114年台師大附中第2次教甄-數學詳解

 國立台灣師範大學附屬高級中學114學年度第2次專任教師甄

一． 選填題（每題 5 分，共 80 分。填在答案卡上，須以最簡分數、最簡根式作答，否則不予計分）

解答:$$y=f(x)={2^x+2^{-x} \over 2^x-2^{-x}} ={2^{2x}+1\over 2^{2x}-1} ={4^x+1\over 4^x-1}=1+{2\over 4^x-1} \Rightarrow x=\log_4 {y+1\over y-1} \\ \Rightarrow \cases{a= \log_4 \displaystyle {4+1\over 4-1} =\log_4 {5\over 3} \\ b=\log_4 \displaystyle {3+1\over 3-1}= \log_4 2} \Rightarrow a+b=\log_4 {10\over 3} \Rightarrow f(a+b)=f(\log_4 {10\over 3}) ={10/3+1\over 10/3-1}= \bbox[red, 2pt]{13\over 7}$$

解答:

$$\cases{x\gt 0\\ 1-y^2 \gt 0} 且\cases{x+y\gt 1且 \sqrt{1-y^2}\gt x\\ 0\lt x+y\lt 1 且\sqrt{1-y^2}\lt x} \Rightarrow 所圍區域如上圖S與T \\ \Rightarrow \cases{S=\pi/4-1/2\\ T=1-\pi/8} \Rightarrow S+T= \bbox[red, 2pt]{{1\over 2}+{1\over 8}\pi}$$

解答:$$P\in C \Rightarrow P(2\cos \theta+3, 2\sin \theta+4) \Rightarrow \overline{PA}^2+ \overline{PB}^2   = \\(2\cos \theta+4)^2+ (2\sin \theta+4)^2 +(2\cos \theta+2)^2+ (2\sin \theta+4)^2 =60+24\cos  \theta+32\sin \theta \\=60+40 \sin(\alpha+\theta) \Rightarrow \cases{M=60+40=100\\ m=60-40=20} \Rightarrow M+m= \bbox[red, 2pt]{120}$$

解答:$$假設\triangle ABC面積為S\Rightarrow S ={1\over 2}bc \sin A \Rightarrow \sin A={2S\over bc}, 又\cos A={b^2+c^2-a^2\over 2bc} \\ 因此\tan A= {4S\over b^2+c^2-a^2}, 同理\tan B={4S\over c^2+a^2-b^2}, \tan C={4S\over a^2+b^2-c^2} \\ \Rightarrow {\tan C\over \tan A}+{\tan C\over \tan B}={b^2+c^2-a^2\over a^2+b^2-c^2} +{c^2+a^2-b^2\over a^2+b^2-c^2} ={2c^2 \over a^2+b^2-c^2}={2c^2\over 8c^2-c^2} =\bbox[red, 2pt]{2\over 7}$$

解答:$$\cases{\alpha^3=4\alpha+2 \\\beta^3=4\beta+2 \\\gamma^3=4\gamma+2 } \Rightarrow \alpha^6+\beta^6+\gamma^6=(4\alpha+2)^2+(4\beta+2)^2 +(4\gamma+2)^2 \\=16(\alpha^2+ \beta^2+ \gamma^2)+16(\alpha+\beta+\gamma)+12=16[(\alpha+\beta+ \gamma)^2-2(\alpha\beta+ \beta\gamma+ \gamma\alpha)] +0+12 \\=16\cdot 8+12= \bbox[red, 2pt]{140}$$

解答:$$本題更正為\bbox[cyan,2pt]{送分}$$

解答:$$\cases{A\in L_1 \Rightarrow A(2s,3s-1,s-3) \\ B\in L_2 \Rightarrow B(4t-1,-2t-4,-t-2)} \Rightarrow \cases{\overrightarrow{PA}=(2s-1,3s-3,s-2) \\ \overrightarrow{PB} =(4t-2,-2t-6,-t-1)} \\ \overrightarrow{PA} \parallel \overrightarrow{PB} \Rightarrow {2s-1\over 4t-2}={3s-3\over -2t-6}={s-2\over -t-1} ={1\over k} \Rightarrow \cases{s=7/3\\ t=-3/5\\ k=-6/5} \\ \Rightarrow |\overrightarrow{PB}| =|k||\overrightarrow{PA}| \Rightarrow {\overline{PA} \over \overline{PB}}={1\over |k|}= \bbox[red, 2pt]{5\over 6}$$

解答:$$\cases{A(4,-1,0) \\B(2,1,2) \\C(0,1,4)} \Rightarrow \cases{P=\overline{AB}中點=(3,0,1) \\Q=\overline{BC}中點=(1,1,3) \\R=\overline{AC}中點=(2,0,2)  \\ \overrightarrow{AB} =(-2,2,2) \\ \overrightarrow{AC}=(-4,2,4) \\ \overrightarrow{BC} =(-2,0,2)}  \Rightarrow \cases{\overline{AB}中垂平面E_1:-(x-3)+y+(z-1)=0 \\\overline{AC}中垂平面E_2:-2(x-2)+y+2(z-2)=0 \\\overline{BC}中垂平面E_2:-(x-1)+(z-3)=0 } \\ \Rightarrow Q=E_1\cap E_2\cap E_3 = \bbox[red, 2pt]{(1,-4,3)}$$

解答:

$$L:2x+y-1=0與x軸交於P({1\over 2},0) \Rightarrow L以P為轉軸,逆時針旋轉\tan^{-1}2變為x軸\\ V(-1,3)以P為轉軸,逆時針旋轉\tan^{-1}2變為V'({1\over 2}-{15\over 2\sqrt 5},0) \\ A(3,3)以P為轉軸,逆時針旋轉\tan^{-1}2變為A'({1\over 2}-{7\over 2\sqrt 5},{8\over \sqrt 5})\\ 由頂點V' 可得旋轉後的拋物線方程式: y^2=4c(x-{1\over 2}+{15\over 2\sqrt 5})\\再將A'代入\Rightarrow c={4\sqrt 5\over 5} \Rightarrow 正焦弦長4c=\bbox[red, 2pt]{16\sqrt 5\over 5}$$

解答:

$$假設\cases{\overline{BI}=a =\overline{DP}\\ \overline{IF}=b\\ \overline{FC}=c =\overline{PE}} \Rightarrow \cases{\triangle BFG \sim \triangle BCA \Rightarrow \displaystyle {a+b\over a+b+c} ={d\over 494} \cdots(1)\\ \triangle CIH \sim \triangle CBA \Rightarrow \displaystyle {b+c\over a+b+c} ={d\over 380} \cdots(2)\\ \displaystyle {\overline{DE} \over \overline{BC}} ={a+c\over a+b+c}={d\over 520} \cdots(3)}\\ \Rightarrow (1)+(2)+(3)=2=d \left( {1\over 494} +{1\over 380} +{1\over 520}\right)  \Rightarrow {d\over 152}=2 \Rightarrow d= \bbox[red, 2pt]{304}$$

解答:$$取\cases{x'=x\\ 2y'=y} 則4x^2+y^2=72 \Rightarrow 4x'^2+4y'^2=72 \Rightarrow x'^2+y'^2=18 \Rightarrow P(-3,6) \Rightarrow P'(-3,3) \\ \Rightarrow \cases{P'逆時針旋轉60^\circ 得Q'(-{3+3\sqrt 3\over 2},{3-3\sqrt 3\over 2}) \Rightarrow Q(-{3+3\sqrt 3\over 2},3-3\sqrt 3)\\ P'順時針旋轉60^\circ得R'({3\sqrt 3-3\over 2},{3\sqrt 3+3\over 2}) \Rightarrow R({3\sqrt 3-3\over 2},{3\sqrt 3+3 })} \\ \Rightarrow \overline{QR} =\sqrt{(3\sqrt 3)^2+(6\sqrt 3)^2}= \sqrt{135}= \bbox[red, 2pt]{3\sqrt{15}}$$

解答:$$|z^{20}|=\sqrt{18^2+(10\sqrt 7)^2} =\sqrt{1024} =2^5 \Rightarrow |z|=2^{1/4} \\ 在半徑為2^{1/4}的圓周上,正20邊形的圓心角為18^\circ, 而\cos (4\times 18^\circ) =\cos72^\circ ={(2^{1/4})^2+ (2^{1/4})^2-a^2 \over 2\cdot 2^{1/4} \cdot 2^{1/4}} \\ \Rightarrow {\sqrt 5-1\over 4}={2\sqrt 2-a^2\over 2\sqrt 2} \Rightarrow 兩頂點距離a=\sqrt{10\sqrt 2-2\sqrt{10}\over 4} \lt \sqrt{1+2\sqrt 2-\sqrt 3}\\\cos 90^\circ =0 \Rightarrow a^2=2\sqrt 2 \Rightarrow a=\sqrt{2\sqrt 2} \gt \sqrt{1+2\sqrt 2-\sqrt 3}\\ 也就是說:任一頂點與其左右四個頂點之距離符合要求，因此機率={ 4\cdot 2\cdot 20/2\over c^{20}_2}= \bbox[red, 2pt]{8\over 19}$$

解答:

$$y=kx^2代入圓C \Rightarrow k^2x^4+(1-6k)x^2+1=0 \Rightarrow 判別式: (1-6k)^2-4k^2=0\\ \Rightarrow 32k^2-12k+1 =0 \Rightarrow k={1\over 4} \quad (k={1\over 8}代回原式\Rightarrow 無解) \\ \Rightarrow {1\over 16}x^4-{1\over 2}x^2+1=0 \Rightarrow ({1\over 4}x^2-1)^2=0 \Rightarrow x=\pm 2 \Rightarrow 切點\cases{P(2,1) \\Q(-2,1)}\\ \Rightarrow R= \pi \int_{-2}^2 (3-\sqrt{8-x^2})^2\,dx -\pi \int_{-2}^2 ({1\over 4}x^2)^2 \,dx =({116\over 3}\pi-12\pi^2)-{4\over 5}\pi \\=-12\pi^2+{568\over 15}\pi \Rightarrow 3a+b+c=-36+{568\over 15}+0= \bbox[red, 2pt]{28\over 15}$$

解答:$$假設a\le b\le c \Rightarrow abc=2(a+b+c) \le 2(c+c+c)=6c \Rightarrow ab\le 6 \\ \Rightarrow \cases{(a,b)= (1,1) \Rightarrow c=4+2c \Rightarrow \times\\(a,b)=(1,2) \Rightarrow 2c=6+2c \Rightarrow \times\\(a,b)=(1,3) \Rightarrow 3c=8+2c \Rightarrow c=8\\ (a,b)=(1,4) \Rightarrow 4c=10+2c \Rightarrow c=5 \\ (a,b)=(1,5) \Rightarrow 5c=12+2c \Rightarrow c=4 \not \ge b\\(a,b)=(1,6) \Rightarrow 6c=14+2c \Rightarrow \times\\ (a,b)=(2,2)\Rightarrow 4c=8+2c \Rightarrow c=4\\(a,b)=(2,3) \Rightarrow 6c=10+2c \Rightarrow \times} \Rightarrow (a,b,c)=\cases{(1,3,8)排列數6\\ (1,4,5)排列數6\\ (2,2,4)排列數3} \\ \Rightarrow 共有6+6+3=\bbox[red, 2pt]{15}種可能$$

解答:$$分子=S= \sum_{k=1}^{20} k(k+1)(21-k)(22-k) =\sum_{k=1}^{20}(k^4-42k^3+419k^2+462k) \\={1\over 30}(20\cdot 21\cdot 41\cdot (1200+60-1)) -42\cdot ({20\cdot 21\over 2})^2+ 419\cdot {1\over 6}\cdot 20\cdot 21\cdot 41 +462\cdot {1\over 2}\cdot 20\cdot 21 \\=722666-1852200+1202530+97020=170016 \\ 分母=T=\sum_{k=1}^{21} k(k+1)  =\sum_{k=1}^{21}(k^2+k)={1\over 6}\cdot 21\cdot 22\cdot 43 +{1\over 2}\cdot 22\cdot 21 =3542 \\ \Rightarrow {S\over T}={170016\over 3542}= \bbox[red, 2pt]{48}$$

解答:$$x_{k+1}={1\over 3}x_k^2+x_k \Rightarrow 3x_{k+1}=x_k(x_k+3) \Rightarrow {1\over x_k+3} ={x_k\over 3x_{k+1}}  ={x_k^2\over 3 x_kx_{k+1}} ={3x_{k+1} -3x_k \over 3 x_kx_{k+1}}  \\ \Rightarrow {1\over x_k+3}={1\over x_k}-{1\over x_{k+1}} \Rightarrow {1\over x_1+3}+{1\over x_2+3}+\cdots +{1\over x_{2025}+3} \\= \left( {1\over x_1}-{1\over x_2} \right) + \left( {1\over x_2}-{1\over x_3} \right) +\cdots + \left( {1\over x_{2025}}-{1\over x_{2026}} \right) ={1\over x_1}-{1\over x_{2026}}=20-{1\over x_{2026}} \\ 由於0\lt {1\over x_{2026}} \lt 1 \Rightarrow 整數部分為\bbox[red, 2pt]{19}$$

二． 非選題（共 20 分。請用黑色或藍色原子筆寫在作答卷上，須詳述過程，否則酌予扣分）

解答:$$(k+1)^3-k^3=3k^2+3k+1 \Rightarrow \sum_{k=1}^n ((k+1)^3-k^3) =\sum_{k=1}^n(3k^2+3k+1)\\左式:\sum_{k=1}^n ((k+1)^3-k^3) =(2^3-1^3)+(3^3-2^3)+\cdots +((n+1)^3-n^3)=(n+1)^3-1 \\右式: \sum_{k=1}^n(3k^2+3k+1)=3 \sum_{k=1}^n k^2+3\cdot {n(n+1)\over 2}+n \\ 左式=右式\Rightarrow 3 \sum_{k=1}^n k^2 =(n+1)^3-1-3\cdot {n(n+1)\over 2}-n ={n(n+1)(2n+1) \over 2} \\ \Rightarrow \sum_{k=1}^n k^2= {n(n+1)(2n+1) \over 6} \quad \bbox[red, 2pt]{故得證}$$

解答:$$S_n=({n+1\over n})+({n+1\over n})^2+ \cdots+({n+1\over n})^n =-n\left({n+1\over n}-({n+1\over n})^{n+1} \right) \\ \Rightarrow \lim_{n\to \infty}{S_n\over n}= \lim_{n\to \infty} \left(({n+1\over n})^{n+1}-{n+1\over n} \right) =\bbox[red, 2pt]{e-1}$$

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