臺北市立成淵高級中學 114 學年度正式教師甄選
第壹部分：填充題（每題 6 分，共 72 分）
※每格完全答對才給分；答案若用分數呈現，請以最簡分數表示。
解答:$$S_n=n^2+4n-2=(n+2)^2-6 \Rightarrow S_{n-1}=(n+1)^2-6 \Rightarrow a_n=S_n-S_{n-1}=(n+2)^2-(n+1)^2 \\ \Rightarrow a_n=2n+3 \Rightarrow \cases{a_1=3\\ a_n=2n+3, n\ge 2} \Rightarrow \sum_{n=1}^{10} a_n^2 =3^2+\sum_{n=2}^{10} (2n+3)^2 \\= 9+ 4\sum_{n=2}^{10} n^2+ 12\sum_{n=2}^{10}n + 9\sum_{n=2}^{10}1 =9+4 \left( 385-1 \right) + 12\cdot (55-1) +9\cdot (10-1) \\=9+ 1536+648+ 81= \bbox[red, 2pt]{2274}$$
解答:$$f(x) =(6\sin x-8\cos x)(12\sin x-5\cos x)=72\sin^2x-66 \sin x\cos x-40\cos^2x \\=112\sin^2x-33\sin 2x-40=112 \left( {1-\cos 2x\over 2} \right)-33\sin 2x-40 =16-56\cos 2x-33\sin 2x \\=16-65 \sin(2x+\theta) \Rightarrow \cases{最大值16+65=81\\ 最小值16-65=-49} \Rightarrow (M,m)= \bbox[red, 2pt]{(81,-49)}$$
解答:$$假設\cases{C(0,0) \\ 正六邊形邊長為1} \Rightarrow \cases{A(\sqrt 3/2, 3/2) \\B(0,1) \\ C(0,0)\\ Z(26\sqrt 3,1)} \Rightarrow \cases{\overrightarrow{CZ} =(26\sqrt 3,1) \\ \overrightarrow {AB} =(-\sqrt 3/2, -1/2) \\ \overrightarrow {AC}=(-\sqrt 3/2, -3/2)} \\ \Rightarrow \alpha \left( -{\sqrt 3\over 2},-{1\over 2} \right)+ \beta \left( -{\sqrt 3\over 2},-{3\over 2} \right) =(26\sqrt 3,1) \Rightarrow \cases{\alpha+\beta= -52\\ \alpha+3\beta=-2} \Rightarrow (\alpha,\beta) = \bbox[red, 2pt]{(-77,25)}$$
解答:$$\vec u= \overrightarrow{BD}\times \overrightarrow{CE} \Rightarrow |\vec u|=\sqrt{1+20+28}=7 \Rightarrow \vec e={\vec u\over |\vec u|} = \left( {1\over 7},{2\sqrt 5\over 7},{2\sqrt 7\over 7} \right) \\ \Rightarrow \overrightarrow{AF} =-7\vec e =(-1,-2\sqrt 5, -2\sqrt 7) = \bbox[red, 2pt]{\left( -{\sqrt 7\over 7},-{2\sqrt{35}\over 7},-2 \right)}$$
解答:$$y= (3-\log_{\sqrt [3]2 }x)^2+(3-\log_x 2)^2 =(3-3\log_2 x)^2+ (3-{1\over \log_2 x})^2 \\= (3-A)^2+(3-B)^2,其中 A=3\log_2 x, B={1\over \log_2 x},且AB=3 \Rightarrow (A+B)^2=A^2+B^2+6\\ \Rightarrow A+B\ge 2\sqrt{AB}=2\sqrt 3 \\ \Rightarrow y=(A^2+B^2)-6(A+B)+18 = (A+B)^2-6(A+B)+12 =(A+B-3)^2+3 \\\ge (2\sqrt 3-3)^2+3= \bbox[red, 2pt]{24-12\sqrt 3}$$
解答:$$S= z^{70}+ 2z^{69}+ 3z^{68}+ \cdots+70z+71 \Rightarrow zS=z^{71}+ 2z^{70}+ 3z^{69}+ \cdots+70z^2+71z \\ \Rightarrow (z-1)S=z^{71}+z^{70} +z^{69}+\cdots+z-71 = {z(1-z^{71}) \over 1-z}-71 \Rightarrow S={z^{72}-z \over (z-1)^2}-{71\over z-1} \\z={\sqrt 3+i\over 2} =e^{i\pi/6} \Rightarrow z^6=-1 \Rightarrow z^{72}=1 \Rightarrow S={1-z\over (z-1)^2}-{71\over z-1} ={1\over 1-z}-{71\over z-1} \\=-{72\over z-1}=-{72\over {\sqrt 3-2+i\over 2}} =-{144\over \sqrt 3-2+i} =-{144(\sqrt 3-2-i)\over 8-4\sqrt 3} \\=-(36\sqrt 3-72)(2+\sqrt 3)+36(2+\sqrt 3)i =36+72+36\sqrt 3i \Rightarrow (a,b)=\bbox[red, 2pt]{(36,72+36\sqrt 3)}$$
解答:$$先求兩圖形\cases{y=f(x) = \sin x+1\\ y=g(x)= \log x}最右邊的交點,以圖形y=f(x)波峰/谷為起/迄點\\ g(x)=2 \Rightarrow x=100\Rightarrow 在x\lt 100的範圍內x={61\pi\over 2}是最右邊的波峰, x={63\pi \over 2 }是最右邊的波谷\\ \Rightarrow 最右邊的交點位於{61\pi\over 2}\lt x\lt {63\pi\over 2} \\接下來找最左邊的交點, \cases{f(\pi/2)=2 \gt g(\pi/2) \\ f(3\pi/2)=0\lt g(3\pi/2)}\Rightarrow 最左邊的交點位於{\pi\over 2}\lt x\lt {3\pi\over 2} \\ \Rightarrow 共經過\left( {63\pi\over 2}-{3\pi\over 2} \right)+1=31個半週期 \Rightarrow 共有\bbox[red, 2pt]{31}個交點$$
解答:
$$圓:x^2+y^2=R^2 \Rightarrow 四分之圓的重心x坐標 = {\int_0^R x\sqrt{R^2-x^2}\,dx \over {1\over 4}R^2\pi} ={4R\over 3\pi}\\ 本題R^2=6 \Rightarrow R=\sqrt 6 \Rightarrow 在第二象限四分之一圓的重心G坐標 \left( -{4R\over 3\pi} ,{4R\over 3\pi}\right) = \left( -{4\sqrt 6\over 3\pi} ,{4\sqrt 6\over 3\pi}\right) \\ \Rightarrow \cases{四分之一圓面積A=6\pi/4=3\pi/2\\ G至原點距離d= 8\sqrt 3/3\pi} \Rightarrow 旋轉體積V=2\pi d\cdot A ={16\sqrt 3\over 3} \cdot {3\pi\over 2} = \bbox[red, 2pt]{8\sqrt 3\pi}$$
解答:$$將12顆球擺在一起S=\{1,1,2,2,3,3,4,4,5,6,7,8\} \Rightarrow 組合數=12\times 12=144\\ \Rightarrow a=b的組合數\Rightarrow n(a=b)=(2\times 2)\times 4+(1\times 1)\times 4=20 \\ \Rightarrow n(a\gt b)+ n(a=b)+ n(a\lt b) =2n(a\gt b)+n(a=b)=2n(a\gt b)+20=144 \Rightarrow n(a\gt b)=62\\ a\gt b且b=5 \Rightarrow (a,b)=(6-8,5) \Rightarrow 組合數3 \Rightarrow 欲求之機率= \bbox[red, 2pt]{3\over 62}$$
解答:$$路徑1:1\to 2\to 3\to 5: {1\over 3}\times {1\over 3}\times {1\over 6} ={1\over 54} \\ 路徑2: 1\to 2\to 4\to 5: {1\over 3}\times {1\over 9}\times {5\over 18} ={5\over 486}\\ \Rightarrow 機率={1\over 54} +{5\over 486} = \bbox[red, 2pt]{7\over 243}$$
解答:
$$假設\cases{A\to G的期望步數E\\ \{B,D,E\} \to G的期望步數x\\ \{C,F,H\} \to G的期望步數y}, 又\cases{P(A\to \{B,D,E\})=1\\ P(\{B,D,E\} \to A)=1/3\\ P(\{B,D,E\} \to \{C,F,H\})=2/3 \\ P(\{C,F,H\} \to G)=1/3\\ P(\{C,F,H\} \to \{B,D,E\})=2/3} \\ \Rightarrow \cases{E=1+x\\ x=1+{1\over 3}E+{2\over 3}y \\ y=1+{2\over 3}x} \Rightarrow E=\bbox[red, 2pt]{10}$$
解答:$$\cases{A(0,0) \\B(4,4)} \Rightarrow\cases{\overline{AB}=4\sqrt 2 \\ L=\overleftrightarrow{AB}:x-y=0} \Rightarrow h=d(C,L) \Rightarrow \triangle ABC面積={1\over 2}\cdot 4\sqrt 2\cdot h=16 \Rightarrow h=4\sqrt 2\\ C(x,y) \Rightarrow h={|x-y|\over \sqrt 2} =4\sqrt 2 \Rightarrow |x-y|=8 \Rightarrow y-x=8 \Rightarrow y=x+8 (\because x-y\lt 0) \\ \Rightarrow C(x,x+8) \Rightarrow \cases{\overline{AC}= \sqrt{x^2+ (x+8)^2} \\ \overline{BC}=\sqrt{(x-4)^2+(x+4)^2}} \Rightarrow \overline{AC}+ \overline{BC}=16\sqrt 2-\overline{AB} \\ \Rightarrow \sqrt{x^2+(x+8)^2}+\sqrt{(x-4)^2+(x+4)^2}=12\sqrt 2 \Rightarrow x^2+4x-14=0 \Rightarrow x=-2-3\sqrt 2 \\ \Rightarrow y=x+8=6-3\sqrt 2 \Rightarrow C= \bbox[red, 2pt]{(-2-3\sqrt 2, 6-3\sqrt 2)}\\ 答案有誤，應更正為\bbox[cyan,2pt]{(-3\sqrt 2-2,-3\sqrt 2+6)}$$
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解題僅供參考，其他教甄試題及詳解
解答:$$\cases{A(0,0) \\B(4,4)} \Rightarrow\cases{\overline{AB}=4\sqrt 2 \\ L=\overleftrightarrow{AB}:x-y=0} \Rightarrow h=d(C,L) \Rightarrow \triangle ABC面積={1\over 2}\cdot 4\sqrt 2\cdot h=16 \Rightarrow h=4\sqrt 2\\ C(x,y) \Rightarrow h={|x-y|\over \sqrt 2} =4\sqrt 2 \Rightarrow |x-y|=8 \Rightarrow y-x=8 \Rightarrow y=x+8 (\because x-y\lt 0) \\ \Rightarrow C(x,x+8) \Rightarrow \cases{\overline{AC}= \sqrt{x^2+ (x+8)^2} \\ \overline{BC}=\sqrt{(x-4)^2+(x+4)^2}} \Rightarrow \overline{AC}+ \overline{BC}=16\sqrt 2-\overline{AB} \\ \Rightarrow \sqrt{x^2+(x+8)^2}+\sqrt{(x-4)^2+(x+4)^2}=12\sqrt 2 \Rightarrow x^2+4x-14=0 \Rightarrow x=-2-3\sqrt 2 \\ \Rightarrow y=x+8=6-3\sqrt 2 \Rightarrow C= \bbox[red, 2pt]{(-2-3\sqrt 2, 6-3\sqrt 2)}\\ 答案有誤，應更正為\bbox[cyan,2pt]{(-3\sqrt 2-2,-3\sqrt 2+6)}$$
第貳部分：計算證明題（共 28 分）
※請將解題過程書寫於答案卷方框內。若只有答案，沒有詳述原因或推導過程會斟酌扣分。
解答:$$X= \begin{bmatrix} 2& 1\\ 0& 5\end{bmatrix} =\begin{bmatrix} 1& 1/3\\ 0& 1\end{bmatrix} \begin{bmatrix} 2& 0\\ 0& 5\end{bmatrix} \begin{bmatrix} 1& -1/3\\ 0& 1\end{bmatrix} \Rightarrow X^n= \begin{bmatrix} 2& 1\\ 0& 5\end{bmatrix} =\begin{bmatrix} 1& 1/3\\ 0& 1\end{bmatrix} \begin{bmatrix} 2^n& 0\\ 0& 5^n\end{bmatrix} \begin{bmatrix} 1& -1/3\\ 0& 1\end{bmatrix} \\=\left[ \begin{matrix} 2^n & \frac{-2^n+5^n}{3} \\0 & 5^n \end{matrix}\right] \Rightarrow \textbf{(1) }b_4={-2^4+5^4 \over 3} = \bbox[red, 2pt]{203} \Rightarrow \textbf{ (2) }b_n= \bbox[red, 2pt]{{1\over 3}(5^n-2^n)}$$
解答:$$\textbf{(1) }f(a+b)={1\over 3}f(a)f(b) \Rightarrow f(x)=f(x+0)={1\over 3}f(x)f(0) \Rightarrow f(0)=3\\ \Rightarrow \lim_{x\to 0}{(f(x))^2-9\over x} = \lim_{x\to 0}{f(x)f(x)-9\over x}= \lim_{x\to 0}{3f(2x) -9\over x} = 2\lim_{x\to 0}{3f(2x) -3f(0)\over 2x}\\= 6\lim_{x\to 0}{f(2x) -f(0)\over 2x} = 2025\Rightarrow f'(0)={2025\over 6} =\bbox[red, 2pt]{675\over 2} \\\textbf{(2) } \lim_{h\to 0}{f(h+c) -f(c) \over h} = \lim_{h\to 0}{{1\over 3}f(h)f(c) -f(c) \over h}= \lim_{h\to 0}{{1\over 3}f(c)(f(h)-3) \over h}\\\quad = \lim_{h\to 0}{{1\over 3}f(c)(f(h)-f(0)) \over h} ={1\over 3}f(c) \lim_{h\to 0}{f(h)f(0)\over h} ={1\over 3}f(c)\cdot f'(0) ={1\over 3}f(c)\cdot {675\over 2}\\\quad ={225\over 2}f(c)存在\Rightarrow f'(c)存在 \Rightarrow f(x)可微,\forall c \bbox[red, 2pt]{QED}$$
解答:$$三根之積\beta \cdot 2\beta^2 \cdot 3\beta^3=6\beta^6=-d\lt 0 \Rightarrow \beta 為虛數 \Rightarrow \bar \beta為其中一根 \\若\bar \beta=2\beta^2 \Rightarrow |\bar \beta|=2|\beta|^2 \Rightarrow |\beta|=1/2 \Rightarrow 三根之積6\beta^6=6/64 \gt 0矛盾 \\ \Rightarrow \bar \beta=3\beta^3 \Rightarrow 第三根2\beta^2 \in \mathbb R \Rightarrow \beta^2為實數\Rightarrow \beta為虛數 \Rightarrow \bar \beta=-\beta \Rightarrow -\beta=3\beta^2 \Rightarrow \beta^2=-1/3 \\ \Rightarrow \cases{第1根\beta\\ 第2根3\beta^3 = -\beta\\ 第3根2\beta^2 =-2/3} \Rightarrow 三根之和=-2/3=-月\Rightarrow b= \bbox[red, 2pt]{2\over 3}$$
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解題僅供參考，其他教甄試題及詳解
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