115年成淵高中教甄-數學詳解

 臺北市立成淵高級中學 115 學年度正式教師甄選

第壹部分：填充題（每題 6 分，共 72 分）

解答:$$\sqrt{x^2+{2\over 3}x+{1\over 9}}+ \sqrt{x^2-x+{1\over 4}}= \sqrt{ \left( x+{1\over 3} \right)^2} + \sqrt{ \left( x-{1\over 2} \right)^2} =x+{5\over 6} \\ \Rightarrow \left| x+{1\over 3}\right|+\left| x-{1\over 2}\right|=x+{5\over 6} \\ \textbf{Case I }x\ge {1\over 2} \Rightarrow x+{1\over 3}+x-{1\over 2}=x+{5\over 6} \Rightarrow x={5\over 6}+{1\over 2}-{1\over 3}=1 \Rightarrow x=1 \\ \textbf{Case II }-{1\over 3} \le x\le {1\over 2} \Rightarrow x+{1\over 3}-x+{1\over 2}=x+{5\over 6} \Rightarrow x=0 \\ \textbf{Case III }x\le -{1\over 3} \Rightarrow -x-{1\over 3}-x+{1\over 2}=x+{5\over 6} \Rightarrow x=-{2\over 9} \not \le -{2\over 3} \\ \Rightarrow x= \bbox[red, 2pt]{0,1}$$

解答:$$  \cases{xy=64 \Rightarrow \log x+\log y= \log 64=6\log 2 \cdots(1)\\2\log_xy +3\log_y x={2\log y\over \log x}+ {3\log x\over \log y}=7 \cdots(2)} \\取u=\log_x y 代入(2) \Rightarrow 2u+{3\over u}=7 \Rightarrow 2u^2-7u+3=0 \Rightarrow (2u-1)( u-3)=0 \\ \Rightarrow \log_x y =u={1\over 2} \Rightarrow y=\sqrt x 代入(1) \Rightarrow x^{3/2} =64=2^6 \Rightarrow x=2^4=16 \Rightarrow y=4   \\\Rightarrow \log_x y =u=3 \Rightarrow y=x^{3 } 代入(1)\Rightarrow x^{4}=64=2^6 \Rightarrow x= 2^{3/2}=2\sqrt 2 \Rightarrow y=2^{9/2} =16\sqrt 2 \\ 因此(x,y) = \bbox[red, 2pt]{(16,3),(2\sqrt 2, 16\sqrt 2)}$$

解答:$$\textbf{Case I }A點數=B點數\Rightarrow (A,B)=(2,2) ,(3,3),(5,5) \Rightarrow 機率為{3\over 36} \\ \textbf{Case II }A點數\gt B點數 \Rightarrow (A,B)=(3,2),(4,2-3),(5,2-3),(6,2-3),(6,5) \Rightarrow 機率為{8\over 36} \\ \textbf{Case III }A點數\lt B點數\Rightarrow 共有36-3-8=25 \Rightarrow 機率={25\over 36} \\ 期望值=2026\cdot {3\over 36}+115\cdot {8\over 36}+ 0\cdot {25\over 36}= {3499\over 18} ={q\over p}\Rightarrow (p,q) = \bbox[red, 2pt]{(18,3499)}$$

解答:$$柯西不等式: \left[ \left( {3\over 2} \right)^2+ 1^2 \right] \left[ x^2+ \left( \sqrt{4-x^2} \right)^2 \right] \ge \left( {3\over 2}x+ \sqrt{4-x^2} \right)^2 \\ \Rightarrow 13 \ge \left( {3\over 2}x+ \sqrt{4-x^2} \right)^2 \Rightarrow -\sqrt{13} \le {3\over 2}x+ \sqrt{4-x^2} \le \sqrt{13} \Rightarrow 1-\sqrt{13} \le f(x)\le 1+\sqrt{13}\\ 由於\sqrt{4-x^2}\ge 0 \Rightarrow 最小值發生在x=-2 \Rightarrow 最小值=1+{3\over 2}\cdot (-2) =-2\\ \Rightarrow (M,m)= \bbox[red, 2pt]{(1+\sqrt{13},-2)}$$

解答:$$\left( x+{1\over x} \right) \left( x+{2\over x} \right) \left( x+{4\over x} \right) =  \left( x^3+{8\over x^3} \right) +7\left( x+{2\over x} \right) \\ 算幾不等式:\cases{x^3+ \displaystyle {8\over x^3} \ge 2\sqrt{8} =4\sqrt 2 \\ x+ \displaystyle {2\over x}\ge 2\sqrt 2} \Rightarrow \left( x^3+{8\over x^3} \right) +7\left( x+{2\over x} \right) \ge 4\sqrt 2+7\cdot 2\sqrt 2= \bbox[red, 2pt]{18\sqrt 2} \\註: 兩個不等式等號成立的條件相同,即x=\sqrt 2$$

解答:$$\overline{AC}^2= \overline{AB}^2+ \overline{BC}^2 \Rightarrow \angle B=90^\circ \Rightarrow \sin \angle CAB={4\over 5}, \cos \angle CAB={3\over 5}\\ \triangle ACD: \cos \angle CAD ={5^2+6^2-7^2\over 2\cdot 5\cdot 6} ={1\over 5} \Rightarrow \sin \angle CAD ={2\sqrt 6\over 5} \\ \Rightarrow \sin \angle DAB= \sin(\angle CAD+ \angle CAB) = \sin \angle CAD \cos \angle CAB + \cos \angle CAD \sin \angle CAB ={4+6\sqrt 6\over 25} \\ \Rightarrow \triangle ABD面積= {1\over 2} \cdot \overline{AB} \cdot \overline{AD} \sin \angle DAB= {36+54\sqrt 6\over 25} \\ \Rightarrow 四邊形ABCD面積= \triangle ABC+ \triangle ACD= {1\over 2} \cdot 3\cdot 4+ {1\over 2}\cdot 5\cdot 6\cdot {2\sqrt 6\over 5}=6+6\sqrt 6 \\ \Rightarrow \triangle CBD= 四邊形ABCD面積-\triangle ABD=6+6\sqrt 6-{36+54\sqrt 6\over 25} ={114+96\sqrt 6\over 25} \\ \Rightarrow {\overline{AE} \over \overline{CE}} ={ \triangle ABD\over \triangle CBD} = {36+54\sqrt 6\over 114+96\sqrt 6} ={6+9\sqrt 6\over 19+16\sqrt 6} \Rightarrow \overline{AE}= \overline{AC} \times {\triangle ABD\over \triangle ABD+\triangle CBD} \\=5\times {{36+54\sqrt 6\over 25}\over 6+6\sqrt 6} = {1\over 5} \times {6+9\sqrt 6\over 1+\sqrt 6} = \bbox[red, 2pt]{{48\over 25}-{3\over 25}\sqrt 6}$$

解答:

$$假設\cases{a= \overline{BC}=7\\ b=\overline{AC}=6 \\ c=\overline{AB}=5} \Rightarrow s={1\over 2}(a+b+c)=9 \Rightarrow \triangle ABC面積= \sqrt{9(9-7)(9-6)(9-5)} =6\sqrt 6 \\ \Rightarrow 內切圓半徑r={\triangle ABC面積\over s} ={6\sqrt 6\over 9}\Rightarrow \overline{AD}=s-a=9-7=2 \Rightarrow \overline{AK} =\sqrt{\overline{AD}^2+r^2} \\ =\sqrt{4+{24\over 9}} ={2\sqrt{15}\over 3} \Rightarrow t={\overline{AQ} \over \overline{AK}} ={\overline{AK}-r\over \overline{AK}} =1-{r\over \overline{AK}} =1-{2\sqrt 6/3\over 2\sqrt{15}/3} =1-{\sqrt{2} \over \sqrt 5} \\=1-{\sqrt{10}\over 5} =\bbox[red, 2pt]{5-\sqrt{10}\over 5}$$

解答:$$5x^2-6xy+5y^2= \begin{bmatrix}x& y \end{bmatrix} \begin{bmatrix}5&-3\\-3& 5 \end{bmatrix} \begin{bmatrix}x\\y \end{bmatrix} =\begin{bmatrix}x& y \end{bmatrix} P\begin{bmatrix}2&0\\0& 8 \end{bmatrix} P^{-1}\begin{bmatrix}x\\y \end{bmatrix} \\ \Rightarrow 2x'^2+ 8y'^2=72 \Rightarrow {x'^2\over 36}+{y'^2\over 9}=1 \Rightarrow \cases{a=6\\ b=3} \Rightarrow 正焦弦長={2b^2\over a}={2\cdot 9\over 6} =\bbox[red, 2pt]3$$

解答:$$\overline{DP}: \overline{PE}=2:3 \Rightarrow \overrightarrow{OP} ={3\over 5}\overrightarrow{OD}+ {2\over 5} \overrightarrow{OE} ={1\over 5}(3(\vec a+ \vec b) +2(\vec a+ \vec c)) = \vec a+{3 \over 5}\vec b+{2\over 5} \vec c\\ Q在\overleftrightarrow{OP}上 \Rightarrow \overrightarrow{OQ} = k\overrightarrow{OP} = k\vec a+{3 \over 5}k\vec b+{2\over 5}k \vec c \cdots(1)\\ 又Q在平面ACD上\Rightarrow \overrightarrow{OQ} =\alpha \overrightarrow{OA}+ \beta \overrightarrow{OB}+ \gamma \overrightarrow{OD}, 其中\alpha+ \beta+ \gamma=1 \\ \Rightarrow \overrightarrow{OQ} =\alpha \vec a+ \beta \vec c+ \gamma(\vec a+ \vec b)= (\alpha+ \gamma) \vec a+ \gamma \vec b+ \beta \vec c \cdots(2) \\ (1)=(2) \Rightarrow \cases{\alpha+\gamma=k\\ \gamma={3\over 5}k\\ \beta={2\over 5}k} \Rightarrow \alpha+ \beta+ \gamma ={2\over 5}k+{2\over 5}k+{3\over 5}k=1 \Rightarrow k={5\over 7} \Rightarrow \overline{OQ}:\overline{PQ}=5:2 \\ \Rightarrow {\triangle OAQ\over \triangle APQ}={\overline{OQ} \over \overline{PQ}} ={5\over 2} \Rightarrow \triangle OAQ: \triangle APQ= \bbox[red, 2pt]{5:2}$$

解答:$$\alpha,\beta,\gamma 為x^3-2x^2+3x-4=0的三根\Rightarrow \cases{ \alpha+ \beta+ \gamma=2\\ \alpha\beta+ \beta\gamma +\gamma \alpha=3\\ \alpha \beta \gamma=4} \Rightarrow \cases{\alpha +\beta= 2-\gamma\\ \alpha\beta+ \gamma(\alpha+\beta)=3 } \\\Rightarrow \alpha \beta=3-\gamma(\alpha+ \beta)= 3-\gamma(2-\gamma) =\gamma^2-2\gamma+3 \\ 因此\cases{\alpha^2 +\alpha \beta+ \beta^2=(\alpha+\beta)^2-\alpha\beta=(2-\gamma)^2-(\gamma^2-2\gamma+3)= 1-2\gamma\\ \beta^2 +\beta \gamma+ \gamma^2=1-2\alpha\\ \gamma^2+ \gamma \alpha + \alpha^2=1-2\beta}\\ 以1-2\alpha,1-2\beta,1-2\gamma為三根的函數g(x)=x^3+ax^2+bx+c=0 \Rightarrow a=2\alpha-1+2\beta-1+2\gamma-1 \\=2(\alpha+ \beta+\gamma)-3=4-3=1 \Rightarrow b=(1-2\alpha)(1-2\beta)+ (1-2\beta)(1-2\gamma)+ (1-2\gamma) (1-2\alpha) \\=1-2(\alpha+\beta)+ 4\alpha \beta +1-2(\beta +\gamma)+4\beta\gamma+1-2(\alpha+ \gamma)+ 4\gamma\alpha\\ =3-4(\alpha+ \beta+ \gamma)+ 4(\alpha\beta+ \beta \gamma+\gamma \alpha)=3-8+12=7\Rightarrow -c=(1-2\alpha)(1-2\beta)(1-2\gamma) \\=1-2(\alpha+ \beta+ \gamma)+4(\alpha\beta +\beta \gamma+ \gamma \alpha)-8\alpha\beta \gamma=1-4+12-32=-23 \Rightarrow c=23 \\ \Rightarrow (a,b,c)= \bbox[red, 2pt]{(1,7,23)}$$

解答:$$\omega=\cos{2\pi\over 5}+i \sin {2\pi\over 5}=e^{2\pi i/5 } \Rightarrow P_k= \omega^k,\; k=0,1,\dots,4  \Rightarrow 1+\omega+ \omega^2+ \omega^3+ \omega^4=0\\ \Rightarrow z={1\over 3}(P_1+P_3+ P_4)={1\over 3} (\omega+ \omega^3+ \omega^4) \Rightarrow |z|={1\over 3}|\omega+ \omega^3+ \omega^4| ={1\over 3}|-1-\omega^2|={1\over 3}|1+\omega^2| \\={1\over 3 } \left|1+ \cos{4\pi\over 5}+i\sin {4\pi\over 5} \right| ={1\over 3 } \left|2\cos^2{2\pi\over 5}+i2\sin {2\pi\over 5} \cos{2\pi\over 5}\right| ={2\over 3}\left|\cos {2\pi\over 5}\right| |\omega|={2\over 3}\left|\cos {2\pi\over 5}\right| \\={2\over 3}\cdot {\sqrt 5-1\over 4} = \bbox[red, 2pt]{\sqrt 5-1\over 6}$$

解答:$$f(x)=x^3+ ax^2+bx+c ,已知\lim_{x\to 0}{f(x)\over x} =-1存在\Rightarrow \cases{f(0)=c=0\\f'(0)=b=-1} \Rightarrow f(x)=x(x^2+ax-1) \\ \Rightarrow f(x)=0的三根為0,\alpha,\beta \Rightarrow \cases{\alpha+\beta =-a\\ \alpha\beta=-1} \Rightarrow \alpha,\beta 為一正一負，可假設\alpha\lt 0\lt \beta\\ \Rightarrow \cases{\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=a^2+2\\ \alpha^3+\beta^3=(\alpha+ \beta)(\alpha^2+\beta^2)-\alpha \beta(\alpha+\beta) =-a^3-3a\\ \alpha^4+ \beta^4=(\alpha^2+ \beta^2)^2-2(\alpha\beta)^2 =a^4+4a^2+2}\\ f(3)=24+9a\lt 0 \Rightarrow a\lt -{8\over 3} \\ 兩個封閉面積和= \int_\alpha^0 f(x)\,dx+ \int_0^\beta (-f(x)) \,dx= -{1\over 4}(\alpha^4+\beta^4)-{a\over 3}(\alpha^3+ \beta^3)+{1\over 2}(\alpha^2+\beta^2) \\=-{1\over 4}(a^4+4a^2+2)-{a\over 3}(-a^3-3a)+{1\over 2}(a^2+2) ={1\over 12}(a^4+6a^2+6) ={253\over 2} \\ \Rightarrow a^4+6a^2-1512=0 \Rightarrow (a^2-36)(a^2+42)=0 \Rightarrow a^2=36\Rightarrow a=-6 (\because a\lt -{8\over 3}) \\ \Rightarrow f(3)=24+9a= \bbox[red, 2pt]{-30}$$

第貳部分：計算證明題（共 28 分）

解答:$$\textbf{(1)} 假設身高依序為a_1\lt a_2\lt a_3\lt a_4\lt a_5\lt a_6,並令\cases{S=a_1+a_2+\cdots+a_6\\ 平均值\mu=S/6\\ 中位數M_e= (a_3+a_4)/2= \mu} \\ \Rightarrow {S\over 6}={a_3+a_4\over 2} \Rightarrow S=3(a_3+a_4) \Rightarrow S是3的倍數, \bbox[red, 2pt]{故得證} \\ \textbf{(2) }已知四數\cases{a_1=158\\ 163\\173\\a_6=178},並假設另外兩數為x\lt y \Rightarrow S=672+x+y=6\mu \Rightarrow x+y=6\mu-672\cdots(1) \\ \textbf{Case I }163\lt x\lt y\lt 173 \Rightarrow \mu={x+y\over 2} \Rightarrow x+y=2\mu代入(1) \Rightarrow 4\mu=672 \Rightarrow \mu=168 \\ \textbf{Case II }158\lt x\lt 163\lt y\lt 173\lt 178 \Rightarrow \mu={163+y\over 2} \Rightarrow y=2\mu-163 代入(1) \Rightarrow x=4\mu-509 \\ \qquad \Rightarrow \cases{y=164 \Rightarrow \mu=163.5 \Rightarrow x=145 \not \gt 158\\   \cdots\\ y=170 \Rightarrow \mu=166.5\Rightarrow x=157\not \gt 158\\ y=171 \Rightarrow \mu=167\Rightarrow x=159 \\ y=172 \Rightarrow \mu=167.5 \Rightarrow x=161} \\ \textbf{Case III }163\lt x\lt 173\lt y\lt 178 \Rightarrow \mu={x+173\over 2} \Rightarrow x=2\mu-173代入(1) \Rightarrow y=4\mu-499 \\ \qquad \Rightarrow \cases{x=164 \Rightarrow \mu=168.5 \Rightarrow y=175\\ x=165 \Rightarrow \mu=169 \Rightarrow y=177\\ x\ge 166 \Rightarrow y\ge 179 \gt a_6} \\ \textbf{Case IV }158\lt x\lt 163\lt 173\lt y\lt 178 \Rightarrow \mu=168 \Rightarrow x+y=336 \\ \qquad \Rightarrow \cases{x\in \{159,160,161,162\} \\y \in\{174,175,176,177\}} \Rightarrow x+y= 159+177 =160+176= 161+175= 162+174=336 \\ \Rightarrow 因此符合條件的平均\mu = \bbox[red, 2pt]{167,167.5,168,168.5,169}$$

解答:$$\textbf{(1) }使用數學歸納法\\ n=1 \Rightarrow a_1-2=2-2=0為5的倍數\\ 假設n=k時,a_k-2為5的倍數\\當n=k+1時, a_{k+1}-2=a_k^2+a_k+1-2=a_k^2+a_k-1 =(a_k-2)^2+5a_k-5 \\ \qquad =(a_k-2)^2+5(a_k-2)+5 為5的倍數 (\because a_k-2是5的倍數) \\ 因此,由數學歸納法可知，對於所有的正整數 n,a_n-2是5的倍數 \;\bbox[red, 2pt]{故得證} \\\textbf{(2) }使用數學歸納法\\ n=1 \Rightarrow a_1^2+1=2^2+1=5為5^1的倍數\\ 假設n=k時,a_k^2+1為5^k的倍數, 假設a_k^2+1= m\cdot 5^k, m為整數\\當n=k+1時, a_{k+1}^2+1 =(a_k^2+a_k+1)^2+1=  \left[ (a_k^2+1)+a_k \right]^2+1 \\= (a_k^2+1)^2 +2(a_k^2+ 1)a_k+ a_k^2+1 =(a_k^2+1) \left[ a_k^2+2a_k+2 \right]=m\cdot5^k [a_k^2+2a_k+2] \\= m\cdot 5^k \left[ (a_k-2)^2+ 5a_k+ (a_k-2) \right] \\ 由上題可知a_k-2是5的倍數，因此 [(a_k-2)^2+ 5a_k+ (a_k-2)] 是5的倍數 \\ 因此m\cdot 5^k \left[ (a_k-2)^2+ 5a_k+ (a_k-2) \right] 是5^{+1}的倍數,即n=k+1是亦成立\\ 因此,由數學歸納法可知，對於所有的正整數 n,a_n^2+1是5^n的倍數 \;\bbox[red, 2pt]{故得證}$$

====================== END ==========================
解題僅供參考，其他教甄試題及詳解