107年全國高中職教甄聯招-數學科詳解

教育部受託辦理107學年度
公立高級中等學校教師甄選
數學科試題

第一部分：選擇題

一、單選題

解：

$$\cases{A(0,0,0)\\ B(2,0,0)\\ C(2,2,0)\\ D(0,2,0) \\ E(1,0,0)\\ F(0,1,0)\\ G(2,2,1)} \Rightarrow \cases{\overrightarrow{GE} =(-1,-2,-1) \\\overrightarrow{GF} =(-2,-1,-1)} \Rightarrow \vec n=\overrightarrow{GE} \times \overrightarrow{GF} =(1,1,-3)\\ 平面EFG: (x-1)+y-3z=0 \Rightarrow x+y-3z-1=0\\ \Rightarrow D至平面EFG的距離={0+2+0-1 \over \sqrt{1^2+1^2+3^2}} ={1\over \sqrt{11}}，故選\bbox[red, 2pt]{(A)}$$

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解：

$$\cases{\log_8 a+\log_4 b=3 \\ \log_8 b+\log_4 a=7 } \Rightarrow \cases{{1\over 3}\log_2 a+{1\over 2}\log_2 b=3 \\ {1\over 3} \log_2 b+ {1\over 2}\log_2 a=7 } ,兩式相加\Rightarrow {5\over 6}(\log_2 a+\log_2 b)=10 \\ \Rightarrow \log_2 ab=12 \Rightarrow ab=2^{12}= 4096，故選\bbox[red, 2pt]{(D)}$$

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解：

$$b為偶數\Rightarrow (a,b,c)共有6\times 3\times 6=108種組合;\\ \begin{vmatrix} a & b\\ b& c\end{vmatrix} >0  \Rightarrow ac > b^2 \Rightarrow \begin{array}{} b(偶數)& a &c & 數量\\\hline 2 & 1 & 5-6 & 2 \\ & 2 & 3-6 & 4 \\ & 3-4 & 2-6 & 10 \\ & 5-6 &1-6 & 12 \\\hdashline  4 & 3 & 6 & 1\\ & 4 & 5-6 & 2\\ & 5 & 4-6 & 3 \\ & 6 & 3-6 & 4 \\\hline  & & 共& 38 \end{array} \\ \Rightarrow 機率為{38\over 108} \approx 0.352，故選\bbox[red, 2pt]{(C )}$$

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解：

$$令\cases{f(a,b)= \sqrt{2a+1}+\sqrt{3b+2} \\g(a,b)=a+b-2},由\text{Lagrange }算子可得\cases{{\partial \over \partial a}f=\lambda{\partial \over \partial a} g\\ {\partial \over \partial b}f=\lambda{\partial \over \partial b}g \\g=0} \\ \Rightarrow \cases{{1\over \sqrt{2a+1}}=\lambda \cdots(1)\\  {3\over 2\sqrt{3b+2}}=\lambda \cdots(2)\\a+b=2 \cdots(3)},由(1)及(2) \Rightarrow {1\over \sqrt{2a+1}}={3\over 2\sqrt{3b+2}} \Rightarrow {1\over 2a+1} ={9\over 12b+8} \\ \Rightarrow 18a-12b+1=0 \cdots(4) \\ 由(3)及(4) \Rightarrow \cases{a=23/30\\ b=37/30} \Rightarrow f(23/30,37/30)= \sqrt{{46\over 30}+1 } +\sqrt{{111\over 30}+2} = \sqrt{76\over 30} +\sqrt{171\over 30} \\ = {2\sqrt{19} \over \sqrt{30}} + {3\sqrt{19} \over \sqrt{30}}= {5\sqrt{19} \over \sqrt{30}}= {5\sqrt{570} \over 30} ={\sqrt{570} \over 6}，故選\bbox[red, 2pt]{(B )}$$

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解：

$$紅球R,藍球B,黃球Y，現有RRRBBY分給三人a,b,c;\\\Rightarrow \begin{array}{}a & b & c\\\hline RR & RB & BY\\ & RY & BB \\ & BB & RY\\ & BY & RB \\\hdashline RB & RR & BY \\ & RB & RY \\ & RY & RB \\ & BY & RR \\ \hdashline RY & RR & BB \\ & RB & RB \\ & BB & RR\\ \hdashline BB & RR & RY \\ & RY & RR\\\hdashline BY & RR & RB \\ & RB & RR\\\hline\end{array}\\ \Rightarrow 共15種分法，故選\bbox[red, 2pt]{(B )}$$

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解：

$$\cases{a+b+c+d+e=10 \\a^2+b^2+c^2+d^2+e^2=205/4} \Rightarrow \cases{a+b+d+e=10-c \\ a^2+b^2+d^2+e^2=205/4-c^2}\\ 柯西不等式:(a^2+b^2+d^2+e^2)(1^2+1^2+1^2+1^2) \ge (a+b+d+e)^2 \\ \Rightarrow ({205\over 4}-c^2) \times 4 \ge  (10-c)^2 \Rightarrow 205-4c^2 \ge c^2-20c +100 \\ \Rightarrow 5c^2-20c-105 \le 0 \Rightarrow c^2-4c-21 \le 0 \Rightarrow (c-7)(c+3) \le 0 \\ \Rightarrow -3 \le c \le 7 \Rightarrow \cases{k=7 \\ t=-3}，故選\bbox[red, 2pt]{(A )}$$

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解：

$$\cases{6\le x+y \le 8 \\ -2 \le 2x+y \le 0}\Rightarrow 封閉區域頂點\cases{P(16,-8) \\ Q(18,-10) \\ R(14,-8) \\ S(12,-6)};\\ \vec A \cdot \vec B = |\vec A||\vec B|\cos 60^\circ =1\times 2\times {1\over 2}=1 \\\Rightarrow \vec u\cdot \vec v= (\vec A+\vec B) \cdot (x\vec A+ y\vec B) =x|\vec A|^2+ (x+y)\vec A\cdot \vec B +y|\vec B|^2 \\ =x + (x+y)+4y= 2x+5y;\\ 令f((x,y))=2x+5y \Rightarrow \cases{f(P)=32-40=-8 \\ f(Q)=36-50=-14\\ f(R)=28-40=-12 \\ f(S)=24-30=-6} \Rightarrow \vec u\cdot \vec v 最大值為-6，故選\bbox[red, 2pt]{(C )}$$

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解：

$$正弦定理:{\overline{AC} \over \sin \angle B} ={\overline{AB} \over \sin \angle C} = 2R \Rightarrow {25/13\over \sin \angle B} ={ 3\over \sin \angle C} =2\times {5\over 2} =5 \Rightarrow \cases{\sin \angle B=5/13 \\ \sin \angle C=3/5} \\ \Rightarrow \cases{\cos \angle B=12 /13 \\ \cos \angle C=-4/5}\Rightarrow \overline{BC} = \overline{AB}\cdot \cos \angle B + \overline{AC}\cdot \cos \angle C = {36\over 13} -{20\over 13} ={16\over 13}\\ 再由餘弦定理: \cos \angle BAC= {3^2+(25/13)^2 - (16/13)^2\over 2\cdot 3\cdot (25/13)} ={9+{369\over 169} \over {150\over 13}} ={1890\over 169}\times {13\over 150} ={63\over 65}\\，故選\bbox[red, 2pt]{(D )}$$

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二、複選題

解：

$${a-c\over 2b} ={a-b\over 2c} \Rightarrow  {ac-c^2\over 2bc} ={ab-b^2\over 2bc} \Rightarrow ac-c^2=ab-b^2 \Rightarrow b^2-c^2 =ab-ac \\\Rightarrow \cases{b=c \cdots(1)\\ b+c=a  \Rightarrow {a\over a+b+c}={a\over 2a}={1\over 2}} \\ 同理,{b+c\over 2a} ={a-c\over 2b} \Rightarrow b^2+bc=a^2-ac \Rightarrow c(a+b)=(a+b)(a-b) \\\Rightarrow \cases{a=-b \cdots(2)\\ b+c=a}，由(1)及(2) \Rightarrow \cases{a=-b\\ c=b} \Rightarrow {a\over a+b+c}={-b\over b}=-1\\ 因此{a\over a+b+c}=-1,{1\over 2}，故選\bbox[red, 2pt]{(BC )}$$

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解：

$$f(x)=x^3-kx = x(x^2-k)=0 \Rightarrow x=-\sqrt k,0,\sqrt k\\ \Rightarrow \int_0^1|f(x)|\;dx =|\int_0^\sqrt k f(x)\;dx | + |\int_\sqrt k^1 f(x)\;dx | = \left | \left. \left[ {1\over 4}x^4-{k\over 2}x^2\right] \right|_0^\sqrt k\right| +\left | \left. \left[ {1\over 4}x^4-{k\over 2}x^2\right] \right|_\sqrt k^1\right| \\ = {1\over 4}k^2+{1\over 4}(k-1)^2 ={1\over 4}(2(k-{1\over 2})^2+{1\over 2}) \Rightarrow k={1\over 2}時，\int_0^1|f(x)|\;dx有最小值:{1\over 4}\cdot {1\over 2}={1\over 8} \\ \Rightarrow \cases{a=1/2\\ b=1/8}，故選\bbox[red, 2pt]{(AC )}$$

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解：

$$(A)\bigcirc: {b_n\over b_{n-1} } ={2^{a_n} \over 2^{a_{n-1}}} =2^{a_n-a_{n-1}} =2^d \Rightarrow < b_n > 為公比2^d的等比數列\\ (B)\times: d>0 \Rightarrow r=2^d > 1\\ (C)\bigcirc: \cases{b_1=2^{a_1}=2^1=2 \\ r=2^d= \sqrt 2 } \Rightarrow b_n > 4096 \Rightarrow b_1r^{n-1} = 2\cdot 2^{(n-1)/2} =2^{(n+1)/2} > 4096 =2 ^ {12} \\ \qquad \Rightarrow {n+1\over 2} > 12 \Rightarrow n > 23 \Rightarrow n=24 \\ (D) \bigcirc: \sum_{i=1}^n a_i = {2a_1+(n-1)d \over 2}\cdot n= {2+(n-1)/2 \over 2}\cdot n =22 \Rightarrow (n+11)(n-8)=0 \Rightarrow n=8 \\\qquad \Rightarrow \sum_{i=1}^8 b_i = {b_1-b_1r^8\over 1-r} = {2-2\cdot 2^{4} \over 1-\sqrt 2} =(2^5-2)(\sqrt 2+1) =30(1+\sqrt 2)\\，故選\bbox[red, 2pt]{(ACD )}$$

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解：

$$(A)\bigcirc: 迴歸直線必經(50,M) \Rightarrow M={3\over 4}\times 50+20 = 57.5 \\ (B)\bigcirc: 相關係數r \Rightarrow {3\over 4}= r\times {\sigma_y \over \sigma_x} \Rightarrow r={3\over 4}\times {8\over 12} = 0.5 \\ (C)\times: 迴歸直線只是推估，不能保證\\ (D) \bigcirc: \sigma_y > \sigma_x \Rightarrow 物理成績較分散\\，故選\bbox[red, 2pt]{(ABD )}$$

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第二部份：綜合題

一、填充題

解：

$$(\sin^2 63^\circ-3\sin^2 27^\circ)(\sin^29^\circ-3\cos^2 171^\circ) =(\cos^2 27^\circ-3(1-\cos^2 27^\circ))(\sin^29^\circ- 3\cos^2 9^\circ) \\=(-3+4\cos^2 27^\circ )( \sin^2 9^\circ-3(1-\sin^2 9^\circ)) =(-3+4\cos^2 27^\circ ) (-3+4\sin^2 9^\circ) \\ ={\color{blue}{\cos 27^\circ}(-3+4\cos^2 27^\circ ) \color{blue}{\sin 9^\circ}(-3+4\sin^2 9^\circ) \over \color{blue}{\cos 27^\circ\sin 9^\circ}} ={-\cos 81^\circ \sin 27^\circ\over \cos 27^\circ\sin 9^\circ} \left(\because \cases{\cos (3x) =4\cos^3x-3\cos x\\ \sin(3x)=-4\sin ^3x +3\sin x} \right) \\ ={-\sin 9^\circ \sin 27^\circ\over \cos 27^\circ\sin 9^\circ} = -\tan 27^\circ = \tan(360^\circ-27^\circ)=\tan 333^\circ \Rightarrow \theta =\bbox[red,2pt]{333^\circ}$$

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解：

$$\begin{array}{} & a_2 & = & 2+1=3=a_1+3 \\ & a_3 & = & 3+2\sqrt 4+1=8 = a_2+5 \\  & a_4 & = & 8+2\sqrt 9+1=15 = a_3+7 \\ & \cdots & & \cdots\\ +&a_n &=& a_{n-1}+(2n-1)\\\hline & a_n & = & 3+ 5+7+\cdots + 2n-1 =\sum_{k=2}^n (2k-1)\end{array} \\ \Rightarrow a_{30} =\sum_{k=2}^{30} (2k-1) = 32\times 29-29 = \bbox[red, 2pt]{899}$$

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解：

$$令g(x)=xf(x)\Rightarrow \cases{g(0)=0\\ g(t)=tf(t)={1\over 2},t=1-2019} \Rightarrow g(t)=a(t-1)(t-2)\cdots (t-2019)+{1\over 2}\\ \Rightarrow g(0)=a \times -2019!+{1\over 2}=0 \Rightarrow a={1\over 2\times 2019!} \Rightarrow g(2020)= a\times 2019!+{1\over 2} ={1\over 2} +{1\over 2}=1 \\ \Rightarrow f(2020)={1\over 2020}\times g(2020) =\bbox[red, 2pt]{1\over 2020}$$

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解：

$${1\over 2\times 3\times 4\times 5} +{1\over 3\times 4\times 5\times 6} +{1\over 4\times 5\times 6\times 7} + {1\over 5\times 6\times 7\times 8} + \cdots\\ ={1\over 3}({1\over 2\times 3\times 4} -{1\over 3\times 4\times 5}) +{1\over 3}({1\over 3\times 4\times 5} -{1\over 4\times 5\times 6}) +{1\over 3}({1\over 4\times 5\times 6} -{1\over 5\times 6\times 7}) \\\qquad \quad+{1\over 3}({1\over 5\times 6\times 7} -{1\over 6\times 7\times 8}) +\cdots\\ = {1\over 3}\times {1\over 2\times 3\times 4} = \bbox[red, 2pt]{1\over 72}$$

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解：

$$A=\begin{bmatrix} 1& 0 \\ -1 & 2\end{bmatrix} \Rightarrow A^2=\begin{bmatrix} 1& 0 \\ -1 & 2\end{bmatrix}\begin{bmatrix} 1& 0 \\ -1 & 2\end{bmatrix} =\begin{bmatrix} 1& 0 \\ -3 & 4\end{bmatrix} \\ \Rightarrow A^4= \begin{bmatrix} 1& 0 \\ -3 & 4\end{bmatrix} \begin{bmatrix} 1& 0 \\ -3 & 4\end{bmatrix} =\begin{bmatrix} 1& 0 \\ -15 & 16\end{bmatrix} \Rightarrow A^8 =\begin{bmatrix} 1& 0 \\ -15 & 16\end{bmatrix} \begin{bmatrix} 1& 0 \\ -15 & 16\end{bmatrix} \\ =\begin{bmatrix} 1& 0 \\ 255 & 256\end{bmatrix} =\begin{bmatrix} 1& 0 \\ 0 & 1\end{bmatrix}+255 \begin{bmatrix} 0& 0 \\ -1 & 1\end{bmatrix} \Rightarrow (a,b)=\bbox[red, 2pt]{(1,255)}$$

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解：

$$\cases{P(x,-x^2+5x+1) \\ Q(x,-x+6)} \Rightarrow \overline{PQ} =-x^2+5x+1+x-6 =-(x^2-6x+9)+4 \\ =-(x-3)^2+4 \le 4 \Rightarrow 最大值為\bbox[red, 2pt]{4}$$

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解：

$$\begin{array}{} n & {1\over 2}\log_2 n-1 & [{1\over 2}\log_2 n-1] & 小計\\\hline 1-3 & [-1,0) & -1 & -3\\ 4-15 & [0,1) & 0 & 0\\ 16-63 & [1,2) & 1 & 48 \\ 64-100 & [2,3) & 2 & 74 \\\hdashline & &  & 119\end{array}\Rightarrow \bbox[red, 2pt]{119}$$

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解：

$$(\sqrt{m+\sqrt{m^2-n}} +\sqrt{m-\sqrt{m^2-n}})^2 =6^2 \Rightarrow 2m+2\sqrt{n}=36 \Rightarrow n=(m-18)^2\\ 由m^2 \ge n \Rightarrow n=(m-18)^2 = m^2-36m+18^2 \Rightarrow -36m+18^2 \le 0 \Rightarrow m\ge 9\\ \Rightarrow \sum n= 9^2+8^2+\cdots +1^2= {9\times 10 \times 19\over 6} =\bbox[red, 2pt]{285}$$

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解：

$$a_n=a_{n-1}-a_{n-2},n\ge 3\\ \Rightarrow \begin{array}{l} a_1=a_1\\ a_2=a_2\\a_3= a_2-a_1\\ a_4=a_3-a_2= a_2-a_1-a_2=-a_1 \\ a_5=a_4-a_3 =-a_1-(a_2-a_1) =-a_2 \\a_6=a_5-a_4=-a_2-(-a_1) =-a_2+a_1 \\\hdashline a_7 =a_6-a_5=-a_2+a_1-(-a_2)=a_1 \\ a_8=a_7-a_6= a_1-(-a_2+a_1) =a_2 \\ a_9=a_8-a_7=a_2-a_1\\ \cdots\end{array} \Rightarrow \cases{a_m=a_n,\text{if }(m \mod 6)=(n \mod 6)\\ \sum_{k=n}^{n+5} a_k=0,k\ge 1}\\ \cases{\sum_{n=1}^{40}a_n=30 \\ \sum_{n=1}^{80}a_n= 78} \Rightarrow \cases{a_{37}+ a_{38}+a_{39}+a_{40} =30\\ a_{79}+a_{80}=78}  \Rightarrow \cases{a_{1}+ a_{2}+a_{3}+a_{4} =30\\ a_{1}+a_{2}=78} \\ \Rightarrow \cases{a_{1}+ a_{2}+(a_{2}-a_1)+(-a_{1}) =30\\ a_{1}+a_{2}=78} \Rightarrow \cases{2a_{2} -a_{1} =30\\ a_{1}+a_{2}=78} \Rightarrow \cases{a_1=42 \\ a_2=36}\\ \sum_{n=1}^{123}a_n= a_{121}+ a_{122}+a_{123}= a_1+a_2+a_3= 2a_2 = \bbox[red, 2pt]{72}$$

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二、計算證明題

解：

$$f(x)=x^3-4x^2+2x+4 = (x-2)(x^2-2x-2)，因此f(x)=0 \Rightarrow x=0,1\pm \sqrt{3} \\ \Rightarrow \cases{\alpha =1-\sqrt 3\\ \beta =2 \\ \gamma =1+\sqrt 3} \Rightarrow  {1\over x-\alpha}+ {1\over x-\beta} +{1\over x-\gamma} ={ 3x^2-2(\alpha+ \beta +\gamma)x +\alpha\beta+\beta\gamma +\gamma\alpha\over (x-\alpha)(x-\beta)(x-\gamma)} \\= {3x^2-8x+2 \over (x-1+\sqrt 3)(x-2)(x-1-\sqrt 3)}  ={(x-{4+\sqrt{10}\over 3})(x-{4-\sqrt{10}\over 3}) \over (x-1+\sqrt 3)(x-2)(x-1-\sqrt 3)};\\因此 {1\over x-\alpha}+ {1\over x-\beta} +{1\over x-\gamma}>0 \\\Rightarrow (x-{4+\sqrt{10}\over 3})(x-{4-\sqrt{10}\over 3})(x-1+\sqrt 3)(x-2)(x-1-\sqrt 3)>0 \\ \Rightarrow \bbox[red, 2pt]{\cases{x > 1+\sqrt 3\\ 2< x < {4+\sqrt{10}\over 3}\\ 1-\sqrt 3< x <{4-\sqrt{10} \over 3} }}$$

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解：

$$|z-1-i|-|z+1+i|=2 \Rightarrow |(x-1)+(y-1)i|-|(x+1)+(y+1)i|=2 \\ \Rightarrow (x-1)^2+(y-1)^2 =4+4\sqrt{(x+1)^2+(y+1)^2} +(x+1)^2+(y+1)^2 \\ \Rightarrow -x-y-1=\sqrt{(x+1)^2+(y+1)^2} \Rightarrow (x+y+1)^2=(x+1)^2+(y+1)^2 \\ \Rightarrow x^2+y^2 +2xy+2x+2y+1 =x^2+y^2 +2x+2y+2 \Rightarrow \bbox[red, 2pt]{xy=1/2}$$

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解：

$$\lim_{n\to \infty}\left(1+{1\over n} \right)^n = \lim_{n\to \infty} 1+{n\over 1!}\cdot {1\over n} +{n(n-1)\over 2!}\cdot {1\over n^2}  +{n(n-1)(n-2)\over 3!}\cdot {1\over n^3} +\cdots \\ = \lim_{n\to \infty} 1+{1\over 1!}+{1\over 2!}(1-{1\over n})+ {1\over 3!}(1-{1\over n})(1-{2\over n})+ \cdots \\ =1+{1\over 1!} +{1\over 2!}+{1\over 3!}+\cdots =\sum_{n=1}^\infty {1\over n!}，故得證$$

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