114年板橋高中教甄-數學詳解

新北市立板橋高級中學114學年度第一次教師甄選

一、填充題(每題6分，10題，共60分)

解答:$$假設\cases{O(0,0,0) \\ P(a,b,c)\\ A(-3,-4, -12)} \Rightarrow \cases{\overline{OA}= \sqrt{3^2+4^2+12^2}=13\\ 原式:\overline{OP}+ \overline{PA}= 13} \Rightarrow P \in \overline{OA} \Rightarrow P(-3t,-4t, -12t),0\lt t\lt 1 \\ \Rightarrow 欲求{6a+3b+2c\over \sqrt{a^2+b^2+c^2}} +{20a+ 5b+4c+128\over \sqrt{(a+3)^2+ (b+4)^2 +(c+12)^2}} \\= {-54t\over 13t}+{-128t+ 128\over 13(1-t)} = {128-54\over 13} = \bbox[red, 2pt]{74\over 13}$$

解答:$$\tan 82^\circ = \tan(94^\circ-12^\circ) = {\tan 94^\circ -\tan 12^\circ \over 1+ \tan 94^\circ \tan 12^\circ } \\\Rightarrow \tan 82^\circ+ \tan82^\circ \tan 94^\circ \tan 12^\circ =\tan 94^\circ -\tan 12^\circ \\\Rightarrow \tan 12^\circ \tan 94^\circ \tan 82^\circ =\tan 94^\circ-\tan82^\circ -\tan 12^\circ 代回原式 \Rightarrow -\tan 82^\circ =\tan k^\circ \\ \Rightarrow \tan k^\circ = \tan(180^\circ-82^\circ) =\tan 98^\circ =\tan (98^\circ+180^\circ) =\tan 278^\circ \Rightarrow k= \bbox[red, 2pt]{278}$$

解答:$$對任一個人而言，至少拿到一顆球的機率=1- \left( {5\over 6} \right)^{10}\\ 因此總獎金的期望值=2\times 10+6\times \left[ 1- \left( {5\over 6} \right)^{10} \right] = \bbox[red, 2pt]{26-{5^{10} \over 6^9}}$$

解答:$$f_1(x)={2x-7\over x+1}  \Rightarrow f_2(x)= {2 ({2x-7\over x+1} )-7\over {2x-7\over x+1} +1} ={x+7\over 2-x} \Rightarrow f_3(x)={{2x-7\over x+1} +7\over 2-{2x-7\over x+1} } =x \\ \Rightarrow 循環數3 \Rightarrow \cases{13 \equiv 1{\mod 3} \\ 23 \equiv 2 {\mod 3} \\ 36 \equiv 0 {\mod 3}} \Rightarrow \cases{f_{13}=f_1\\ f_{23}=f_2 \\ f_{36}=f_3} \Rightarrow f_1+f_3=2f_2 \\ \Rightarrow {2x-7\over x+1}+x=2 \left( {x+7\over 2-x} \right) \Rightarrow x^3+3x^2+3x+28=0 \Rightarrow (x+4)(x^2-x+7)=0 \\ \Rightarrow x=-4 \Rightarrow \cases{f_1=5\\ f_2=1/2\\ f_3=-4} \Rightarrow f_1\times f_2\times f_3=-10 \\ \Rightarrow \log \left( f_1\times f_2\times f_3 \times \cdots \times f_{115} \right) =\log \left( (-10)^{38}\times f_{115} \right)=\log \left( 10^{38}\times f_1 \right)=38+\log 5 \\=38+1-0.301=38.699 \approx \bbox[red, 2pt]{39}$$

解答:$$假設切線斜率為m \Rightarrow y'=2x=m \Rightarrow x={m\over 2} \Rightarrow 切點Q({m\over 2}, {m^2\over 4}) \\ \Rightarrow 切線方程式:y=m(x-{ m\over 2})+{m^2\over 4} \Rightarrow m^2-4xm+4y=0 \Rightarrow m=m_1,m_2 \\ \Rightarrow \cases{m_1+m_2=4x\\ m_1m_2=4y} \Rightarrow (m_1-m_2)^2=16x^2-16y \Rightarrow |m_1-m_2|= 4\sqrt{x^2-y}\\ \Rightarrow \tan \angle QPR=4= \left|{m_1-m_2\over 1+m_1m_2} \right|={4\sqrt{x^2-y} \over 1+4y} \Rightarrow (1+4y)^2=x^2-y \\\Rightarrow x^2+0xy-16y^2+0x-9y+1=0 \Rightarrow (B,C,D,E,F) = \bbox[red, 2pt]{(0,-16,0,-9,1)}$$

解答:$$x^2+y^2=4 \Rightarrow \cases{x= 2\cos \theta\\ y=2\sin \theta} \Rightarrow {4\over x^2}+{6y\over x}={4\over 4\cos^2\theta} +{12\sin \theta \over 2\cos \theta} =\sec^2\theta+6\tan \theta \\=\tan^2\theta+1+6\tan \theta =(\tan+3)^2-8 \ge \bbox[red, 2pt]{-8}$$

解答:$$\cases{2^1 \equiv 2 {\mod 10} \\2^2 \equiv 4 {\mod 10} \\2^3 \equiv 8 {\mod 10} \\2^4 \equiv 6 {\mod 10} \\2^5 \equiv 2 {\mod 10} } \Rightarrow 循環數4 \\ 取f(x)=(1+x)^{2025} = \sum _{k=0}^{2025}C^{2025}_kx^k \Rightarrow f'(x)=2025(1+x)^{2024} = \sum_{k=0}^{2025} kC^{2025}_kx^{k-1} \\ \Rightarrow g(x) =xf'(x)=2025x(1+ x)^{2024} = \sum_{k=0}^{2025} kC^{2025}_kx^{k}  \\\Rightarrow g'(x)= 2025(1 +x)^{2024} +2025\cdot 2024x(1+x)^{2023} = \sum_{k =0}^{2025} k^2C^{2025}_kx^{k-1} \\ \Rightarrow g'(1)=2025\cdot 2^{2024}+ 2025\cdot 2024\cdot 2^{2023} =\sum_{k =0}^{2025} k^2C^{2025}_k \\ \Rightarrow S=\sum_{k=1}^{2025} k^2C^{2025}_k =2025\cdot 2^{2024}+ 2025\cdot 2024\cdot 2^{2023} =8205300\cdot 2^{2022}\\ 2022 \equiv 2 {\mod 4} \Rightarrow 2^{2022} \equiv 2^2 \equiv {\mod 10}=4 \\ \Rightarrow S \equiv 8205300\cdot 4 {\mod 1000} = \bbox[red, 2pt]{200}$$

解答:$$直線L:\cases{x=z\\ y=0}的斜率為1 \Rightarrow \Gamma: z=\sqrt{x^2+y^2}與z軸對稱\Rightarrow 球心C(0,0,c), 球半徑R \\ \Rightarrow \cases{d(C, \Gamma) =R \Rightarrow c\sin 45^\circ =R \Rightarrow R={\sqrt 2c\over 2} \\ d(C,E)=R \Rightarrow R={66-6c\over \sqrt{50}}} \Rightarrow {\sqrt 2c\over 2}={66-6c\over \sqrt{50}} \Rightarrow c=6 \\\Rightarrow R={\sqrt 2\over 2}\cdot 6= \bbox[red, 2pt]{3\sqrt 2}$$

解答:

$$\cases{z\le 10-2x^2 \\ z\le 10-5y^2} \Rightarrow \cases{x^2 \le {10-z\over 2} \\y^2\le {10-z\over 5}} \Rightarrow \cases{-\sqrt{10-z\over 2} \le x\le \sqrt{10-z\over 2} \\ -\sqrt{10-z\over 5}\le y\le \sqrt{10-z\over 5}} \\ \Rightarrow 對任意高度z, 截面是一個矩形 \Rightarrow 截面積A(z)= \left( 2\sqrt{10-z\over 2} \right) \times \left( 2\sqrt{10-z\over 5} \right) \\ \Rightarrow A(z)={4(10-z) \over \sqrt{10}} \Rightarrow 體積V=\int_0^{10} {4(10-z) \over \sqrt{10}}\,dz= \bbox[red, 2pt]{20\sqrt{10}}$$

解答:$$假設\gcd(a,c)=x \Rightarrow \cases{a=kx\\ c=mx \\ \gcd(k,m)=1} \Rightarrow (kx)b=(mx)d \Rightarrow kb=md \\ 由於\gcd(k,m)=1 \Rightarrow \cases{k整除d\\ m整除b} \Rightarrow \cases{b=my\\ d=ky} \Rightarrow a+b+c+d =xk+my+xm+ky=203 \\ \Rightarrow (x+y)(m+k)=203=7\times 29 \Rightarrow \cases{\cases{x+y=7\\ k+m=29} \Rightarrow 共6\times 28=168組\\ \cases{x+y=29\\ k+m=7} \Rightarrow 也是168組} \Rightarrow 共\bbox[red, 2pt]{336}組$$

二、計算證明題(每題10分，4題題，共40分)

解答:$$f(x)=(a-1)(\log_5 x)^2-8a(\log_5 x)+3a+1 = \left( (\log_5 x)^2-8\log_5x+3 \right)a-(\log_5 x)^2+1 =g(x) \\ \Rightarrow \cases{g(0)= -(\log_5 x)^2+1 \gt 0\\ g(1)=-8\log_5x +4\gt 0 } \Rightarrow \cases{1/5\lt x\lt 5\\ x\lt \sqrt 5} \Rightarrow \bbox[red, 2pt]{{1\over 5}\lt x\lt \sqrt 5}$$

解答:$$假設f(x)=a_nx^2+ \cdots+ a_3x^3+ x^2+x+1=0的根為x_1, x_2, \dots, x_n,\\ \Rightarrow g(x)=x^n+x^{n-1}+x^{n-2} +a_3x^{n-3}+ \cdots+a_{n-1}x+a_n=0的根為{1\over x_1}, {1\over x_2}, \dots,{1\over x_n}\\ 依據根與係數關係 \Rightarrow \cases{\sum {1\over x_i}=-1 \\ \sum {1\over x_ix_j}=1}\\ 恆等式: \left( \sum{1\over x_i} \right)^2 = \sum \left( {1\over x_i} \right)^2+ 2\sum {1\over x_ix_j} \Rightarrow (-1)^2=\sum \left( {1\over x_i} \right)^2+2\cdot 1 \\ \Rightarrow \sum \left( {1\over x_i} \right)^2=-1 \Rightarrow 至少有一虛根, 又虛根成對存在，因此至少有兩個虛根 \bbox[red, 2pt]{故得證}$$

解答:$$S=1-{1\over 5}+{1\over 7}-{1\over 11}+{1\over 13}+ \cdots = \int_0^1 \left( 1-x^4+x^6-x^{10} + x^{12}- \cdots +x^{6n-6} -x^{6n-2}+ \cdots\right)\,dx \\=\int_0^1 \left(  (1-x^4)+x^6(1-x^4) + x^{12}(1-x^4)+ \cdots+x^{6n-6}(1-x^4)+ \cdots\right)\,dx \\= \int_0^1 (1-x^4)(1+x^6+x^{12} +\cdots)\,dx =\int_0^1 {1-x^4\over 1-x^6}\,dx =\int_0^1 {1+x^2\over 1+x^2+x^4}\,dx \\={1\over 2} \int_0^1 {dx\over x^2-x+1}+{1\over 2} \int_0^1 {dx\over x^2+x+1}={1\over 2} \left. \left[ {2\over \sqrt 3}\tan^{-1}{2x-1\over \sqrt 3} \right] \right|_0^1 +{1\over 2} \left. \left[ {2\over \sqrt 3}\tan^{-1}{2x+1\over \sqrt 3} \right] \right|_0^1 \\={1\over 2} \left( {2\pi\over 3\sqrt 3} \right) +{1\over 2} \left( {\pi\over 3\sqrt 3} \right) = \bbox[red, 2pt]{\sqrt 3\pi \over 6}$$

解答:$$ F={\tan \alpha+\tan \beta+\tan \gamma\over \tan \alpha \tan \gamma}={1\over \tan \gamma}+{\tan \beta \over \tan \alpha \tan \gamma}+{1\over \tan \alpha} =\cot \gamma+\cot \alpha \cot \gamma \tan \beta+ \cot \alpha \\假設\cases{\overleftrightarrow{BC}在x軸上\\ A在y軸上 \\ \overline{BD}= \overline{DE} =\overline{EC}=k} \Rightarrow 取\cases{A(0,h) \\B(x-k,0)\\ D(x,0)\\ E(x+k,0)\\ C(x+2k,0)} \Rightarrow \cases{\cot \alpha= (h^2+x^2-kx)/hk\\ \cot \beta =(h^2+x^2+kx)/hk\\ \cot \gamma=(h^2+x^2+3kx+2k^2)/hk} \\ \Rightarrow \cases{\cot \alpha=\cot \beta-2x/h \\ \cot \gamma =\cot \beta+(2x+2k)/h } \Rightarrow  \cot \beta +\cot \gamma=2\cot \beta+{2k\over h} \\ \Rightarrow \cot\alpha \cot \gamma = \left( \cot \beta-{2x\over h} \right) \left( \cot \beta+{2x+2k\over h} \right) =(\cot \beta)^2+{2k\over h} \cos \beta-{4(x^2+kx)\over h^2} \\  =(\cot \beta)^2+{2k\over h} \cos \beta-{4(hk \cot \beta-h^2)\over h^2} =(\cot \beta)^2-{2k\over h} \cot \beta+4 \\ \Rightarrow F= 2\cot \beta+{2k\over h}+ \left( (\cot \beta)^2-{2k\over h} \cot \beta+4 \right) \tan \beta =\bbox[red, 2pt]{3\cot \beta+ 4\tan \beta}$$

 

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