113年台師大附中教甄-數學詳解

國立臺灣師範大學附屬高級中學 113 學年度
第 1 次專任教師甄選數學科筆試

一、 選填題：（每題 5 分，共 90 分。 填在答案卡上， 分數或根式須以最簡形式回答，否則不予計分）

解答:$$x=8^{\log_2 x}-9^{\log_3 x}-4^{\log_2 x}+\log_{0.5}0.25 =2^{\log_2 x^3}-3^{\log_3 x^2}-2^{\log_2 x^2}+2 \\\qquad =x^3-x^2-x^2+2 \Rightarrow x^3-2x^2-x+2=0 \Rightarrow (x^2-1)(x-2)=0\\ \Rightarrow x=1,2 (x=-1不合,因為x\gt 0) \Rightarrow 所有解之和=\bbox[red, 2pt]3$$

解答:$$(1-2x)^5(1+ 4x^2)^5 (1+2x)^5 =(1-4x^2)^5(1+4x^2)^5 = (1-16x^4)^5 = \sum_{k =0}^5 C^5_k(-16)^kx^{4k} \\ \Rightarrow \cases{k=3 \Rightarrow a=C^5_3(-16)^3 =-10\cdot 16^3\\ k=4 \Rightarrow b=C^5_4 (-16)^4 =5\cdot 16^4} \Rightarrow {b \over a}=-{5\times 16^4\over 10\times 16^3} =\bbox[red, 2pt]{-8}$$

解答:$$E(X)=(a_1+\cdots+a_{12})\div 12=13 \Rightarrow a_1+\cdots+a_{12}=13\times 12=156 \\ Var(X)=E(X^2)-(E(X))^2 \Rightarrow 5^2=E(X^2)-13^2 \Rightarrow E(X^2)=194\\ \Rightarrow a_1^2+\cdots+a_{12}^2 =194\cdot 12=2328\\ \sum_{i=1}^{12} a_ib_i= \sum_{i=1}^{12} a_i\left( {a_i-13\over 5} \right)= {1\over 5}\sum_{i=1}^{12}\left( a_i^2-13a_i\right)\\ ={1\over 5}(a_1^2+\cdots+ a_{12}^2)-{ 13\over 5}(a_1+\cdots +a_{12}) ={1\over 5}\cdot 2328-{13\over 5}\cdot 156 ={300\over 5}=\bbox[red, 2pt]{60}$$

解答:

$$x^2+y^2=10 \Rightarrow 圓半徑r=\sqrt{10}=\overline{OP}  \Rightarrow \overline{AO}= \sqrt{6^2+7^2} =\sqrt{85} \Rightarrow \overline{AP}= \sqrt{85-10} =5\sqrt 3 \\ 令\angle PAO=\theta \Rightarrow \cases{\sin \theta=\sqrt{10}/\sqrt{85}  \\ \cos \theta =5\sqrt 3/\sqrt{85}} \Rightarrow \sin 2\theta=2\sin \theta\cos \theta={2\sqrt{30}\over 17}\\ \triangle APB: {\overline{BP} \over \sin 2\theta} ={\overline{BA} \over \sin \angle APB} \Rightarrow {\overline{BA} \over \overline{BP}} ={\sin \angle APB\over \sin 2\theta} ={\sin \angle APB\over 2\sqrt{30}/17} =k\\k要最大 \Rightarrow \sin \angle APB=1, 此時k={17\over 2\sqrt{30}} =\bbox[red, 2pt]{17\sqrt{30} \over 60}$$

解答:$$此題相當環形著色問題:環形m=6塊,顏色n=6,代公式:(n-1)^m+(-1)^m(n-1)\\ =5^6+5=15630=30\times 521 \Rightarrow k=\bbox[red, 2pt]{521}$$

解答:

$$\cases{A(1,7)\\ B(7,-1)\\O(0,0)\\ P(x,y)} \Rightarrow \cases{\overrightarrow{PA}=(x-1,y-7)\\ \overrightarrow{PB}=(x-7,y+1)} \Rightarrow \overrightarrow{PA} \cdot \overrightarrow{PB} =(x-1)(x-7)+ (y-7)(y+1)=-19\\ \Rightarrow (x-4)^2+(y-3)^2=6 為一圓,圓心Q(4,3),圓半徑r=\sqrt 6 \\\Rightarrow \overline{OP}最大值= \overline{OQ}+r = \bbox[red, 2pt]{5+\sqrt 6}$$

解答:$$\left[(3a-2b+1)^2+ (2a+b-2)^2+ (4a-5b-3)^2 \right] \left[ (-2)^2+1^2+1^2 \right] \\ \qquad \qquad \ge \left[ -2(3a-2b+1) +(2a+b-2)+(4a-5b-3)\right]^2 \\ \Rightarrow \left[(3a-2b+1)^2+ (2a+b-2)^2+ (4a-5b-3)^2 \right]\cdot 6\ge (-7)^2 \\ \Rightarrow \left[(3a-2b+1)^2+ (2a+b-2)^2+ (4a-5b-3)^2 \right]\ge {49\over 6} \Rightarrow 最小值=\bbox[red, 2pt]{49\over 6}$$

解答:$$P\in L \Rightarrow P(t+1,-t+2,t) \Rightarrow \cases{\overline{PA}= \sqrt{(t+1)^2 +(-t-1)^2 (t-3)^2} \\ \overline{PB}=\sqrt{t^2+(-t-1)^2+ (t+2)^2}} \\ \Rightarrow \cases{\overline{PA}=\sqrt 3\cdot \sqrt{(t-1)^2+8/3} \\ \overline{PB}=\sqrt 3\cdot \sqrt{(t+1)^2+2/3}} \Rightarrow \overline{PA}+\overline{PB}= \sqrt 3(\overline{QC}+ \overline{QD}),其中\cases{Q(t,0)\\ C(1,2\sqrt 6/3)\\ D(-1,-\sqrt 6/3) } \\ \overline{QC}+\overline{QD}的最小值=\overline{CD}=\sqrt{10} \Rightarrow \overline{PA}+ \overline{PB}的最小值 =\sqrt 3\cdot \sqrt{10}=\sqrt{\bbox[red, 2pt]{30}}$$

解答:$$\cfrac{H^9_3\left({4!\over 2!2!} +{5!\over 2!3!}+{6!\over 2!4!}\right)}{H^4_4H^9_3} ={5115\over 5775} =\bbox[red,2pt] {31\over 35}\\ 先紅球再藍球的組合數:3紅5藍,最後一球是藍,前面有3紅,有4藍要插4空格有組合數:H^4_4\\ 依序紅藍綠的組合數:最後一球是綠,前面已有3紅5藍有9空格要插3綠球,組合數H^9_3\\ 因此要依序完成紅藍綠的組合數=H^4_4H^9_3=5775\\ 再考量最後一個紅球前已有2藍:即2紅2藍的組合數{4!\over 2!2!}剩下就是9個空格插3綠球H^9_3\\最後一個紅球前已有3藍:即2紅3藍的組合數{5!\over 2!3!}剩下就是9個空格插3綠球H^9_3\\最後一個紅球前已有4藍:即2紅4藍的組合數{6!\over 2!4!}剩下就是9個空格插3綠球H^9_3\\ 所以最後的機率就是\cfrac{H^9_3\left({4!\over 2!2!} +{5!\over 2!3!}+{6!\over 2!4!}\right)}{H^4_4H^9_3}$$

解答:$$假設六個兩面角為\alpha,\alpha,\beta,\beta,\gamma,\gamma,則\cos \alpha +\cos \beta+\cos \gamma=1 \Rightarrow \sum_{i=1}^6 \cos \theta_i =2 \\ \Rightarrow \sum_{i=1}^6 2\cos \theta_i =4 \Rightarrow \sum_{i=\color{blue}2}^6 \cos \theta_i =4-2\cos 120^\circ =\bbox[red, 2pt] 5\\ \href{https://web.math.sinica.edu.tw/mathmedia/HTMLarticle18.jsp?mID=40106}{公式來源}$$

解答:$$\lim_{x\to \infty} \left(\sqrt[5]{x^5+3x^4+4x^3+3x} -\sqrt[3]{x^3+3x^2+4x+1} \right)\\ = \lim_{x\to \infty} \left(x\left( \sqrt[5]{1+{3\over x}+{4\over x^2}+ {3\over x^4}}\right) -x \left( \sqrt[3]{1+{3\over x}+{4\over x^2}+{1\over x^3}} \right) \right) \\=\lim_{x\to \infty} \cfrac{ \sqrt[5]{1+{3\over x}+{4\over x^2}+ {3\over x^4}}-  \sqrt[3]{1+{3\over x}+{4\over x^2}+{1\over x^3}}}{1\over x} =\lim_{x\to \infty} \cfrac{ \left(\sqrt[5]{1+{3\over x}+{4\over x^2}+ {3\over x^4}}-  \sqrt[3]{1+{3\over x}+{4\over x^2}+{1\over x^3}} \right)'}{ ({1\over x})' } \\=\lim_{x\to \infty}  \left({1\over 5}(1+{3\over x}+{4\over x^2}+{3\over x^4})^{-4/5} (3+{8\over x}+{12\over x^3})\\ \qquad \qquad -{1\over 3}(1+{3\over x}+{4\over x^2}+ {1\over x^3})^{-2/3} (3+{8\over x}+{3\over x^2}) \right) \\={3\over 5}-1=\bbox[red, 2pt]{-2\over 5}$$

解答:$$(x-1)f(x)= 4\int_1^x f(t)\,dt \Rightarrow \frac{\text{d} }{\text{d}x}( (x-1)f(x)) = \frac{\text{d} }{\text{d}x}\left(  4\int_1^x f(t)\,dt\right) \\ \Rightarrow f(x)+(x-1)f'(x)=4f(x) \Rightarrow f'(x)-{3\over x-1}f(x)=0 \\ 一階微分方程, 取積分因子I(x)=e^{\int -{3\over x-1}\,dx} ={1\over (x-1)^3}\\ \Rightarrow I(x)f'(x)-{3\over x-1 }I(x) f(x)=0 \Rightarrow {1\over (x-1)^3} f'(x) -{3\over (x-1)^4}f(x)=0\\ \Rightarrow \left({1\over (x-1)^3} f(x) \right)'=0 \Rightarrow {1\over (x-1)^3} f(x)= C \Rightarrow f(x)=C(x-1)^3 \\\Rightarrow f(0)=-C=-2 \Rightarrow C=2 \Rightarrow f(x)=2(x-1)^3 \Rightarrow f(5)=2\cdot 4^3=\bbox[red, 2pt]{128}$$

解答:$$\omega=t(-\sqrt 3+4i)+(2-2t)i =-\sqrt 3 t+(2+2t)i \; 相當於直線L:2x+\sqrt 3y=2\sqrt 3\\ 　z^8=-{1\over 2}+{\sqrt 3\over 2}i = \cos({2\pi \over 3}+2k\pi) +i\sin ({2\pi \over 3}+2k\pi) \\ \Rightarrow z_i=\cos \left( {3k+1\over 12}\pi \right)+ i\sin \left( {3k+1\over 12}\pi \right), i=0,1,\dots,7\\ 此題相當於單位圓上8個點 P_i(\cos \left( {3k+1\over 12}\pi \right), \sin \left( {3k+1\over 12}\pi \right)),i=0-7,與L最近的距離\\考量L特性,只要計算第一象限的點就可以,即i＝0，1\\ \cases{ P_0(\cos(\pi/12), \sin(\pi/12))=((\sqrt 6+\sqrt 2)/4,(\sqrt 6-\sqrt 2)/4 \\ P_1(\cos(\pi/3), \sin(\pi /3)) =(1/2,\sqrt 3/2)} \\\left| (\sqrt 6+5\sqrt 2)/4-2\sqrt 3\right| \gt \left| 5/2-2\sqrt 3\right|\Rightarrow 最短矩離=d(P_1,L) =\cfrac{\left| 5/2-2\sqrt 3\right|}{ \sqrt 7} \\ =\cfrac{2\sqrt 3- 5/2 }{ \sqrt 7} =\bbox[red, 2pt]{{4\sqrt{21}-5\sqrt 7 \over 14}}$$

解答:$$13x^2-10xy+13y^2-6x-42y-27=[x,y]\begin{bmatrix}13 & -5 \\-5 & 13   \end{bmatrix} \begin{bmatrix}x \\y \end{bmatrix} +[-6,-42] \begin{bmatrix}x \\y \end{bmatrix}-27 \\ A=\begin{bmatrix}13 & -5 \\-5 & 13   \end{bmatrix}  =\begin{bmatrix}\sqrt 2/2 & -\sqrt 2/2 \\\sqrt 2/2 & \sqrt 2/2 \end{bmatrix} \begin{bmatrix}8 & 0 \\0 & 18 \end{bmatrix} \begin{bmatrix}\sqrt 2/2 & \sqrt 2/2 \\-\sqrt 2/2 & \sqrt 2/2 \end{bmatrix} \\ \Rightarrow [-6,-42] \begin{bmatrix}\sqrt 2/2 & -\sqrt 2/2 \\\sqrt 2/2 & \sqrt 2/2 \end{bmatrix} =[-24\sqrt 2,-18\sqrt 2] \\ \Rightarrow 8x'^2+18y'^2-24\sqrt 2x'-18\sqrt 2y'-27=0 \Rightarrow 8(x'-3/\sqrt 2)^2+ 18(y'-1/\sqrt 2)^2=72 \\ \Rightarrow {(x-3/\sqrt 2)^2 \over 9}+{(y'-1/\sqrt 2)^2 \over 4}=1 \Rightarrow \cases{a=3\\b=2} \Rightarrow 正焦弦長＝{2b^2 \over a}=\bbox[red, 2pt]{8\over 3}$$

解答:$$P(X=k)=0.8^{k-1}\times 0.2 \Rightarrow \cases{P(X=10)=0.8^{9}\cdot 0.2\gt 0.02 \\ P(X=11)=0.8^{10}\cdot 0.2\gt 0.02 \\ P(X=12)=0.8^{11}\cdot 0.2 \approx 0.017} \\ \Rightarrow k=\bbox[red, 2pt]{12}\\ \log(0.8^{10}\cdot 0.2)=10\log 0.8+\log 0.2=10(3\log 2-1)+\log 2-1 =31\log 2-11\\\qquad =31\times 0.301-11=-1.669\\ \log 0.02=\log 2-2=0.301-2=-1.699\\ 因此\log(0.8^{10}\cdot 0.2) \gt \log 0.02 \Rightarrow 0.8^{10}\cdot 0.2 \gt 0.02$$

解答:$$\cases{x(y+z-x)=39-2x^2\\ y(z+x-y)=52-2y^2\\ z(x+y-z)=78-2z^2} \Rightarrow \cases{xy+xz=39-x^2 \\ yz+xy=52-y^2\\ xz+yz=78-z^2} \\三式相加 \Rightarrow 2(xy+yz+zx)=169-(x^2+y^2+z^2) \\\Rightarrow x^2+y^2+z^2-2(xy+yz+zx)=(x+y+z)^2 =169 \Rightarrow x+y+z=13 \\ \Rightarrow \cases{y+z-x=13-2x\\ z+x-y=13-2y\\ x+y-z=13-2z} 代回原式 \Rightarrow \cases{13x-2x^2=39-2x^2\\ 13y-2y^2=52-2y^2\\ 13z-2z^2=78-2z^2} \Rightarrow \cases{x=3\\y =4\\z= 6} \\ \Rightarrow abc=3 \cdot 4\cdot 6 =\bbox[red, 2pt]{72}$$

解答:

$$\cases{A(0,a)\\ B(\sqrt 3a,0)\\ C(0,0)\\ P(\cos \theta, \sin \theta) } \Rightarrow \cases{\overline{PA}^2=\cos^2\theta+(\sin\theta-a)^2=2\\ \overline{PB}^2= (\cos\theta-\sqrt 3a)^2+ \sin^2\theta= 10} \Rightarrow \cases{-2a\sin \theta+a^2=1\\ -2\sqrt 3a\cos \theta+3a^2=9} \\ \Rightarrow \cases{\sin\theta =(a^2-1)/2a\\ \cos\theta =(3a^2-9)/2\sqrt 3a} \Rightarrow \sin^2 \theta+\cos^2 \theta= {(a^2-1)^2 \over 4a^2} +{(3a^2-9)^2 \over 12a^2} =1 \\ \Rightarrow 3(a^2-1)^2+(3a^2-9)^2=12a^2 \Rightarrow a^4-6a^2+7=0 \Rightarrow a^2=3+\sqrt 2\\ \Rightarrow \triangle ABC={1\over 2} \cdot \overline{CA}\cdot \overline{CB}={1\over 2}\sqrt 3a^2 = \bbox[red, 2pt]{{3\sqrt 3+\sqrt 6\over 2}}$$

解答:

$$|\vec a|=1 \Rightarrow 假設A(1,0); \\又|\vec a-\vec b|={1\over 2} \Rightarrow B在以A為圓心,半徑為1/2的圓上\\ |5\vec a-\vec c|=1 \Rightarrow C在以O_1(5,0)為圓心,半徑為1的圓上\\ \vec a和\vec d的夾角為{\pi\over 4}\Rightarrow D在直線L: x＝y上\\ 欲求\overline{BD}+\overline{CD}的最小值,因此將L為對稱軸,圓O_2為圓O_1的對稱圓,\\則\overline{AO_2}與圓A的交點即為B,與圓O_2的交點即為C的對稱點C'\\ 因此\overline{BD}+\overline{CD}的最小值=\overline{AO_2}-\overline{O_2C'}-\overline{AB}=\sqrt{26}-1-{1\over 2}= \bbox[red, 2pt]{-{3\over 2}+\sqrt{26}}$$

二、 證明題：（共 10 分。請用黑色或藍色原子筆寫在作答卷上，須詳細過程，否則酌予扣分）

解答:$$\left({1\over a^3(b+c)} +{1\over b^3(a+c)} +{1\over c^3(a+b)}  \right) (a(b+c)+ b(a+c)+c(a+b)) \ge ({1\over a}+{1\over b}+{1\over c})^2 \\ \Rightarrow \left({1\over a^3(b+c)} +{1\over b^3(a+c)} +{1\over c^3(a+b)}  \right) (2(ab+bc+ca) \ge ({ab+bc+ca \over abc} )^2 \\ \Rightarrow {1\over a^3(b+c)} +{1\over b^3(a+c)} +{1\over c^3(a+b)}\ge {ab+bc+ca\over 2} \ge {3\over 2} \left( \because {ab+bc+ca\over 3} \ge \sqrt[3]{abc}=1\right) \\ \Rightarrow {1\over a^3(b+c)} +{1\over b^3(a+c)} +{1\over c^3(a+b)}\ge {3\over 2}\;\bbox[red, 2pt]{QED}$$

============== END ===============

解題僅供參考,其他歷年試題及詳解