rate is \(30\) mph, and the southbound rate is \(120\) mph. For equal distances, the round-trip average is \[ \frac{2(30)(120)}{30+120}=48\text{ mph}. \] The distance \(m\) cancels.

Therefore, the correct answer is B.

25 答案 E

由换底恒等式，\[ \log_kx\cdot\log_5k=\log_5x\text{。}\]因此 \(\log_5x=3\)，所以 \(x=5^3=125\)。

所以正确答案为 E。

English

By the change-of-base identity, \[ \log_kx\cdot\log_5k=\log_5x. \] Hence \(\log_5x=3,\) so \(x=5^3=125.\)

Thus, the correct answer is E.

26 答案 B

原数 \(u\) 变为 \[ 5(u+20)-20=5u+80\text{。}\]因而对原来的 \(n\) 个数求和可得 \[ 5s+80n\text{。}\]

所以正确答案为 B。

English

An original number \(u\) becomes \[ 5(u+20)-20=5u+80. \] Summing over the \(n\) original numbers therefore gives \[ 5s+80n. \]

Thus, the correct answer is B.

27 答案 A

经过前两个点的直线斜率为 \[ \frac{3-(-3)}{4-2}=3\text{。}\]因此 \[ \frac{\frac{k}{2}-3}{5-4}=3\text{，}\]得到 \(\frac{k}{2}=6\) 和 \(k=12\)。

所以正确答案为 A。

English

The slope through the first two points is \[ \frac{3-(-3)}{4-2}=3. \] Thus \[ \frac{\frac{k}{2}-3}{5-4}=3, \] which gives \(\frac{k}{2}=6\) and \(k=12.\)

Therefore, the correct answer is A.

28 答案 B

每次取出充分混合的散热器内容物的四分之一，因此留下当时水量的四分之三。四次更换后，水所占的比例为 \[ \left(\frac34\right)^4=\frac{81}{256}\text{。}\]

所以正确答案为 B。

English

Each removal takes one-fourth of the well-mixed radiator contents, so it leaves three-fourths of the water then present. After four replacements, the water fraction is \[ \left(\frac34\right)^4=\frac{81}{256}. \]

Therefore, the correct answer is B.

29 答案 E

由三角形 \(AEC\) 得 \[ x+y+w+n=180^\circ\text{。}\]由三角形 \(BED\) 得 \[ m+a+b+w=180^\circ\text{。}\]令两式左边相等并消去 \(w\)，得到 \[ x+y+n=a+b+m\text{。}\]

所以正确答案为 E。

English

Triangle \(AEC\) gives \[ x+y+w+n=180^\circ. \] Triangle \(BED\) gives \[ m+a+b+w=180^\circ. \] Equating the left sides and cancelling \(w\) yields \[ x+y+n=a+b+m. \]

Thus, the correct answer is E.

30 答案 C

由于 \(xy=b\)，\[ a=\frac{x^2+y^2}{x^2y^2} =\frac{x^2+y^2}{b^2}\text{，}\]所以 \(x^2+y^2=ab^2\)。因此 \[ \begin{aligned} (x+y)^2 &=x^2+y^2+2xy\\ &=ab^2+2b\\ &=b(ab+2) \end{aligned}\text{。}\]

所以正确答案为 C。

English

Since \(xy=b,\) \[ a=\frac{x^2+y^2}{x^2y^2} =\frac{x^2+y^2}{b^2}, \] so \(x^2+y^2=ab^2.\) Therefore \[ \begin{aligned} (x+y)^2 &=x^2+y^2+2xy\\ &=ab^2+2b\\ &=b(ab+2). \end{aligned} \]

Therefore, the correct answer is C.

31 答案 B

设每条腰长为 \(a\)，底边的一半为 \(b\)。由周长得 \(a+b=16\)。高平分底边，所以 \[ a^2-b^2=8^2\text{。}\]因而 \((a-b)(a+b)=64\)，得到 \(a-b=4\)。于是 \(b=6\)，底边长为 \(12\)。面积为 \[ \frac12(12)(8)=48\text{。}\]

所以正确答案为 B。

English

Let each equal side be \(a\) and half the base be \(b.\) The perimeter gives \(a+b=16.\) The altitude bisects the base, so \[ a^2-b^2=8^2. \] Thus \((a-b)(a+b)=64,\) giving \(a-b=4.\) Hence \(b=6,\) so the base is \(12.\) The area is \[ \frac12(12)(8)=48. \]

Therefore, the correct answer is B.

32 答案 E

购买方程为 \[ 25s+26c=1000\text{。}\]对 \(25\) 取模得 \(c\equiv0\pmod{25}\)。由于 \(c\) 为正且 \(26c<1000\)，唯一可能是 \(c=25\)。于是 \[ s=\frac{1000-26(25)}{25}=14\text{。}\]只有一个解，并且 \(c>s\)。

所以正确答案为 E。

English

The purchase equation is \[ 25s+26c=1000. \] Reducing modulo \(25\) gives \(c\equiv0\pmod{25}.\) Since \(c\) is positive and \(26c<1000,\) the only possibility is \(c=25.\) Then \[ s=\frac{1000-26(25)}{25}=14. \] There is one solution, and \(c>s.\)

Thus, the correct answer is E.

33 答案 B

设两根为 \(q\) 和 \(2q\)。韦达定理给出 \[ 3q=-\frac ba,\qquad 2q^2=\frac ca\text{。}\]第一个关系式平方后得 \(9q^2=\frac{b^2}{a^2}\)。将其除以第二个关系式，得到 \[ \frac92=\frac{b^2}{ac}\text{，}\]即 \(2b^2=9ac\)。

所以正确答案为 B。

English

Let the roots be \(q\) and \(2q.\) Vieta’s formulas give \[ 3q=-\frac ba,\qquad 2q^2=\frac ca. \] Squaring the first relation yields \(9q^2=\frac{b^2}{a^2}.\) Dividing this by the second relation gives \[ \frac92=\frac{b^2}{ac}, \] or \(2b^2=9ac.\)

Therefore, the correct answer is B.

34 答案 A

所需比较给出 \[ 6x+1\gt7-4x\text{，}\]所以 \(10x\gt6\)，且 \(x\gt\frac{3}{5}\)。与 \(-2\le x\le2\) 取交集得到 \[ \frac35\lt x\le2\text{。}\]

所以正确答案为 A。

English

The required comparison gives \[ 6x+1\gt7-4x, \] so \(10x\gt6\) and \(x\gt\frac{3}{5}.\) Intersecting this with \(-2\le x\le2\) gives \[ \frac35\lt x\le2. \]

Thus, the correct answer is A.

35 答案 D

将一个顶点平移到原点后，把另两个顶点写成 \((a,c)\) 和 \((b,d)\)，其中所有坐标均为整数。面积为 \[ \frac12|ad-bc|\text{。}\]由于 \(ad-bc\) 是整数，面积是整数或半整数，两种情况下都是有理数。

所以正确答案为 D。

English

After translating one vertex to the origin, write the other two as \((a,c)\) and \((b,d),\) with all coordinates integers. The area is \[ \frac12|ad-bc|. \] Since \(ad-bc\) is an integer, the area is an integer or a half-integer, and in either case it is rational.

Therefore, the correct answer is D.

36 答案 D

设与长度为 \(30\) 的边相邻的线段为 \(x\)。若高为 \(h\)，则 \[ \begin{aligned} 30^2-x^2&=h^2,\\ h^2&=70^2-(80-x)^2 \end{aligned}\text{。}\]消去 \(x^2\) 并求解得 \(x=15\)。另一段为 \(80-15=65\)，它是较长的一段。

所以正确答案为 D。

English

Let \(x\) be the segment adjacent to the side of length \(30.\) If the altitude has length \(h,\) then \[ \begin{aligned} 30^2-x^2&=h^2,\\ h^2&=70^2-(80-x)^2. \end{aligned} \] Cancelling \(x^2\) and solving gives \(x=15.\) The other segment is \(80-15=65,\) which is the larger one.

Thus, the correct answer is D.

37 答案 A

末项为 \[ k^2+1+2k=(k+1)^2\text{。}\]首项与末项的平均数为 \(k^2+k+1\)。因此总和为 \[ \begin{aligned} S&=(2k+1)(k^2+k+1)\\ &=k^3+(k+1)^3 \end{aligned}\text{。}\]

所以正确答案为 A。

English

The last term is \[ k^2+1+2k=(k+1)^2. \] The average of the first and last terms is \(k^2+k+1.\) Therefore the sum is \[ \begin{aligned} S&=(2k+1)(k^2+k+1)\\ &=k^3+(k+1)^3. \end{aligned} \]

Therefore, the correct answer is A.

38 答案 D

由于 \(r^2=x^2+y^2\)，\[ s^2+c^2=\frac{y^2+x^2}{r^2}=1\text{。}\]因此 \(s^2-c^2=2s^2-1\)。因为 \(0\le s^2\le1\)，该表达式的取值从 \(-1\) 到 \(1\)，包括两个端点。

所以正确答案为 D。

English

Since \(r^2=x^2+y^2,\) \[ s^2+c^2=\frac{y^2+x^2}{r^2}=1. \] Thus \(s^2-c^2=2s^2-1.\) Because \(0\le s^2\le1,\) this expression ranges from \(-1\) through \(1,\) with both endpoints included.

Thus, the correct answer is D.

39 答案 C

令 \(u=|x|\ge0\)。则 \[ \begin{aligned} u^2+u-6&=0,\\ (u+3)(u-2)&=0 \end{aligned}\text{。}\]非负解为 \(u=2\)，所以 \(|x|=2\)，且 \(x=\pm2\)。两根之和为 \(0\)。

所以正确答案为 C。

English

Let \(u=|x|\ge0.\) Then \[ \begin{aligned} u^2+u-6&=0,\\ (u+3)(u-2)&=0. \end{aligned} \] The nonnegative solution is \(u=2,\) so \(|x|=2\) and \(x=\pm2.\) Their sum is \(0.\)

Therefore, the correct answer is C.

40 答案 B

当 \(n=1\) 时，\[ 3^2-(1)a_2=-1\text{，}\]所以 \(a_2=10\)。当 \(n=2\) 时，\[ 10^2-3a_3=1\text{，}\]得到 \(a_3=33\)。

所以正确答案为 B。

English

For \(n=1,\) \[ 3^2-(1)a_2=-1, \] so \(a_2=10.\) For \(n=2,\) \[ 10^2-3a_3=1, \] which gives \(a_3=33.\)

Thus, the correct answer is B.

41 答案 C

对原二次方程，\[ r+s=-\frac BA,\qquad rs=\frac CA\text{。}\]因此 \[ \begin{aligned} r^2+s^2 &=(r+s)^2-2rs\\ &=\frac{B^2-2AC}{A^2} \end{aligned}\text{。}\]系数 \(p\) 是这个和的相反数，所以 \[ p=\frac{2AC-B^2}{A^2}\text{。}\]

所以正确答案为 C。

English

For the original quadratic, \[ r+s=-\frac BA,\qquad rs=\frac CA. \] Hence \[ \begin{aligned} r^2+s^2 &=(r+s)^2-2rs\\ &=\frac{B^2-2AC}{A^2}. \end{aligned} \] The coefficient \(p\) is the negative of this sum, so \[ p=\frac{2AC-B^2}{A^2}. \]

Therefore, the correct answer is C.

42 答案 E

由于 \(E\) 在 \(BC\) 和 \(AD\) 上，三角形 \(AEC\) 与 \(ACD\) 共有角 \(A\)。又因 \(\angle ACE=\angle ACB\) 所对弦为 \(AB\)，而 \(\angle ADC\) 所对弦为 \(AC\)。由于 \(AB=AC\)，两角相等。因此 \[ \triangle AEC\sim\triangle ACD\text{。}\]对应边给出 \[ \frac{AE}{AC}=\frac{AC}{AD}\text{。}\]所以 \(\frac{8}{12}=\frac{12}{AD}\)，从而 \(AD=18\)。

所以正确答案为 E。

English

Because \(E\) lies on \(BC\) and \(AD,\) triangles \(AEC\) and \(ACD\) share angle \(A.\) Also \(\angle ACE=\angle ACB\) subtends chord \(AB,\) while \(\angle ADC\) subtends chord \(AC.\) Since \(AB=AC,\) these angles are equal. Thus \[ \triangle AEC\sim\triangle ACD. \] Corresponding sides give \[ \frac{AE}{AC}=\frac{AC}{AD}. \] Therefore \(\frac{8}{12}=\frac{12}{AD},\) so \(AD=18.\)

Thus, the correct answer is E.

43 答案 D

令 \(a=BC\)、\(b=CA\)、\(c=AB\)。由于直角在 \(C\)，\(c^2=a^2+b^2\)。中线公式给出 \[ \begin{aligned} 4(7^2) &=a^2+4b^2,\\ 4(4^2) &=4a^2+b^2 \end{aligned}\text{。}\]解得 \(a^2=4\) 和 \(b^2=48\)。因此 \[ \begin{aligned} AB=c &=\sqrt{a^2+b^2}\\ &=\sqrt{52}=2\sqrt{13} \end{aligned}\text{。}\]

所以正确答案为 D。

English

Let \(a=BC,\) \(b=CA,\) and \(c=AB.\) Since the right angle is at \(C,\) \(c^2=a^2+b^2.\) The median formulas give \[ \begin{aligned} 4(7^2) &=a^2+4b^2,\\ 4(4^2) &=4a^2+b^2. \end{aligned} \] Solving yields \(a^2=4\) and \(b^2=48.\) Therefore \[ \begin{aligned} AB=c &=\sqrt{a^2+b^2}\\ &=\sqrt{52}=2\sqrt{13}. \end{aligned} \]

Thus, the correct answer is D.

44 答案 E

令 \(P\) 表示 \(a>b\)，\(Q\) 表示 \(c>d\)，\(S\) 表示 \(c<d\)，\(R\) 表示 \(e>f\)。前提为 \[ P\Rightarrow Q,\qquad S\Rightarrow R\text{。}\]命题 \(Q\) 与 \(S\) 不能同时成立。两个前提的逆否命题分别为 \(\neg Q\Rightarrow\neg P\) 和 \(\neg R\Rightarrow\neg S\)。四个给出的蕴含都不能由此推出。例如，已知 \(P\) 可得 \(Q\)，进而得到 \(\neg S\)，但无法判断 \(R\)；已知 \(\neg R\) 只能得到 \(\neg S\)，不能得到 \(P\)。

所以正确答案为 E。

English

Let \(P\) mean \(a>b,\) \(Q\) mean \(c>d,\) \(S\) mean \(c<d,\) and \(R\) mean \(e>f.\) The premises are \[ P\Rightarrow Q,\qquad S\Rightarrow R. \] The statements \(Q\) and \(S\) cannot both hold. The contrapositives are \(\neg Q\Rightarrow\neg P\) and \(\neg R\Rightarrow\neg S.\) None of the four proposed implications follows. For example, knowing \(P\) gives \(Q\) and hence \(\neg S,\) but says nothing about \(R;\) knowing \(\neg R\) gives \(\neg S,\) not \(P.\)

Therefore, the correct answer is E.

45 答案 B

以美分计，错误金额减正确金额为 \[ \begin{aligned} &(100y+x)-(100x+y)\\ &\qquad=99(y-x) \end{aligned}\text{。}\]由于 \(\$17.82\) 等于 \(1782\) 美分，\[ 99(y-x)=1782\text{，}\]所以 \(y-x=18\)。两位数 \(x=18,\ y=36\) 满足此关系，并且 \(y=2x\)。因此这种相等关系可以发生。

所以正确答案为 B。

English

In cents, the incorrect amount minus the correct amount is \[ \begin{aligned} &(100y+x)-(100x+y)\\ &\qquad=99(y-x). \end{aligned} \] Since \(\$17.82\) is \(1782\) cents, \[ 99(y-x)=1782, \] so \(y-x=18.\) The two-digit values \(x=18,\ y=36\) satisfy this relation and have \(y=2x.\) Thus that equality can occur.

Therefore, the correct answer is B.

46 答案 E

令 \(t=x-1\)。则 \(-5\lt t\lt0\)，且表达式变为 \[ \frac{t^2+1}{2t} =\frac12\left(t+\frac1t\right)\text{。}\]当 \(t\lt0\) 时，\(t+\frac{1}{t}\le-2\)，等号在 \(t=-1\) 时成立。此值允许且对应 \(x=0\)。因此表达式至多为 \(-1\)，其最大值为 \(-1\)。

所以正确答案为 E。

English

Set \(t=x-1.\) Then \(-5\lt t\lt0,\) and the expression becomes \[ \frac{t^2+1}{2t} =\frac12\left(t+\frac1t\right). \] For \(t\lt0,\) \(t+\frac{1}{t}\le-2,\) with equality when \(t=-1.\) That value is allowed and corresponds to \(x=0.\) Hence the expression is at most \(-1,\) and its maximum is \(-1.\)

Thus, the correct answer is E.

47 答案 D

由于 \(AF\perp BD\) 且 \(PQ\perp AF\)，有 \(PQ\parallel BD\)。又因 \(PS\perp BD\)，所以 \(PS\parallel AF\)。因此四边形 \(QPSF\) 是矩形，且 \[ PS=QF\text{。}\]令 \(T=PQ\cap AC\)。矩形的两条对角线与边 \(AB\) 所成角相等，所以 \(\angle PAT=\angle APT\)，从而 \(AT=PT\)。直角三角形 \(ATQ\) 和 \(PTR\) 在 \(T\) 处有相同的角，故相似。由于它们的斜边 \(AT\) 和 \(PT\) 相等，\(AQ=PR\)。因此 \[ PR+PS=AQ+QF=AF\text{。}\]

所以正确答案为 D。

English

Since \(AF\perp BD\) and \(PQ\perp AF,\) we have \(PQ\parallel BD.\) Also \(PS\perp BD,\) so \(PS\parallel AF.\) Thus quadrilateral \(QPSF\) is a rectangle, and \[ PS=QF. \] Let \(T=PQ\cap AC.\) The diagonals of a rectangle make equal angles with side \(AB,\) so \(\angle PAT=\angle APT,\) giving \(AT=PT.\) The right triangles \(ATQ\) and \(PTR\) are similar because they share the angle at \(T.\) Since their hypotenuses \(AT\) and \(PT\) are equal, \(AQ=PR.\) Therefore \[ PR+PS=AQ+QF=AF. \]

Thus, the correct answer is D.

48 答案 E

取 \(O=(0,0)\)、\(C=(-1,0)\)、\(D=(1,0)\)，以及满足 \(u^2+v^2=25\) 的 \(P=(u,v)\)。则 \[ \begin{aligned} CP^2&=26+2u,\\ DP^2&=26-2u \end{aligned}\text{。}\]因此 \[ \begin{aligned} (CP+DP)^2 &=52\\ &\quad+2\sqrt{676-4u^2} \end{aligned}\text{。}\]当 \(u=0\) 时该式最大，此时恰有 \(CP=DP\)。

所以正确答案为 E。

English

Place \(O=(0,0),\) \(C=(-1,0),\) \(D=(1,0),\) and \(P=(u,v)\) with \(u^2+v^2=25.\) Then \[ \begin{aligned} CP^2&=26+2u,\\ DP^2&=26-2u. \end{aligned} \] Therefore \[ \begin{aligned} (CP+DP)^2 &=52\\ &\quad+2\sqrt{676-4u^2}. \end{aligned} \] This is largest when \(u=0,\) exactly when \(CP=DP.\)

Therefore, the correct answer is E.

49 答案 D

每个不同项都是 \(a^ib^jc^k\)，其中非负整数满足 \[ i+j+k=10\text{。}\]由隔板法，这样的三元组个数为 \[ \binom{10+3-1}{3-1}=\binom{12}{2}=66\text{。}\]

所以正确答案为 D。

English

Each distinct term is \(a^ib^jc^k\) for nonnegative integers satisfying \[ i+j+k=10. \] By stars and bars, the number of such triples is \[ \binom{10+3-1}{3-1}=\binom{12}{2}=66. \]

Thus, the correct answer is D.

50 答案 C

两条标有数字的线段平行，因此经过固定交点的投影给出 \(x\) 与 \(y\) 的线性关系。图示端点对应为 \((x,y)=(3,5)\) 和 \((x,y)=(4,1)\)。斜率为 \(\frac{1-5}{4-3}=-4\)，所以 \(y=-4x+17\)。若 \(x=a\)，则 \[ \begin{aligned} x+y &=a+(-4a+17)\\ &=17-3a \end{aligned}\text{。}\]

所以正确答案为 C。

English

The two numbered segments are parallel, so projection through the fixed intersection point gives a linear relation between \(x\) and \(y.\) The endpoint associations shown are \((x,y)=(3,5)\) and \((x,y)=(4,1).\) The slope is \(\frac{1-5}{4-3}=-4,\) so \(y=-4x+17.\) If \(x=a,\) then \[ \begin{aligned} x+y &=a+(-4a+17)\\ &=17-3a. \end{aligned} \]

Therefore, the correct answer is C.