1969 AMC 12 真题（P版）

1 答案 B

由 \[ \frac{a+x}{b+x}=\frac cd \] 得 \(ad+dx=bc+cx\)。因此 \(x(d-c)=bc-ad\)，所以 \[ x=\frac{bc-ad}{d-c}=\frac{ad-bc}{c-d}\text{。}\]

因此，正确答案是 B。

English

From \[ \frac{a+x}{b+x}=\frac cd \] we obtain \(ad+dx=bc+cx.\) Hence \(x(d-c)=bc-ad,\) so \[ x=\frac{bc-ad}{d-c}=\frac{ad-bc}{c-d}. \]

Therefore, the correct answer is B.

2 答案 A

若成本为 \(C\)，则 \(x=0.85C\) 且 \(y=1.15C\)。因此 \[ \begin{aligned} y:x&=1.15C:0.85C\\ &=115:85=23:17 \end{aligned}\text{。}\]

因此，正确答案是 A。

English

If the cost is \(C,\) then \(x=0.85C\) and \(y=1.15C.\) Therefore \[ \begin{aligned} y:x&=1.15C:0.85C\\ &=115:85=23:17. \end{aligned} \]

Therefore, the correct answer is A.

3 答案 E

从 \(11000_2\) 中减去 \(1\)，需要依次向三个末尾的零借位。因此 \[ 11000_2-1_2=10111_2\text{。}\]

因此，正确答案是 E。

English

Subtracting \(1\) from \(11000_2\) requires borrowing through the three trailing zeros. Thus \[ 11000_2-1_2=10111_2. \]

Therefore, the correct answer is E.

4 答案 E

第一个有序对为 \((3-0,2+0)=(3,2)\)。第二个为 \((x-3,y+2)\)。第一坐标相等给出 \(x-3=3\)，所以 \(x=6\)。

因此，正确答案是 E。

English

The first pair is \((3-0,2+0)=(3,2).\) The second is \((x-3,y+2).\) Equality of the first coordinates gives \(x-3=3,\) so \(x=6.\)

Therefore, the correct answer is E.

5 答案 B

将 \(N-\frac{4}{N}=R\) 乘以 \(N\)，得到 \[ N^2-RN-4=0\text{。}\]由韦达定理，两根之和为 \(R\)。因为两根之积为 \(-4\)，所以两根都非零。

因此，正确答案是 B。

English

Multiplying \(N-\frac{4}{N}=R\) by \(N\) gives \[ N^2-RN-4=0. \] The sum of its two roots is \(R\) by Vieta’s formulas. Both roots are nonzero because their product is \(-4.\)

Therefore, the correct answer is B.

6 答案 C

设两个半径为 \(R\) 与 \(r\)。由圆环面积可得 \[ \pi(R^2-r^2)=\frac{25}{2}\pi\text{。}\]过切点的半径平分该弦。若弦长为 \(L\)，则 \[ \left(\frac L2\right)^2+r^2=R^2\text{。}\]因此 \(L=2\sqrt{R^2-r^2}\)，所以 \(L=2\sqrt{\frac{25}{2}}=5\sqrt2\)。

因此，正确答案是 C。

English

Let the radii be \(R\) and \(r.\) The ring area gives \[ \pi(R^2-r^2)=\frac{25}{2}\pi. \] The radius to the tangent point bisects the chord. If its length is \(L,\) then \[ \left(\frac L2\right)^2+r^2=R^2. \] Thus \(L=2\sqrt{R^2-r^2},\) so \(L=2\sqrt{\frac{25}{2}}=5\sqrt2.\)

Therefore, the correct answer is C.

7 答案 A

有 \(y_1=a+b+c\) 且 \(y_2=a-b+c\)。因此 \(y_1-y_2=2b=-6\)，所以 \(b=-3\)。

因此，正确答案是 A。

English

We have \(y_1=a+b+c\) and \(y_2=a-b+c.\) Therefore \(y_1-y_2=2b=-6,\) so \(b=-3.\)

Therefore, the correct answer is A.

8 答案 D

由弧度数之和得 \[ \begin{aligned} 6x+78&=360,\\ x&=47 \end{aligned}\text{。}\]因此三条弧的度数为 \(122^\circ\)、\(119^\circ\)、\(119^\circ\)。所对弧为 \(122^\circ\) 的圆周角为 \(61^\circ\)。

因此，正确答案是 D。

English

The arc sum gives \[ \begin{aligned} 6x+78&=360,\\ x&=47. \end{aligned} \] The arcs then measure \(122^\circ,\) \(119^\circ,\) and \(119^\circ.\) The angle intercepting the \(122^\circ\) arc measures \(61^\circ.\)

Therefore, the correct answer is D.

9 答案 C

这些整数从 \(2\) 到 \(2+51=53\)。它们的平均数是首项与末项的平均数：\[ \frac{2+53}{2}=\frac{55}{2}=27\frac12\text{。}\]

因此，正确答案是 C。

English

The integers run from \(2\) through \(2+51=53.\) Their mean is the average of the first and last: \[ \frac{2+53}{2}=\frac{55}{2}=27\frac12. \]

Therefore, the correct answer is C.

10 答案 C

设圆心为 \(O\)，半径为 \(r\)。到两条平行切线距离相等的点必在它们的中线上，而该中线经过 \(O\)。若该点到 \(O\) 的距离为 \(t\)，则它到任一切线的距离为 \(r\)，到圆的距离为 \(\lvert t-r\rvert\)。因此 \(\lvert t-r\rvert=r\)，得到 \(t=0\) 或 \(t=2r\)。当 \(t=0\) 时有一个点，当 \(t=2r\) 时有两个点，共 \(3\) 个。

因此，正确答案是 C。

English

Let the circle have center \(O\) and radius \(r.\) A point equidistant from the two parallel tangents must lie on their midline, which passes through \(O.\) If its distance from \(O\) is \(t,\) its distance from either tangent is \(r,\) while its distance from the circle is \(\lvert t-r\rvert.\) Thus \(\lvert t-r\rvert=r,\) giving \(t=0\) or \(t=2r.\) There is one point with \(t=0\) and two with \(t=2r,\) for a total of \(3.\)

Therefore, the correct answer is C.

11 答案 B

由三角不等式，\(PR+RQ\ge PQ\)，当 \(R\) 在线段 \(PQ\) 上时取等号。\(PQ\) 的斜率为 \(\frac{2-(-2)}{4-(-1)}=\frac{4}{5}\)。从 \(x=-1\) 移到 \(x=1\) 时，\(y\) 坐标增加 \((\frac{4}{5})(2)=\frac{8}{5}\)，所以 \[ m=-2+\frac85=-\frac25\text{。}\]

因此，正确答案是 B。

English

By the triangle inequality, \(PR+RQ\ge PQ,\) with equality when \(R\) lies on segment \(PQ.\) The slope of \(PQ\) is \(\frac{2-(-2)}{4-(-1)}=\frac{4}{5}.\) Moving from \(x=-1\) to \(x=1\) raises the \(y\)-coordinate by \((\frac{4}{5})(2)=\frac{8}{5},\) so \[ m=-2+\frac85=-\frac25. \]

Therefore, the correct answer is B.

12 答案 A

有 \[ F=x^2+\frac83x+\frac m2\text{。}\]要使它成为平方，必须等于 \((x+\frac{4}{3})^2\)，其常数项为 \(\frac{16}{9}\)。因此 \(\frac{m}{2}=\frac{16}{9}\)，所以 \(m=\frac{32}{9}\)，它位于 \(3\) 与 \(4\) 之间。

因此，正确答案是 A。

English

We have \[ F=x^2+\frac83x+\frac m2. \] For this to be a square, it must equal \((x+\frac{4}{3})^2,\) whose constant term is \(\frac{16}{9}.\) Hence \(\frac{m}{2}=\frac{16}{9},\) so \(m=\frac{32}{9},\) which lies between \(3\) and \(4.\)

Therefore, the correct answer is A.

13 答案 B

面积条件为 \[ \pi R^2=\frac ab\pi(R^2-r^2)\text{。}\]因此 \(bR^2=aR^2-ar^2\)，所以 \(ar^2=(a-b)R^2\)。从而 \[ \frac Rr=\sqrt{\frac a{a-b}}\text{，}\]且 \(R:r=\sqrt a:\sqrt{a-b}\)。

因此，正确答案是 B。

English

The area condition is \[ \pi R^2=\frac ab\pi(R^2-r^2). \] Thus \(bR^2=aR^2-ar^2,\) so \(ar^2=(a-b)R^2.\) Therefore \[ \frac Rr=\sqrt{\frac a{a-b}}, \] and \(R:r=\sqrt a:\sqrt{a-b}.\)

Therefore, the correct answer is B.

14 答案 A

将式子因式分解为 \[ \frac{(x-2)(x+2)}{(x-1)(x+1)}\text{。}\]在 \(-2\)、\(-1\)、\(1\)、\(2\) 处作符号表，可知它在 \[ (-\infty,-2)\cup(-1,1)\cup(2,\infty) \] 上为正。端点 \(\pm2\) 使式子为零，而 \(\pm1\) 处无定义。

因此，正确答案是 A。

English

Factor the expression as \[ \frac{(x-2)(x+2)}{(x-1)(x+1)}. \] A sign chart at \(-2,\) \(-1,\) \(1,\) and \(2\) shows it is positive on \[ (-\infty,-2)\cup(-1,1)\cup(2,\infty). \] The endpoints \(\pm2\) give zero and \(\pm1\) are undefined.

Therefore, the correct answer is A.

15 答案 D

三角形 \(OAB\) 为等边三角形。因为 \(OM\perp AB\)，所以 \(M\) 是 \(AB\) 的中点，从而 \(AM=\frac{r}{2}\) 且 \(\angle OAM=60^\circ\)。在直角三角形 \(AMD\) 中，斜边为 \(AM=\frac{r}{2}\)，所以 \[ AD=\frac r4,\qquad MD=\frac{r\sqrt3}{4}\text{。}\]因此 \[ \begin{aligned} [MDA]&=\frac12(AD)(MD)\\ &=\frac{r^2\sqrt3}{32} \end{aligned}\text{。}\]

因此，正确答案是 D。

English

Triangle \(OAB\) is equilateral. Since \(OM\perp AB,\) \(M\) is the midpoint of \(AB,\) so \(AM=\frac{r}{2}\) and \(\angle OAM=60^\circ.\) In right triangle \(AMD,\) the hypotenuse is \(AM=\frac{r}{2},\) giving \[ AD=\frac r4,\qquad MD=\frac{r\sqrt3}{4}. \] Hence \[ \begin{aligned} [MDA]&=\frac12(AD)(MD)\\ &=\frac{r^2\sqrt3}{32}. \end{aligned} \]

Therefore, the correct answer is D.

16 答案 E

第二项与第三项之和为 \[ -na^{n-1}b+\frac{n(n-1)}2a^{n-2}b^2\text{。}\]代入 \(a=kb\)，再除以非零量 \(na^{n-2}b\)，得 \[ -k+\frac{n-1}{2}=0\text{。}\]因此 \(n=2k+1\)。

所以正确答案是 E。

English

The sum of the second and third terms is \[ -na^{n-1}b+\frac{n(n-1)}2a^{n-2}b^2. \] Substituting \(a=kb\) and dividing by the nonzero quantity \(na^{n-2}b\) gives \[ -k+\frac{n-1}{2}=0. \] Therefore \(n=2k+1.\)

Therefore, the correct answer is E.

17 答案 D

令 \(t=2^x\)。于是 \[ \begin{aligned} t^2-8t+12&=0,\\ (t-2)(t-6)&=0 \end{aligned}\text{。}\]根 \(t=6\) 给出 \[ \begin{aligned} x&=\log_2 6\\ &=1+\log_2 3\\ &=1+\frac{\log3}{\log2} \end{aligned}\text{。}\]（方程还有一个解 \(x=1\)，但它不在选项中。）

所以正确答案是 D。

English

Set \(t=2^x.\) Then \[ \begin{aligned} t^2-8t+12&=0,\\ (t-2)(t-6)&=0. \end{aligned} \] The root \(t=6\) gives \[ \begin{aligned} x&=\log_2 6\\ &=1+\log_2 3\\ &=1+\frac{\log3}{\log2}. \end{aligned} \] (The equation is also satisfied by \(x=1,\) which is not a listed choice.)

Therefore, the correct answer is D.

18 答案 B

第一个图形由直线 \(x-y+2=0\)、\(3x+y-4=0\) 组成，第二个图形由直线 \(x+y-2=0\)、\(2x-5y+7=0\) 组成。第一组中的每条直线都与第二组中的每条直线相交。解这四组方程，得到四个互异的点：\[ \begin{gathered} (0,2),\quad(-1,1),\\ (1,1),\quad\left(\frac{13}{17},\frac{29}{17}\right) \end{gathered}\text{。}\]因而共有 \(4\) 个公共点。

所以正确答案是 B。

English

The first graph is the pair \(x-y+2=0,\) and \(3x+y-4=0,\) and the second is \(x+y-2=0,\) and \(2x-5y+7=0.\) Each line in the first pair meets each line in the second pair. Solving the four pairings gives four distinct points: \[ \begin{gathered} (0,2),\quad(-1,1),\\ (1,1),\quad\left(\frac{13}{17},\frac{29}{17}\right). \end{gathered} \] Thus there are \(4\) common points.

Therefore, the correct answer is B.

19 答案 B

令 \(u=x^2y^2\)。则 \[ \begin{aligned} u^2-10u+9&=0,\\ (u-1)(u-9)&=0 \end{aligned}\text{。}\]因为 \(x\) 和 \(y\) 都是正整数，所以 \(xy=1\) 或 \(xy=3\)。这些有序对为 \((1,1)\)、\((1,3)\)、\((3,1)\)，共 \(3\) 个。

所以正确答案是 B。

English

Let \(u=x^2y^2.\) Then \[ \begin{aligned} u^2-10u+9&=0,\\ (u-1)(u-9)&=0. \end{aligned} \] Since \(x\) and \(y\) are positive integers, this means \(xy=1\) or \(xy=3.\) The pairs are \((1,1),\) \((1,3),\) and \((3,1),\) for a total of \(3.\)

Therefore, the correct answer is B.

20 答案 C

两个因数满足 \[ \begin{aligned} 3.6\cdot10^{18}&\lt A\lt3.7\cdot10^{18},\\ 3.4\cdot10^{14}&\lt B\lt3.5\cdot10^{14} \end{aligned}\text{。}\]因此 \[ 12.24\cdot10^{32}\lt AB\lt12.95\cdot10^{32}\text{，}\]从而 \(10^{33}\lt P\lt10^{34}\)。所以 \(P\) 有 \(34\) 位。

所以正确答案是 C。

English

The factors satisfy \[ \begin{aligned} 3.6\cdot10^{18}&\lt A\lt3.7\cdot10^{18},\\ 3.4\cdot10^{14}&\lt B\lt3.5\cdot10^{14}. \end{aligned} \] Hence \[ 12.24\cdot10^{32}\lt AB\lt12.95\cdot10^{32}, \] so \(10^{33}\lt P\lt10^{34}.\) Therefore \(P\) has \(34\) digits.

Therefore, the correct answer is C.

21 答案 E

圆的半径为 \(\sqrt m\)。原点到该直线的距离是 \[ \frac{\lvert-\sqrt{2m}\rvert}{\sqrt{1^2+1^2}}=\sqrt m\text{。}\]因此对每个 \(m\ge0\)，该直线都与圆相切（当 \(m=0\) 时，图形退化为原点）。

所以正确答案是 E。

English

The circle has radius \(\sqrt m.\) The distance from the origin to the line is \[ \frac{\lvert-\sqrt{2m}\rvert}{\sqrt{1^2+1^2}}=\sqrt m. \] Thus the line is tangent for every \(m\ge0\) (with the \(m=0\) case degenerate at the origin).

Therefore, the correct answer is E.

22 答案 C

从 \(x=0\) 到 \(5\)，该区域是面积为 \(\frac12(5)(5)=12.5\) 的三角形。从 \(x=5\) 到 \(8\)，两端的高度分别为 \(5\) 和 \(11\)，所以梯形面积为 \[ \frac12(5+11)(3)=24\text{。}\]因此 \(K=12.5+24=36.5\)。

所以正确答案是 C。

English

From \(x=0\) to \(5,\) the region is a triangle of area \(\frac12(5)(5)=12.5.\) From \(x=5\) to \(8,\) the endpoint heights are \(5\) and \(11,\) so the trapezoid has area \[ \frac12(5+11)(3)=24. \] Thus \(K=12.5+24=36.5.\)

Therefore, the correct answer is C.

23 答案 A

区间中的每个整数都可写成 \(n!+k\)，其中 \(2\le k\le n-1\)。因为 \(n!\) 是 \(k\) 的倍数，所以 \(n!+k\) 也是它的倍数，并且 \(n!+k\gt k\)。因此每个这样的整数都是合数。（当 \(n=2\) 时，该区间为空。）所以区间中没有质数。

所以正确答案是 A。

English

Every integer in the interval has the form \(n!+k\) for some \(2\le k\le n-1.\) Because \(n!\) is divisible by \(k,\) so is \(n!+k,\) and \(n!+k\gt k.\) Thus every such integer is composite. (For \(n=2,\) the interval is empty.) Hence there are no primes in the interval.

Therefore, the correct answer is A.

24 答案 E

根据余数的定义， \[ \begin{aligned} P&\equiv R\pmod D,\\ P'&\equiv R'\pmod D \end{aligned}\text{。}\]两式相乘，得 \(PP'\equiv RR'\pmod D\)。由于每个数都有介于 \(0\) 与 \(D-1\) 之间的唯一余数，所以二者的余数必相等：\(r=r'\)。

所以正确答案是 E。

English

By definition of the remainders, \[ \begin{aligned} P&\equiv R\pmod D,\\ P'&\equiv R'\pmod D. \end{aligned} \] Multiplying gives \(PP'\equiv RR'\pmod D.\) Since each has a unique remainder between \(0\) and \(D-1,\) their remainders must be equal: \(r=r'.\)

Therefore, the correct answer is E.

25 答案 D

对数条件给出 \(\log_2(ab)\ge6\)，所以 \(ab\ge64\)。由算术平均值与几何平均值不等式， \[ a+b\ge2\sqrt{ab}\ge2\sqrt{64}=16\text{。}\]当 \(a=b=8\) 时取等号。

所以正确答案是 D。

English

The logarithm condition gives \(\log_2(ab)\ge6,\) so \(ab\ge64.\) By AM-GM, \[ a+b\ge2\sqrt{ab}\ge2\sqrt{64}=16. \] Equality is attained at \(a=b=8.\)

Therefore, the correct answer is D.

26 答案 B

将 \(M\) 置于原点，并让跨度落在 \(x\) 轴上。拱门的方程为 \(y=16-ax^2\)。由于 \((20,0)\) 在抛物线上，\(0=16-400a\)，所以 \(a=\frac{1}{25}\)。当 \(x=5\) 时， \[ y=16-\frac{25}{25}=15\text{。}\]

所以正确答案是 B。

English

Place \(M\) at the origin with the span on the \(x\)-axis. The arch has equation \(y=16-ax^2.\) Since \((20,0)\) lies on it, \(0=16-400a,\) so \(a=\frac{1}{25}.\) At \(x=5,\) \[ y=16-\frac{25}{25}=15. \]

Therefore, the correct answer is B.

27 答案 E

在第 \(n\) 英里中，\(v_n=\frac{k}{n-1}\)。因为路程为一英里，所需时间为 \[ T_n=\frac1{v_n}=\frac{n-1}{k}\text{。}\]条件 \(T_2=2\) 给出 \(\frac{1}{k}=2\)。因此 \(T_n=2(n-1)\)。

所以正确答案是 E。

English

For the \(n\)th mile, \(v_n=\frac{k}{n-1}.\) Since the distance is one mile, the time is \[ T_n=\frac1{v_n}=\frac{n-1}{k}. \] The condition \(T_2=2\) gives \(\frac{1}{k}=2.\) Therefore \(T_n=2(n-1).\)

Therefore, the correct answer is E.

28 答案 E

令 \(P=(x,y)\)，并取直径端点为 \((-1,0)\) 和 \((1,0)\)。条件化为 \[ \begin{aligned} PA^2+PB^2&=3,\\ 2x^2+2y^2+2&=3 \end{aligned}\text{。}\]因此 \(x^2+y^2=\frac{1}{2}\)。这是一整个位于给定单位圆内部、半径为 \(\frac{1}{\sqrt2}\) 的圆，包含无穷多个点。

所以正确答案是 E。

English

Let \(P=(x,y)\) and take the diameter endpoints as \((-1,0)\) and \((1,0).\) The condition becomes \[ \begin{aligned} PA^2+PB^2&=3,\\ 2x^2+2y^2+2&=3. \end{aligned} \] Thus \(x^2+y^2=\frac{1}{2}.\) This is an entire circle of radius \(\frac{1}{\sqrt2},\) lying inside the given unit circle. It contains infinitely many points.

Therefore, the correct answer is E.

29 答案 C

将两个定义相除，得到 \[ \frac yx=t^{\frac{t-1}{t-1}}=t\text{。}\]此外， \[ y=t^{\frac{t}{t-1}}=\left(t^{\frac{1}{t-1}}\right)^t=x^t\text{。}\]代入 \(t=\frac{y}{x}\)，得 \(y=x^{\frac{y}{x}}\)。两边取 \(x\) 次幂，得到 \(y^x=x^y\)。

所以正确答案是 C。

English

Dividing the definitions gives \[ \frac yx=t^{\frac{t-1}{t-1}}=t. \] Also, \[ y=t^{\frac{t}{t-1}}=\left(t^{\frac{1}{t-1}}\right)^t=x^t. \] Substituting \(t=\frac{y}{x}\) gives \(y=x^{\frac{y}{x}}.\) Raising both sides to the power \(x\) yields \(y^x=x^y.\)

Therefore, the correct answer is C.

30 答案 D

取 \(A=(-a,0)\)、\(B=(a,0)\)、\(C=(0,a)\)、\(P=(p,0)\)。这涵盖了斜边所在直线上的每一点。于是 \[ \begin{aligned} s&=(p+a)^2+(p-a)^2\\ &=2p^2+2a^2,\\ 2CP^2&=2(p^2+a^2)\\ &=2p^2+2a^2 \end{aligned}\text{。}\]因而对 \(P\) 的每个位置都有 \(s=2CP^2\)。

所以正确答案是 D。

English

Take \(A=(-a,0),\) \(B=(a,0),\) \(C=(0,a),\) and \(P=(p,0).\) This describes every point on the hypotenuse line. Then \[ \begin{aligned} s&=(p+a)^2+(p-a)^2\\ &=2p^2+2a^2,\\ 2CP^2&=2(p^2+a^2)\\ &=2p^2+2a^2. \end{aligned} \] Thus \(s=2CP^2\) for every position of \(P.\)

Therefore, the correct answer is D.

31 答案 D

各顶点的像为 \[ \begin{aligned} O&\mapsto(0,0),\\ A&\mapsto(1,0),\\ B&\mapsto(0,2),\\ C&\mapsto(-1,0) \end{aligned}\text{。}\]边 \(OA\) 映为从 \((0,0)\) 到 \((1,0)\) 的线段，\(CO\) 映为从 \((-1,0)\) 到 \((0,0)\) 的线段。在 \(AB\) 上，\(x=1\)，所以 \(u=1-\frac{v^2}{4}\)；在 \(BC\) 上，\(y=1\)，所以 \(u=\frac{v^2}{4}-1\)。它们是连接 \((\pm1,0)\) 与 \((0,2)\) 的两段上方抛物线弧。

所以正确答案是 D。

English

The vertices map as \[ \begin{aligned} O&\mapsto(0,0),\\ A&\mapsto(1,0),\\ B&\mapsto(0,2),\\ C&\mapsto(-1,0). \end{aligned} \] Side \(OA\) maps to the segment from \((0,0)\) to \((1,0),\) and \(CO\) maps to the segment from \((-1,0)\) to \((0,0).\) On \(AB,\) \(x=1,\) so \(u=1-\frac{v^2}{4};\) on \(BC,\) \(y=1,\) so \(u=\frac{v^2}{4}-1.\) These are the two upper parabolic arcs joining \((\pm1,0)\) to \((0,2).\)

Therefore, the correct answer is D.

32 答案 C

对任意多项式 \(p(n)\)，其系数和为 \(p(1)\)。这里该多项式表示 \(u_n\)，而初始条件给出 \(u_1=5\)。因此系数和为 \(5\)。

所以正确答案是 C。

English

For any polynomial \(p(n),\) the sum of its coefficients is \(p(1).\) Here the polynomial represents \(u_n,\) and the initial condition gives \(u_1=5.\) Therefore the coefficient sum is \(5.\)

Therefore, the correct answer is C.

33 答案 A

对等差数列而言，第十一项是前 \(21\) 项的平均数。因此两个数列的第十一项分别为 \(\frac{S_{21}}{21}\) 与 \(\frac{T_{21}}{21}\)。它们的比为 \[ \frac{S_{21}}{T_{21}} =\frac{7(21)+1}{4(21)+27} =\frac{148}{111} =\frac43\text{。}\]

所以正确答案是 A。

English

For an arithmetic sequence, the eleventh term is the average of the first \(21\) terms. Thus the respective eleventh terms are \(\frac{S_{21}}{21}\) and \(\frac{T_{21}}{21}.\) Their ratio is \[ \frac{S_{21}}{T_{21}} =\frac{7(21)+1}{4(21)+27} =\frac{148}{111} =\frac43. \]

Therefore, the correct answer is A.

34 答案 B

写成 \(R(x)=ax+b\)。由于除式为 \((x-1)(x-2)\)，在它的两个根处代入多项式除法恒等式，得到 \[ R(1)=1,\qquad R(2)=2^{100}\text{。}\]因此 \[ \begin{aligned} R(x)&=1+(2^{100}-1)(x-1)\\ &=2^{100}(x-1)-(x-2) \end{aligned}\text{。}\]

所以正确答案是 B。

English

Write \(R(x)=ax+b.\) Since the divisor is \((x-1)(x-2),\) evaluating the division identity at its roots gives \[ R(1)=1,\qquad R(2)=2^{100}. \] Therefore \[ \begin{aligned} R(x)&=1+(2^{100}-1)(x-1)\\ &=2^{100}(x-1)-(x-2). \end{aligned} \]

Therefore, the correct answer is B.

35 答案 B

左侧交点满足 \[ L(m)=-\sqrt{6+m}\text{。}\]因此 \[ \begin{aligned} r&=\frac{-\sqrt{6-m}+\sqrt{6+m}}{m}\\ &=\frac{2}{\sqrt{6+m}+\sqrt{6-m}} \end{aligned}\text{。}\]当 \(m\) 趋近 \(0\) 时，该式趋近于 \(\frac{2}{2\sqrt6}=\frac{1}{\sqrt6}\)。

所以正确答案是 B。

English

The left intersection satisfies \[ L(m)=-\sqrt{6+m}. \] Hence \[ \begin{aligned} r&=\frac{-\sqrt{6-m}+\sqrt{6+m}}{m}\\ &=\frac{2}{\sqrt{6+m}+\sqrt{6-m}}. \end{aligned} \] As \(m\) approaches \(0,\) this approaches \(\frac{2}{2\sqrt6}=\frac{1}{\sqrt6}.\)

Therefore, the correct answer is B.