國立馬祖高級中學114學年度第1次專任教師甄選
單選題(一題3分，共75分)
$$假設\overline{AD}與\overline{BC}交於P,及\cases{\overline{A_4P}=x\\ \overline{A_5P} =y}, 如上圖\\ \triangle PAB \sim \triangle PDC (AAA) \Rightarrow \cases{\overline{BP}:\overline{PC}=a:b\\ \overline{AP}:\overline{PD}=a:b}\\ 又\overline{A_4A_5} \parallel \overline{CD} \Rightarrow \cases{ \overline{PA_5}: \overline{CD} =\overline{PB}:\overline{BC} \\ \overline{A_4P}: \overline{CD} =\overline{AP}:\overline{AD}} \Rightarrow \cases{y:b=a:a+b\\ x:b=a:a+b} \Rightarrow x=y={ab\over a+b} \\ \Rightarrow \overline{A_4B_4}=x+y={2ab\over a+b}，故選\bbox[red, 2pt]{(D)}$$
解答:$$令\cases{m=a^{1/3} \\n=b^{1/3}} \Rightarrow \cases{m^2+n^2=4\\ x=m^3+3mn^2\\ y=n^3+3m^2n} \Rightarrow \cases{x+y=m^3+n^3+3mn(m+n) =(m+n)^3\\x-y=m^3-n^3+3mn(n-m) =(m-n)^3} \\ \Rightarrow (x+y)^{2/3} +(x-y)^{2/3}=(m+n)^2+(m-n)^2 =2(m^2+n^2) =2\cdot 4=8，故選\bbox[red, 2pt]{(A)}$$
解答:$$a^2-2ab-9b^2=0 \Rightarrow \cases{a^2=2ab+9b^2\\ a=(\sqrt{10}+1)b} \Rightarrow \log(a^2+ab-6b^2)-\log(a^2+4ab+15b^2) \\=\log(3ab+3b^2)-\log(6ab+24b^2) =\log{3b(a+b)\over 6b(a+4b)} =\log\left( {1\over 2}\cdot {a+b\over a+4b}\right) =\log\left( {1\over 2}\cdot {(\sqrt {10}+2)b\over (\sqrt{10}+5)b} \right) \\=\log\left( {1\over 2}\cdot {(\sqrt {10}+2)(5-\sqrt{10})\over (\sqrt{10}+5)(5-\sqrt{10})} \right) =\log\left( {1\over 2}\dot{3\sqrt{10}\over 15} \right) =\log{\sqrt{10}\over 10}=-{1\over 2}，故選\bbox[red, 2pt]{(E)}$$
解答:$$z=x+yi, 其中x,y\in \mathbb R \Rightarrow {z\over z-1}=1+{1\over z-1} =1+{1\over (x-1)+yi} \\=1+{x-1-yi\over (x-1)^2+y^2} =1+{x-1\over (x-1)^2+y^2}-{yi\over (x-1)^2+y^2} 為純虛數 \Rightarrow 1+{x-1\over (x-1)^2+y^2}=0 \\ \Rightarrow (x-1)^2+y^2+x-1=0 \Rightarrow (x-{1\over 2})^2+y^2={1\over 4} 為一圓，故選\bbox[red, 2pt]{(A)}$$
解答:$$符合要求的樣本:DCBA, DBCA, DBAC, BDCA, BDAC, BADC，共六種，機率為{6\over 4!}={1\over 4}\\，故選\bbox[red, 2pt]{(C)}$$
解答:$$\cos \angle C={7^2+8^2-13^2\over 2\cdot 7\cdot 8}=-{1\over 2} \Rightarrow \angle C=120^\circ，故選\bbox[red, 2pt]{(D)}$$
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解答:$$\cases{L_1: y=-{4\over 3}(x-a) \\L_2: y=-{3\over 2}(x-a)} \Rightarrow \cases{L_1與y軸交於A(0,4a/3) \\L_2與y軸交於B(0,3a/2)} \Rightarrow \overline{AB}={a\over 6} \\ \Rightarrow \triangle PAB={1\over 2}\cdot {a\over 6}\cdot a=6 \Rightarrow a^2=72 \Rightarrow a=6\sqrt 2，故選\bbox[red, 2pt]{(C)}$$
解答:$$a^2-2ab-9b^2=0 \Rightarrow \cases{a^2=2ab+9b^2\\ a=(\sqrt{10}+1)b} \Rightarrow \log(a^2+ab-6b^2)-\log(a^2+4ab+15b^2) \\=\log(3ab+3b^2)-\log(6ab+24b^2) =\log{3b(a+b)\over 6b(a+4b)} =\log\left( {1\over 2}\cdot {a+b\over a+4b}\right) =\log\left( {1\over 2}\cdot {(\sqrt {10}+2)b\over (\sqrt{10}+5)b} \right) \\=\log\left( {1\over 2}\cdot {(\sqrt {10}+2)(5-\sqrt{10})\over (\sqrt{10}+5)(5-\sqrt{10})} \right) =\log\left( {1\over 2}\dot{3\sqrt{10}\over 15} \right) =\log{\sqrt{10}\over 10}=-{1\over 2}，故選\bbox[red, 2pt]{(E)}$$
解答:$$符合要求的樣本:DCBA, DBCA, DBAC, BDCA, BDAC, BADC，共六種，機率為{6\over 4!}={1\over 4}\\，故選\bbox[red, 2pt]{(C)}$$
解答:
$$\angle B=2\angle C \Rightarrow \angle A+\angle B+\angle C=135^\circ+3\angle C=180^\circ \Rightarrow \angle C=15^\circ \Rightarrow \angle B=30^\circ\\ 正弦定理:{\overline{AC} \over \sin \angle B} ={\overline{BC} \over \sin \angle A} \Rightarrow {\overline{AC} \over 1/2} ={4\over \sqrt 2/2 } \Rightarrow \overline{AC}=2\sqrt 2\\ 餘弦定理: \cos \angle C=\cos 15^\circ ={\sqrt 6+\sqrt 2\over 4} ={2^2+(2\sqrt 2)^2-\overline{AM}^2\over 2\cdot 2\cdot 2\sqrt 2} ={12-\overline{AM}^2\over 8\sqrt 2} \\ \Rightarrow \overline{AM}^2= 8-2\sqrt{12} \Rightarrow \overline{AM}=\sqrt 6-\sqrt 2，故選\bbox[red, 2pt]{(E)}$$
解答:$$S=p+2p^2+3p^3+ \cdots \Rightarrow pS=p^2+2p^3+3p^4 +\cdots \\ \Rightarrow S-pS=p+p^2+p^3+\cdots \Rightarrow (1-p)S={p\over 1-p} \Rightarrow S={p\over (1-p)^2} ={1/3\over 4/9} ={3\over 4}，故選\bbox[red, 2pt]{(A)}$$
解答:$$\begin{bmatrix}a &b & c \\d & e & f \\g & h & i\end{bmatrix}^{101} =\begin{bmatrix}-1 &-2 & -2 \\1 & 2 & 1 \\-1 & -1 & 0\end{bmatrix} = \begin{bmatrix}-1 & -1 & 2 \\1 & 0 & -1 \\0 & 1 & 1\end{bmatrix} \begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & -1\end{bmatrix} \begin{bmatrix} \frac{1}{2} & \frac{3}{2} & \frac{1}{2} \\\frac{-1}{2} & \frac{-1}{2} & \frac{1}{2} \\\frac{1}{2} & \frac{1}{2} & \frac{1}{2}\end{bmatrix} \\ \Rightarrow \begin{bmatrix}a &b & c \\d & e & f \\g & h & i\end{bmatrix} =\begin{bmatrix}-1 &-2 & -2 \\1 & 2 & 1 \\-1 & -1 & 0\end{bmatrix}^{1/101} = \begin{bmatrix}-1 & -1 & 2 \\1 & 0 & -1 \\0 & 1 & 1\end{bmatrix} \begin{bmatrix}1^{1/101} & 0 & 0 \\0 & 1^{1/101} & 0 \\0 & 0 & (-1)^{1/101}\end{bmatrix} \begin{bmatrix}\frac{1}{2} & \frac{3}{2} & \frac{1}{2} \\\frac{-1}{2} & \frac{-1}{2} & \frac{1}{2} \\\frac{1}{2} & \frac{1}{2} & \frac{1}{2}\end{bmatrix} \\ = \begin{bmatrix}-1 & -1 & 2 \\1 & 0 & -1 \\0 & 1 & 1\end{bmatrix} \begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & -1\end{bmatrix} \begin{bmatrix} \frac{1}{2} & \frac{3}{2} & \frac{1}{2} \\\frac{-1}{2} & \frac{-1}{2} & \frac{1}{2} \\\frac{1}{2} & \frac{1}{2} & \frac{1}{2}\end{bmatrix} =\begin{bmatrix}-1 &-2 & -2 \\1 & 2 & 1 \\-1 & -1 & 0\end{bmatrix} \\ \Rightarrow a+b+\cdots +i=-1-2-2+1+2+1-1-1=-3，故選\bbox[red, 2pt]{(D)}$$
解答:$$假設三根為{a\over r},a,ar \Rightarrow 三根之積=a^3= 8 \Rightarrow a=2 \Rightarrow f(2)=0\\ \Rightarrow 8-28+2c-8=0 \Rightarrow c=14，故選\bbox[red, 2pt]{(B)}$$
解答:$$4=\sqrt{16} =\sqrt{6+10} =\sqrt{6+2\sqrt{25}}=\sqrt{6+2\sqrt{7+18}} =\sqrt{6+2\sqrt{7+3\sqrt{36}}} \\=\sqrt{6+2\sqrt{7+3\sqrt{8+ 28}}} = \cdots，故選\bbox[red, 2pt]{(D)}$$
解答:$$令\cases{A(x_1,y_1) \\B(x_2,y_2)\\ L斜率為m} \Rightarrow \cases{2x_1^2-5y_1^2=20\\ 2x_2^2-5y_2^2=20} \Rightarrow 兩式相減:2(x_2^2-x_1^2)=5(y_2^2-y_1^2) \\ \Rightarrow {y_2^2-y_1^2\over x_2^2-x_1^2} ={y_2+y_1\over x_2+x_1}\cdot {y_2-y_1\over x_2-x_1} ={y_2+y_1\over x_2+x_1}\cdot m={2\over 5} \cdots(1)\\ M=(A+B)/2 \Rightarrow \cases{x_1+x_2=10\\ y_1+y_2 =2} 代入(1) \Rightarrow {2\over 10}\cdot m={2\over 5} \Rightarrow m=2 \\ \Rightarrow L: y=2(x-5)+1 \Rightarrow y=2x-9，故選\bbox[red, 2pt]{(C)}$$
解答:$$A(3,1,4)對稱\pi的對稱點A'(-1,-1,2) \Rightarrow \overleftrightarrow{A'B} \cap \pi=(2,1,0)，故選\bbox[red, 2pt]{(D)}$$
解答:$$假設三根為{a\over r},a,ar \Rightarrow 三根之積=a^3= 8 \Rightarrow a=2 \Rightarrow f(2)=0\\ \Rightarrow 8-28+2c-8=0 \Rightarrow c=14，故選\bbox[red, 2pt]{(B)}$$
解答:$$4=\sqrt{16} =\sqrt{6+10} =\sqrt{6+2\sqrt{25}}=\sqrt{6+2\sqrt{7+18}} =\sqrt{6+2\sqrt{7+3\sqrt{36}}} \\=\sqrt{6+2\sqrt{7+3\sqrt{8+ 28}}} = \cdots，故選\bbox[red, 2pt]{(D)}$$
$$\tan \angle B=\tan 2\theta={\overline{AC} \over \overline{BC}} ={{3\over 4}} ={2\tan \theta \over 1-\tan^2 \theta}\Rightarrow \tan \theta ={1\over 3} ={r\over 40-5r} \Rightarrow r=5，故選\bbox[red, 2pt]{(E)}$$
解答:$$\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} \begin{bmatrix}a & b\\c & d\end{bmatrix} \begin{bmatrix}1 & 0\\0 & -2\end{bmatrix} = \begin{bmatrix}c & d\\a & b\end{bmatrix} \begin{bmatrix}1 & 0\\0 & -2\end{bmatrix} = \begin{bmatrix}c & -2d\\a & -2b\end{bmatrix} = \begin{bmatrix}3 & -4\\-9 & -7\end{bmatrix} \\ \Rightarrow \cases{c =3\\ -2b=-7} \Rightarrow c+2b=3+7=10，故選\bbox[red, 2pt]{(D)}$$
解答:$$\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} \begin{bmatrix}a & b\\c & d\end{bmatrix} \begin{bmatrix}1 & 0\\0 & -2\end{bmatrix} = \begin{bmatrix}c & d\\a & b\end{bmatrix} \begin{bmatrix}1 & 0\\0 & -2\end{bmatrix} = \begin{bmatrix}c & -2d\\a & -2b\end{bmatrix} = \begin{bmatrix}3 & -4\\-9 & -7\end{bmatrix} \\ \Rightarrow \cases{c =3\\ -2b=-7} \Rightarrow c+2b=3+7=10，故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{L_1: y=-{4\over 3}(x-a) \\L_2: y=-{3\over 2}(x-a)} \Rightarrow \cases{L_1與y軸交於A(0,4a/3) \\L_2與y軸交於B(0,3a/2)} \Rightarrow \overline{AB}={a\over 6} \\ \Rightarrow \triangle PAB={1\over 2}\cdot {a\over 6}\cdot a=6 \Rightarrow a^2=72 \Rightarrow a=6\sqrt 2，故選\bbox[red, 2pt]{(C)}$$
解答:$$a_n=10^n \Rightarrow b=\sum_{n=1}^3\log_{a_n}a_{n+1}= \sum_{n=1}^3{\log a_{n+1}\over \log a_n} ={\log 10^2 \over \log 10} +{\log 10^3 \over \log 10^2} +{\log 10^4 \over \log 10^3} \\=2+{3\over 2}+{4\over 3} =4+{5\over 6} \Rightarrow 4\lt b\le 5，故選\bbox[red, 2pt]{(C)}$$
解答:$$(A) 奇數有50個\\ (B)質數有25個\\ (C)7的倍數有\lfloor 100/7 \rfloor =14個\\ (D) \cases{2的倍數50個\\ 3的倍數33個\\ 5的倍數20個\\ 6的倍數16個\\ 10的倍數10個\\ 15的倍數6個\\ 30的倍數3個} \Rightarrow 不是2,3,5的倍數有100-(50+33+20)+(16+10+6)-3=26 \\(E)號碼不大於15的有15個\\ \Rightarrow 分母最小的是(C)，故選\bbox[red, 2pt]{(C)}$$
解答:$$令\cases{\vec a=(1,0,0)\\ \vec b=(1,1,0) \\ \vec c=(0,1,0) \\ \vec d=(1,0,1) \\ \vec e=(1,1,1) \\ \vec f=(0,1,1) \\\vec g=(0,0,1)} \Rightarrow \cases{\vec a\cdot(\vec b+\vec c+\cdots+ \vec g)=3\\ \vec b\cdot (\vec c+\vec d+\cdots+ \vec g)=5\\ \vec c \cdot (\vec d +\vec e +\cdots+ \vec g)=2\\ \vec d \cdot(\vec e+ \vec f +\vec g)=4\\ \vec e \cdot (\vec f+ \vec g)=3\\ \vec f\cdot \vec g=1} \Rightarrow 合計18 \Rightarrow 期望值{18\over C^7_2} ={6\over 7}，故選\bbox[red, 2pt]{(C)}$$
解答:
解答:$$(A)\times: 3a_2=2+1 \Rightarrow a_2=1 \ne 2\\ (B)\bigcirc: b_2=1-{2\over 2}+{3\over 4} ={3\over 4} \\(C) \times: 3a_{n+1}=a_n+n \Rightarrow 3a_n=a_{n-1}+n-1 \Rightarrow a_n={1\over 3}a_{n-1}+{n-1\over 3 }\\ \qquad \Rightarrow a_n-{n\over 2}+{3\over 4}={1\over 3}(a_{n-1}-{n-1\over 2}+{3\over 4}) \Rightarrow b_n={1\over 3}b_{n-1} \Rightarrow 公比為{1\over 3}\ne {2\over 3}\\(D) \bigcirc: a_n={1\over 3}a_{n-1}+{n-1\over 3} ={1\over 3}(({1\over 3}a_{n-2}+{n-2\over 3}) +{n-1\over 3}) ={1\over 3^2}a_{n-2}+{(n-2)+(n-1)\over 3^2} \\\quad =\cdots ={1\over 3^{n-1}}a_1+{1+2+\cdots+ (n-1)\over 3^{n-1}} \Rightarrow 3^na_n=3a_1+3(1+2+\cdots+(n-1))為正整數\\ (E)\times: b_n={1\over 3}b_{n-1} ={1\over 3^2}b_{n-2} =\cdots ={1\over 3^{n-1}}b_1={1\over 3^{n-1}}\cdot {}{9\over 4} ={1\over 4} \cdot {1\over 3^{n-3}} \Rightarrow b_{10}= {1\over 4}\cdot {1\over 3^7} \\\qquad \Rightarrow \log b_{10}=-(2\log 2+7\log 3)\approx -3.94 \gt -4 \Rightarrow b_{10} \gt 10^{-4}\\，故選\bbox[red, 2pt]{(BD)}$$
解答:$$(A)\bigcirc: \cases{a_n=81 =1+(n-1)d \\ a_m=9=1+(m-1)d} \Rightarrow \cases{80=(n-1)d\\ 8=(m-1)d} \Rightarrow 10={n-1\over m-1} \Rightarrow 10m=n+9 \\\qquad \Rightarrow 10m是偶數 \Rightarrow n+9是偶數\Rightarrow n是奇數 \\(B) \bigcirc:n是奇數 \Rightarrow a_1與a_n的中間項a_{(n+1)/2}=a_1+\left({n+1\over 2}-1 \right)d =a_1+{n-1\over 2}d= 1+{1\over 2}\cdot 80 \\\qquad =41 \Rightarrow 41也在數列上 \\(C) \times: d={1\over 2}仍符合要求\\ (D)\times: d=1,2, {1\over 2},\dots 有無窮多個都符合要求 \\(E)\times: n=7 \Rightarrow d=40/3 \Rightarrow 41=1+(4-1)d \Rightarrow a_4=41仍符要求\\，故選\bbox[red, 2pt]{(AB)}$$
解答:$$假設球半徑R,各點球坐標\cases{A(R,0,\pi/6)\\ B(R,\pi, \pi/6) \\ C(R,0,\pi/3)\\ D(R,\pi,\pi/3)\\E(R,0,\pi/2) \\北極點P(R,0,0)} \\(A)\bigcirc: \stackrel{\Large \frown}{AP}=R\cdot \pi/6 =\stackrel{\Large \frown}{BP} \\(B)\times: \cases{\stackrel{\Large \frown}{AB}=R\sin{\pi\over 6}\cdot \pi={1\over 2}\pi R\\\stackrel{\Large \frown}{CD} =R \sin{\pi\over 3}\cdot \pi={\sqrt 3\over 2}\pi R} \Rightarrow \stackrel{\Large \frown}{AB} \ne\stackrel{\Large \frown}{CD} \\(C)\bigcirc: A,C,E在同一經度上 \\(D) \bigcirc: \cases{\stackrel{\Large \frown}{CD} ={\sqrt 3\over 2}\pi R \\\stackrel{\Large \frown}{CPD} ={2\over 3}\pi R} \Rightarrow 經過北極路徑比較短 \\(E)\times:\cases{ \stackrel{\Large \frown}{EP} ={1\over 2}\pi R\\ \stackrel{\Large \frown}{CD} ={\sqrt 3\over 2}\pi R } \Rightarrow \stackrel{\Large \frown}{EP}: \stackrel{\Large \frown}{CD}=1:\sqrt 3\\，故選\bbox[red, 2pt]{(ACD)}$$
解答:$$(A) 奇數有50個\\ (B)質數有25個\\ (C)7的倍數有\lfloor 100/7 \rfloor =14個\\ (D) \cases{2的倍數50個\\ 3的倍數33個\\ 5的倍數20個\\ 6的倍數16個\\ 10的倍數10個\\ 15的倍數6個\\ 30的倍數3個} \Rightarrow 不是2,3,5的倍數有100-(50+33+20)+(16+10+6)-3=26 \\(E)號碼不大於15的有15個\\ \Rightarrow 分母最小的是(C)，故選\bbox[red, 2pt]{(C)}$$
解答:$$令\cases{\vec a=(1,0,0)\\ \vec b=(1,1,0) \\ \vec c=(0,1,0) \\ \vec d=(1,0,1) \\ \vec e=(1,1,1) \\ \vec f=(0,1,1) \\\vec g=(0,0,1)} \Rightarrow \cases{\vec a\cdot(\vec b+\vec c+\cdots+ \vec g)=3\\ \vec b\cdot (\vec c+\vec d+\cdots+ \vec g)=5\\ \vec c \cdot (\vec d +\vec e +\cdots+ \vec g)=2\\ \vec d \cdot(\vec e+ \vec f +\vec g)=4\\ \vec e \cdot (\vec f+ \vec g)=3\\ \vec f\cdot \vec g=1} \Rightarrow 合計18 \Rightarrow 期望值{18\over C^7_2} ={6\over 7}，故選\bbox[red, 2pt]{(C)}$$
解答:$$令\cases{A:只報考數學A\\ B:只報考數學B\\ C:兩科都報考} \Rightarrow \cases{P_A=2/5\\ P_C={1\over 2}(P_A+P_C) \\P_A+P_B+P_C = 1} \Rightarrow \cases{P_A=2/5\\ P_B=2/5\\ P_C=1/5} \Rightarrow {P_C\over P_B+P_C}={1\over 3}，故選\bbox[red, 2pt]{(B)}$$
解答:$$f(x)=x^2+bx+c \Rightarrow \cases{f(x-2)= x^2+(b-4)x+4-2b+c \\ f(-x-2)= x^2+(4-b)x+4-2b+c} \\ f(x-2)=f(-x-2) \Rightarrow b=4 \Rightarrow f(x)=x^2+4x+c =(x+2)^2+c-4 \\ \Rightarrow 在-3\le x\le 1時，\cases{最大值f(1)=5+c\\ 最小值f(-2)=c-4} \Rightarrow 5+c=4(c-4) \Rightarrow c=7 \Rightarrow f(-2)=3\\，故選\bbox[red, 2pt]{(C)}$$
解答:$$f(x)=x^2+bx+c \Rightarrow \cases{f(x-2)= x^2+(b-4)x+4-2b+c \\ f(-x-2)= x^2+(4-b)x+4-2b+c} \\ f(x-2)=f(-x-2) \Rightarrow b=4 \Rightarrow f(x)=x^2+4x+c =(x+2)^2+c-4 \\ \Rightarrow 在-3\le x\le 1時，\cases{最大值f(1)=5+c\\ 最小值f(-2)=c-4} \Rightarrow 5+c=4(c-4) \Rightarrow c=7 \Rightarrow f(-2)=3\\，故選\bbox[red, 2pt]{(C)}$$
解答:
$$\cases{O(0,0)\\ P(6,8)} \Rightarrow \overline{OP}的中垂線L:=-{3\over 4}(x-3)+4 \Rightarrow L與坐標軸交於\cases{ A(25/3,0) \\ B(0,25/4)} \\ \Rightarrow \triangle OAB={1\over 2}\cdot {25\over 3}\cdot {25\over 4} ={625\over 24}，故選\bbox[red, 2pt]{(B)}$$
解答:
$$\cases{\overrightarrow{P_2P_3} =k\overrightarrow{P_3A} \\\overrightarrow{P_5 P_6} = t\overrightarrow{AP_6} },如上圖 \Rightarrow \overrightarrow{P_2P_3}與\overrightarrow{P_5P_6}的夾角= \angle P_3AB =135^\circ \\ \Rightarrow \overrightarrow{P_2P_3} \cdot \overrightarrow{P_5P_6} =2 \cdot 2 \sin 135^\circ={2\sqrt 2}，故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{\vec u-\vec v=(2,-1,0) \\ \vec v-\vec w=(-1,2,3)} \Rightarrow (\vec u-\vec v) \cdot (\vec v-\vec w)=0-0-\vec v\cdot \vec v+0 =-|\vec v|^2= -2-2+0 \Rightarrow |\vec v|=2 \\ \Rightarrow (\vec u-\vec v) \cdot (\vec u-\vec v)=|\vec u|^2-0-0+|\vec v|^2=|\vec u|^2+4=(2,-1,0) \cdot (2,-1,0)=5 \Rightarrow |\vec u|=1\\ \Rightarrow (\vec v-\vec w) \cdot (\vec v-\vec w) =|\vec v|^2-0-0+|\vec w|^2=(-1,2,3) \cdot (-1,2,3)=14 \Rightarrow |\vec w|=\sqrt{10} \\ \Rightarrow 體積=2\times 1\times \sqrt{10}= 2\sqrt{10}，故選\bbox[red, 2pt]{(C)}$$
多選題(一題5分，共25分)
解答:$$(A)\times: \cases{\alpha=1\\ \beta=3\pi/7} \Rightarrow z=e^{3\pi i/7} \Rightarrow \cases{z^{10} =e^{30\pi i/7} =e^{2\pi i/7} \Rightarrow x_{10 \gt 0}\\z^3=e^{9\pi i/7} \Rightarrow x_3 \lt 0} \Rightarrow x_{10} \ne x_3 \\(B) \bigcirc:y_3=0 \Rightarrow \sin 3\beta =0 \Rightarrow 3\beta =k\pi \Rightarrow 6\beta=2k\pi \Rightarrow \sin 6\beta=0 \Rightarrow y_6=0, k\in \mathbb Z \\(C) \times: x_3=1 \Rightarrow \alpha^3\cos 3\beta=1 無法保證 \alpha^6 \cos 6\beta =1 \\(D)\times: \cases{\beta=\pi\\ \alpha=2} \Rightarrow \langle y_n=\alpha^n \sin n\beta =0\rangle 收斂, 但\alpha \not \le 1 \\(E)\bigcirc: \langle x_n= \alpha^n \cos n\beta \rangle 收斂\Rightarrow \alpha \lt 1 \Rightarrow \lim_{n\to \infty} \alpha^n \sin n\beta =0 \Rightarrow \langle y_n\rangle 收斂\\，故選\bbox[red, 2pt]{(BE)}$$
解答:$$(A)\bigcirc: f'(x)=5x^4-15x^2+10x \Rightarrow f'(1)=5-15+10=0\\ (B)\bigcirc: f'(x)=5x^4-15x^2+10x =5x(x-1)^2(x+2) \Rightarrow f'(x)\ge 0, x\in [0,2] \\(C)\times: f''(x)=20x^3-30x+10 \Rightarrow f''({1\over 2}) \lt 0 \Rightarrow 在區間[0,2]並非均凹向上\\ (D)\times: g(x)的週期為6,不是6\pi \\(E)\bigcirc: 由(B)可知:f'(x)\gt 0, x\in [3,4]; 又g'(x)={\pi\over 3}\cos({\pi x\over 3}+{\pi\over 2}) \ge 0, x\in [3,4]\\，故選\bbox[red, 2pt]{(ABE)}$$
解答:$$(A)\bigcirc: f'(x)=5x^4-15x^2+10x \Rightarrow f'(1)=5-15+10=0\\ (B)\bigcirc: f'(x)=5x^4-15x^2+10x =5x(x-1)^2(x+2) \Rightarrow f'(x)\ge 0, x\in [0,2] \\(C)\times: f''(x)=20x^3-30x+10 \Rightarrow f''({1\over 2}) \lt 0 \Rightarrow 在區間[0,2]並非均凹向上\\ (D)\times: g(x)的週期為6,不是6\pi \\(E)\bigcirc: 由(B)可知:f'(x)\gt 0, x\in [3,4]; 又g'(x)={\pi\over 3}\cos({\pi x\over 3}+{\pi\over 2}) \ge 0, x\in [3,4]\\，故選\bbox[red, 2pt]{(ABE)}$$
解答:$$(A)\times: 3a_2=2+1 \Rightarrow a_2=1 \ne 2\\ (B)\bigcirc: b_2=1-{2\over 2}+{3\over 4} ={3\over 4} \\(C) \times: 3a_{n+1}=a_n+n \Rightarrow 3a_n=a_{n-1}+n-1 \Rightarrow a_n={1\over 3}a_{n-1}+{n-1\over 3 }\\ \qquad \Rightarrow a_n-{n\over 2}+{3\over 4}={1\over 3}(a_{n-1}-{n-1\over 2}+{3\over 4}) \Rightarrow b_n={1\over 3}b_{n-1} \Rightarrow 公比為{1\over 3}\ne {2\over 3}\\(D) \bigcirc: a_n={1\over 3}a_{n-1}+{n-1\over 3} ={1\over 3}(({1\over 3}a_{n-2}+{n-2\over 3}) +{n-1\over 3}) ={1\over 3^2}a_{n-2}+{(n-2)+(n-1)\over 3^2} \\\quad =\cdots ={1\over 3^{n-1}}a_1+{1+2+\cdots+ (n-1)\over 3^{n-1}} \Rightarrow 3^na_n=3a_1+3(1+2+\cdots+(n-1))為正整數\\ (E)\times: b_n={1\over 3}b_{n-1} ={1\over 3^2}b_{n-2} =\cdots ={1\over 3^{n-1}}b_1={1\over 3^{n-1}}\cdot {}{9\over 4} ={1\over 4} \cdot {1\over 3^{n-3}} \Rightarrow b_{10}= {1\over 4}\cdot {1\over 3^7} \\\qquad \Rightarrow \log b_{10}=-(2\log 2+7\log 3)\approx -3.94 \gt -4 \Rightarrow b_{10} \gt 10^{-4}\\，故選\bbox[red, 2pt]{(BD)}$$
解答:$$(A)\bigcirc: \cases{a_n=81 =1+(n-1)d \\ a_m=9=1+(m-1)d} \Rightarrow \cases{80=(n-1)d\\ 8=(m-1)d} \Rightarrow 10={n-1\over m-1} \Rightarrow 10m=n+9 \\\qquad \Rightarrow 10m是偶數 \Rightarrow n+9是偶數\Rightarrow n是奇數 \\(B) \bigcirc:n是奇數 \Rightarrow a_1與a_n的中間項a_{(n+1)/2}=a_1+\left({n+1\over 2}-1 \right)d =a_1+{n-1\over 2}d= 1+{1\over 2}\cdot 80 \\\qquad =41 \Rightarrow 41也在數列上 \\(C) \times: d={1\over 2}仍符合要求\\ (D)\times: d=1,2, {1\over 2},\dots 有無窮多個都符合要求 \\(E)\times: n=7 \Rightarrow d=40/3 \Rightarrow 41=1+(4-1)d \Rightarrow a_4=41仍符要求\\，故選\bbox[red, 2pt]{(AB)}$$
解答:$$假設球半徑R,各點球坐標\cases{A(R,0,\pi/6)\\ B(R,\pi, \pi/6) \\ C(R,0,\pi/3)\\ D(R,\pi,\pi/3)\\E(R,0,\pi/2) \\北極點P(R,0,0)} \\(A)\bigcirc: \stackrel{\Large \frown}{AP}=R\cdot \pi/6 =\stackrel{\Large \frown}{BP} \\(B)\times: \cases{\stackrel{\Large \frown}{AB}=R\sin{\pi\over 6}\cdot \pi={1\over 2}\pi R\\\stackrel{\Large \frown}{CD} =R \sin{\pi\over 3}\cdot \pi={\sqrt 3\over 2}\pi R} \Rightarrow \stackrel{\Large \frown}{AB} \ne\stackrel{\Large \frown}{CD} \\(C)\bigcirc: A,C,E在同一經度上 \\(D) \bigcirc: \cases{\stackrel{\Large \frown}{CD} ={\sqrt 3\over 2}\pi R \\\stackrel{\Large \frown}{CPD} ={2\over 3}\pi R} \Rightarrow 經過北極路徑比較短 \\(E)\times:\cases{ \stackrel{\Large \frown}{EP} ={1\over 2}\pi R\\ \stackrel{\Large \frown}{CD} ={\sqrt 3\over 2}\pi R } \Rightarrow \stackrel{\Large \frown}{EP}: \stackrel{\Large \frown}{CD}=1:\sqrt 3\\，故選\bbox[red, 2pt]{(ACD)}$$
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解題僅供參考，其他教甄試題及詳解
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