105年全國高中教甄聯招-數學詳解

 教育部受託辦理105學年度公立高級中等學校教師甄選

一、單選題

解：$$只看每個數字的個位數: \cases{4^5的個位數=4 \\5^5的個位數=5 \\6^5的個位數=6 \\7^5的個位數=7 \\9^5的個位數=9 \\11^5的個位數=1 } \Rightarrow 4+5+6+7+9+1 = 32\\ \cases{(A)12^5的個位數=2 \\(B)13^5的個位數=3 \\(C)14^5的個位數=4 \\(D)15^5的個位數=5  },故選\bbox[red,2pt]{(A)}$$

解：$$(A) \cos^2 25^\circ-\sin^2 25^\circ =\cos 50^\circ\\ (B)\sqrt{1+\sin 340^\circ} -\sqrt{1-\sin 340^\circ} = \sqrt{1-\sin 20^\circ} -\sqrt{1+\sin 20^\circ} \lt 0\\ (C)\sin 23^\circ\cos 112^\circ - \sin 292^\circ\sin 67^\circ =-\sin 23^\circ\cos 68^\circ + \sin 68^\circ\cos 23^\circ =\sin(68^\circ-23^\circ)\\\qquad =\sin 45^\circ =\cos 45^\circ\\ (D){2\tan 67.5^\circ \over 1-\tan^2 67.5^\circ} =\tan(2\times 67^\circ) =\tan 135^\circ =-\tan 45^\circ =-1\\ 故選\bbox[red,2pt]{(C)}$$

解：$${x^2\over 75}-{y^2 \over 9} =1 \Rightarrow  c^2=75+9=84 \\ {x^2\over t^2+1} +{y^2\over 7-t}=1 \Rightarrow t^2+1 = 7-t+c^2 = 7-t+84 \Rightarrow t^2+t-90=0 \Rightarrow (t-9)(t+10)=0 \\ \Rightarrow t=-10 (9不合，違反7-t > 0)，故選\bbox[red,2pt]{(D)}$$

解：$$\sqrt{2016} =12\sqrt{14} =a\sqrt{14} +b\sqrt{14} \Rightarrow \cases{a+b=12 \\ a < b\\ a,b \in N} \Rightarrow (a,b)=(1,11)-(5,6)，共5組，\\ 相當於(x,y)=(1\cdot 14,11^2\cdot 14),(2^2\cdot 14,10^2\cdot 14),\dots, (5^2\cdot 14, 6^2\cdot 14)，故選\bbox[red,2pt]{(C)}$$

解：$$f(x)=a\cdot {(x-2)(x-3)\over (1-2)(1-3)} + b\cdot {(x-1)(x-3)\over (2-1)(2-3)} +c \cdot {(x-1)(x-2)\over (3-1)(3-2)}\\ \Rightarrow \cases{f(1)=a\\ f(2)=b\\ f(3)=c} \Rightarrow y=f(x)經過(1,a),(2,b),(3,b)；\\若a、b、c為等差數列，則公差d有以下可能:\\\cases{d=0:(a,b,c)=(1,1,1),(2,2,2),\dots,(6,6,6)，共6種\\ d=1:(1,2,3),(2,3,4),\cdots, (4,5,6)，共4種\\ d=-1:(6,5,4),(5,4,3),\cdots, (3,2,1)，共4種\\ d=2:(1,3,5),(2,4,6)，共2種\\ d=-2:(6,4,2),(5,3,1)，共2種} \\ 共有6+4+4+2+2= 18種，機率為{18\over 6^3} ={1\over 12}\\由於f(x)為二次函數，a、b、c不得為等差數列\\，因此f(x)為二次函數的機率為1-{1\over 12}={11\over 12}，故選\bbox[red, 2pt]{(B)}。$$

解：$$第一天是A，所以第二天只能是B、C、D、E；第五天是C\\，所以第四天只能是A、B、D、E，如下表: \\\begin{array}{}Day 1& Day 2 & Day 3 & Day 4 & Day 5\\ \hline A & B & ? & A & C \\ & C & ? & B & \\ & D & ? & D & \\ & E & ? & E\\\hline \end{array}\\ 因此只要考慮第二天到第四天的過程：\\ \begin{array}{} Day2 & Day 3 & Day 4 & 數量\\\hline B & C,D,E & A &3\\ & A,C,D,E & B & 4 \\ & A,C,E & D & 3\\ & A,C,D & E & 3 \\\hdashline C & B,D,E & A & 3\\ & A,D,E & B & 3 \\ & A,B,E & D & 3 \\ & A,B,D & E & 3\\ \hdashline D & B,C,E & A & 3 \\ & A,C,E & B & 3\\ & A,B,C,E & D & 4\\ & A,B,C & E & 3\\\hdashline E & B,C,D & A & 3\\ & A,C,D & B & 3\\ & A,B,C & D & 3\\ & A,B,C,D & E& 4\\\hline\end{array} \\ 因此共有(3+4+3+3)\times 4-1 = 51，故選\bbox[red,2pt]{(B)}種$$

解：$$假設兩根為\alpha,\beta，則\cases{\alpha +\beta =4a\\ \alpha\beta =4a^2-4a-3b+9 \\ \alpha-\beta=2\sqrt{101}}, a,b\in \mathbb{N}；\\(\alpha+\beta)^2 -4\alpha\beta = (\alpha-\beta)^2 \Rightarrow 16a^2-4(4a^2-4a-3b+9)=404 \\ \Rightarrow 16a+12b =440 \Rightarrow 4a+3b=110 \Rightarrow a=2,5,\dots, 26 =2+3k,k=0-8\\ \Rightarrow 共九組解，故選\bbox[red,2pt]{(A)}$$

解：$${\bf 定理:}假設(x_0,y_0)為ax+by=c的一組整數解，則其通解為\\\cases{x= x_0+{b\over \gcd(a,b)}t \\ y= y_0-{a\over \gcd(a,b)}t}, t\in \mathbb{Z}\\依此定理，假設(x_0,y_0)為方程式2x+3y=k的一組自然數解，且x_0是所有自然數解中最小的\\，則(x_0+3t,y_0-2t) 為其整數解，其中t\in \mathbb{Z}\\由於x_0是所有自然數解中最小的，更小的下一個是x_0-3\le 0 \Rightarrow 0\lt x_0 \le 3\cdots(1)\\ 122個整數解中的y坐標為y_0,y_0-2,\dots,y_0-2\times 121，最小值y_0-2\times 121 \gt 0 \\ ，更小的下一個y_0-2\times 122 \le 0\Rightarrow 2\times 122 \ge y_0 \gt 2\times 121 \cdots(2) \\ 由(1)及(2)可知\cases{x_0=1,2,3\\ y_0=1,2} \Rightarrow k值有3\times 2=6種可能，故選\bbox[red,2pt]{(B)}$$

解：$$(A) \times:f(x)=3x^3-x^2-2x \Rightarrow \cases{f(1)= 0\\f'(x)=9x^2-2x-2} \Rightarrow f'(1)=9-2-2 =5\\\qquad \Rightarrow (1,f(1))= (1,0),斜率5的直線: y=5(x-1) \Rightarrow y=5x-5\\(B)\bigcirc: |\int_0^1f(x)\;dx| =|\left. \left[ {3\over 4}x^4-{1\over 3}x^3-x^2\right] \right|_0^1| =|-{7\over 12}|={7\over 12}\\(C) \bigcirc:f''(x)=0 \Rightarrow 18x-2=0 \Rightarrow x=1/9 \Rightarrow f(1/9)= -{56\over 243} \\(D)\bigcirc: f(x)=0 \Rightarrow x(x-1)(3x+2)=0 \Rightarrow x=0,1,-2/3三相異實根\\ 故選\bbox[red,2pt]{(BCD)}$$

解：$$(A) \bigcirc:\cases{\sum A=42\\ \sum B= 126 \\ \sum C= 54 \\ \sum D=17 }\Rightarrow \cases{\mu_A=42/6=7 \\ \mu_B= 126/6=21 \\ \mu_C= 54/6=9 \\ \mu_D =17/6 < 3 }\Rightarrow\mu_B > \mu_C > \mu_A > \mu_D \\(B)\times: \cases{Var(A)=(6^2+1^2+1^2+2^2+2^2+2^2)/6 \\ Var(C)=(6^2+1^2+1^2+2^2+2^2+2^2)/6} \Rightarrow \sigma_A=\sigma_C\\ (C)\bigcirc: \cases{A,B,C 遞增\\ D遞減} \Rightarrow \cases{r_1,r_2 > 0\\ r_3 < 0};又\cases{r_1= {\sum(A_i-u_A)(B_i-u_B) \over \sqrt{\sum(A_i-u_A)^2} \sqrt{\sum(B_i-u_B)^2}} ={150\over \sqrt{50} \cdot \sqrt{450}}=1\\ r_2={\sum(A_i-u_A)(C_i-u_C) \over \sqrt{\sum(A_i-u_A)^2} \sqrt{\sum(C_i-u_C)^2}}={50\over \sqrt{50} \cdot \sqrt{50}}=1} \Rightarrow r_1=r_2 \\(D)\bigcirc: \sigma_A=\sqrt{(6^2+1^2+1^2+2^2+2^2+2^2)/6} =\sqrt{50\over 6} ={5\over \sqrt 3} ={5\sqrt 3\over 3}\\ 故選\bbox[red,2pt]{(ACD)}$$

解：$$f(x)=x^5-2px^4 +x^3-3px^2+x-2\\ (A)\bigcirc: f(1)=1-5p \ne 0,\forall p \in \mathbb{Z} \Rightarrow x-1不是因式\\ (B)\bigcirc: f(-1)=0 \Rightarrow -5-5p=0 \Rightarrow p=-1 \Rightarrow f(-2)=-44-44p = 0\\(C) \bigcirc: 若f(x)有整係數一次因式，則f(x)=0, \exists x\in \{\pm 1,\pm 2\}；但\\ \qquad \cases{由(A)知:f(1)\ne 0\\ 由(B)知: f(-1)=-5-5p，若f(-1)\ne 0 \Rightarrow p\ne -1 \Rightarrow f(-2)\ne 0\\ 只剩下f(2)=40-44p \ne 0\;\forall p\in \mathbb{Z}}\\\qquad 因此f(x)沒有整係數一次因式\\ (D)\times: 利用長除法可得f(x)=(x+1)^2(x^3-(p+2)x^2+4px-4) +(-9p+10)x+(-4p+5)\\\qquad (x+1)^2 為其因式，則\cases{ -9p+10=0\\ -4p+5=0}\Rightarrow 無解 \Rightarrow (x+1)^2不可能是因式\\故選\bbox[red, 2pt]{(ABC)}$$

解：$$(A)\times: A=\begin{bmatrix} 4 & -3 \\ 2 & -1\end{bmatrix} \Rightarrow \det(A-\lambda I)=0 \Rightarrow \lambda =1,2 \Rightarrow (A-I)(A-2I)= A^2-3A+2I=0 \\(B) \bigcirc: A^4-7A^3+10A^2 -8A+3I =(A^2-3A+2I)(A^2-4A-4I)-12A+11I \\\qquad = -12A+11I =\begin{bmatrix} -48 & 36 \\ -24 & 12\end{bmatrix} +\begin{bmatrix} 11 & 0 \\ 0 & 11\end{bmatrix} =\begin{bmatrix} -37 & 36 \\ -24 & 23\end{bmatrix} \\(C)\bigcirc: B=\begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2\end{bmatrix} =\begin{bmatrix} 1 & 0 \\ 0 & 2\end{bmatrix} \\(D)\times: A=PBP^{-1} \Rightarrow A^{10}=PB^{10}P^{-1} =P\begin{bmatrix} 1 & 0 \\ 0 & 2^{10}\end{bmatrix}P^{-1}，但P^{-1}=\begin{bmatrix} -2 & 3 \\ 1 & -1\end{bmatrix} \ne\begin{bmatrix} 2 & -3 \\ -1 & 1\end{bmatrix}\\ 故選\bbox[red,2pt]{(BC)}$$

第二部份

一、 填充題

解：$$\Gamma:x^2-y^2=9 \Rightarrow a=b=3 \Rightarrow \cases{\overline{AF_2}-\overline{AF_1}=2a= 6\\ \overline{BF_2}-\overline{BF_1}=2a= 6} \stackrel{兩式相加 }{\Longrightarrow}  \overline{AF_2}+ \overline{BF_2}- (\overline{AF_1} +\overline{BF_1}) =12 \\ =\overline{AF_2}+ \overline{BF_2} -\overline{AB}=12 \Rightarrow \overline{AF_2}+ \overline{BF_2}= 12+15=27 \\ \Rightarrow \triangle ABF_2周長= \overline{AB}+ \overline{AF_2}+ \overline{BF_2} =15+27= \bbox[red,2pt]{42}$$

解：$$令\cases{A(-3,2,2)\\ B(-5,4,-3)\\ P(x,y,0)}  \Rightarrow \cases{P在平面E:z=0\\ A、B在平面E的異側} \\\Rightarrow \overline{PA} +\overline{PB} =\sqrt{(x+3)^2+ (y-2)^2 +(-2)^2} +\sqrt{ (x+5)^2 +(y-4)^2 +3^2}\\ 由於A、B在平面E的異側，所以\overline{PA} +\overline{PB}的最小值= \overline{AB} = \sqrt{2^2+ 2^2 +5^2} = \sqrt{33}；\\此時P為L:\overleftrightarrow {AB}與E的交點，而L:{x+3\over -2} ={y-2\over 2}={z-2\over -5} \equiv (-2t-3,2t+2,-5t+2) \\ \Rightarrow (-2t-3,2t+2,-5t+2)=(x,y,0) \Rightarrow -5t+2=0 \Rightarrow t={2\over 5} \Rightarrow \cases{x=-2\cdot {2\over 5}-3=-{19\over 5} \\ y=2\cdot {2\over 5}+2 ={14\over 5}}\\ 因此最小值為\bbox[red,2pt]{\sqrt{33}}，此時(x,y)= \bbox[red,2pt]{(-{19\over 5},{14 \over 5})}$$

解：

$$\cases{x+7y-4\ge 0\\ 4x-5y+17 \ge 0\\ 5x+2y-20 \le 0} 所圍區域頂點\cases{A(2,5) \\B(-3,1)  \\C(4,0)} \\ 令f(x,y)=ax-y=k，依題意\cases{f(C) \le f(A)\\ f(C) \le f(B)} \Rightarrow \cases{4a \le 2a-5\\ 4a \le -3a-1} \Rightarrow \cases{a\le -5/2\\ a \le -1/7} \\ \Rightarrow \bbox[red,2pt]{a\le -{5\over 2}}$$

解：$$\cases{a_1=2\\ a_{n+1}= 2a_n-1} \Rightarrow a_n=2a_{n-1}-1 = 2(2a_{n-2}-1)-1 = 2^2a_{n-2}-2-1=\cdots \\= 2^{n-1}a_1-2^{n-2}-2^{n-2}-\cdots -1= 2^n-(2^{n-1}-1) =2^n-2^{n-1}+1 = 2^{n-1}+1 > 1000\\ \Rightarrow n=\bbox[red,2pt]{11} (\because 2^{10}=1024)$$

解:

$$\cases{A:垂直積木\\ B:水平積} \quad \Rightarrow \cases{6A:1種\\ 6B:1種\\ 2A4B:4種\\ 4A2B: 5種}\Rightarrow 共有1+1+4+5=\bbox[red,2pt]{11}種$$

解:

$$\cases{\cos \angle PBA={4+a^2-1\over 4a} ={a^2+3\over 4a} \\ \cos \angle PBC ={4+a^2-9\over 4a} ={a^2-5\over 4a} =\sin \angle PBA} \Rightarrow \left( {a^2+3\over 4a}\right)^2 +\left({a^2-5\over 4a} \right)^2=1\\ \Rightarrow a^4-10a^2+17=0 \\ \Rightarrow 面積=a^2= {10+\sqrt{32}\over 2} =\bbox[red,2pt]{5+2\sqrt 2}$$

解：

$$令\cases{\overline{AP}=a \\\overline{PQ}=x} \Rightarrow \cases{\overline{CQ}=1-a\\ \overline{BP}=2-a} \Rightarrow \overline{BQ}=2-(1-a)=1+a\\ \Rightarrow \cases{\cos 60^\circ={1\over 2} =\cfrac{(2-a)^2+(1+a)^2 -x^2}{2(1+a)(2-a)} \cdots(1)\\ \cos 45^\circ={\sqrt 2\over 2}=\cfrac{(2-a)^2+x^2-(1+a)^2}{2(2-a)x}\cdots(2)}\\ 由(1)可得 x^2=3a^2-3a+3 代入(2) \Rightarrow a=2-\sqrt 3(2+\sqrt 3不合,違反a < 2)\\ \frac{\triangle BPQ}{\triangle ABC} =\frac{\overline{BP}\times \overline{BQ}}{\overline{BA}\times \overline{BC}} \Rightarrow \triangle BPQ= \triangle ABC\times {(2-a)(1+a)\over 4} =\sqrt 3\times {(2-a)(1+a)\over 4}\\ =\bbox[red,2pt]{\frac{9-3\sqrt 3}{4}}$$

解：$$令\cases{x'=4x+7y\\ y'=6x+2y} \Rightarrow {|x'|\over 3} +{|y'| \over 4}\le 1 所圍面積= 4\times {1\over 2}\times 3\times 4=24\\ \begin{Vmatrix} 4& 7 \\ 5 & 2 \end{Vmatrix} =27 \Rightarrow {|4x+7y|\over 3} +{|5x+2y| \over 4}\le 1所圍面積={24\over 27} =\bbox[red,2pt]{8\over 9}$$

解：$$\begin{bmatrix} 1 & 2 \\ 3& 2\end{bmatrix}^5 = 4^5\begin{bmatrix} 1/4 & 2/4 \\ 3/4& 2/4\end{bmatrix}^5，由於\begin{bmatrix} 1/4 & 2/4 \\ 3/4& 2/4\end{bmatrix}每一行的和為1，因此為馬可夫矩陣(轉移矩陣)\\ \Rightarrow \begin{bmatrix} 1/4 & 2/4 \\ 3/4& 2/4\end{bmatrix}^5 也是馬可夫矩陣 \Rightarrow 4^5\begin{bmatrix} 1/4 & 2/4 \\ 3/4& 2/4\end{bmatrix}^5\begin{bmatrix} 1 & 2 & 3 \\ 3 & 2 & 1\end{bmatrix}的矩陣元素和 \\=4^5(1+2+3+ 3+2+1) =4^5 \times 12 = 1024 \times 12 = \bbox[red,2pt]{12288}$$

解：

$$(x-6)^2+(y-7)^2 = 10 \Rightarrow \cases{圓心C(6,7)\\ 半徑r=\sqrt{10}} \Rightarrow \overline{OC}= \sqrt{6^2+7^2} = \sqrt{85} \\\Rightarrow \overline{OA}=\overline{OB}= \sqrt{\overline{OC}^2-r^2} =5\sqrt 3；\\因此令\angle COB= \theta，則\cases{ \sin \theta = \sqrt{10}/\sqrt{85} \\ \cos \theta=5\sqrt 3/\sqrt{85}} \Rightarrow \sin \angle AOP= \sin 2\theta = 2\sin\theta \cos \theta = {2\sqrt{30} \over 17}\\ \triangle OAP: {\overline{AP} \over \sin 2\theta} ={\overline{OP} \over \sin \angle OAP} = {\overline{OA} \over \sin \angle APO} \Rightarrow {\overline{PO} \over \overline{PA}} ={\sin \angle OAP \over \sin 2\theta} ={\sin \angle OAP \over {2\sqrt{30}/ 17}}=k\\ k要最大，即\sin \angle OAP=1，此時k={17\over 2\sqrt{30}} =\bbox[red, 2pt]{17\sqrt{30}\over 60}$$

計算題

$$\bbox[blue,2pt]{題目有誤}$$

解：$$\cases{B=P_0\\ C=P_n} \Rightarrow \cases{|\overrightarrow{AP_0}| =|\overrightarrow{AB}|=1 \\ \overrightarrow{P_0P_k} ={k\over n} \overrightarrow{P_0P_n} ={k\over n} \overrightarrow{BC},k=0-n}\\S_n = \sum_{k=0}^{n-1}\overrightarrow{AP_k} \cdot \overrightarrow{AP_{k+1}} =\sum_{k=0}^{n-1} (\overrightarrow{AP_0} +\overrightarrow{P_0P_k}) \cdot (\overrightarrow{AP_0}+\overrightarrow{P_0P_{k+1}}) \\=\sum_{k=0}^{n-1} (|\overrightarrow{AP_0}|^2 + \overrightarrow{AP_0} \cdot \overrightarrow{P_0P_{k+1}} +\overrightarrow{P_0P_k} \cdot \overrightarrow{AP_0}+ \overrightarrow{P_0P_k} \cdot \overrightarrow{P_0P_{k+1}})  \\=\sum_{k=0}^{n-1} (1 + \overrightarrow{AP_0} \cdot \overrightarrow{P_0P_{k+1}} +\overrightarrow{P_0P_k} \cdot \overrightarrow{AP_0}+ {k(k+1)\over n^2}) \\=\sum_{k=0}^{n-1} (1 + \overrightarrow{AP_0} \cdot (\overrightarrow{P_0P_k} +\overrightarrow{P_0P_{k+1}} ) + {k(k+1)\over n^2}) \\ =\sum_{k=0}^{n-1} (1 + \overrightarrow{AB} \cdot ({2k+1\over n}\overrightarrow{BC} ) + {k(k+1)\over n^2}) =\sum_{k=0}^{n-1} (1 + {2k+1\over n}(\overrightarrow{AB} \cdot \overrightarrow{BC} ) + {k(k+1)\over n^2}) \\ = \sum_{k=0}^{n-1} (1 + {2k+1\over n}(-{1\over 2}) + {k(k+1)\over n^2}) = \sum_{k=0}^{n-1} (1 - {2k+1\over 2n} + {k(k+1)\over n^2}) \\ =n-{n^2-n+1 \over 2n} +{n^2-1\over 3n} \\ \Rightarrow \lim_{n\to \infty}{S_n\over n} =\lim_{n\to \infty} \left(1-{n^2-n+1 \over 2n^2} +{n^2-1\over 3n^2} \right) =1-{1\over 2}+{1\over 3}= \bbox[red,2pt]{5\over 6}$$

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