高雄市 112 學年度市立高級中等學校教師甄選
一、 計算證明題 (1 至 12 題每題 6 分, 13 至 16 題每題 7 分, 共 100 分)
解答:{A(5,−1,2)B(−5,−1,−6)⇒¯AB中點C=(0,−1,−2),欲求過C且垂直平面E的直線LL的方向向量→u=E的法向量=(1,2,3)⇒L:x1=y+12=z+23欲求之P=L∩E;若P∈L,則P(t,2t−1,3t−2),t∈R;又P∈E⇒t+2(2t−1)+3(3t−2)−6=0⇒t=1⇒P(1,1,1)解答:α,β,γ為x3−x−3=0之三根⇒{α+β+γ=0αβ+βγ+γα=−1αβγ=3α−1α+1+β−1β+1+γ−1γ+1=1−2α+1+1−2β+1+1−2γ+1=3−2(1α+1+1β+1+1γ+1)=3−2((α+1)(β+1)+(β+1)(γ+1)+(γ+1)(α+1)(α+1)(β+1)(γ+1))=3−2⋅αβ+βγ+γα+2(α+β+γ)+3αβγ+αβ+βγ+γα+1=3−2⋅−1+0+33−1+1=3−43=53
解答:
{|x−1|≤2|x+y|≤1所圍區域頂點坐標{A(3,−4)B(3,−2)C(−1,2)D(−1,0)令f(x,y)=x−2y⇒{f(A)=11f(B)=7f(C)=−5f(D)=−1⇒最大值為11
解答:(√10+√a−√10−√a)2=20−2√100−a⇒√10+√a−√10−√a=√2⋅√10−√100−a令L=√10+√1+√10+√2+⋯+√10+√99√10−√1+√10−√2+⋯+√10−√99⇒L−1=(√10+√1−√10−√1)+(√10+√2−√10−√2)+⋯+(√10+√99−√10−√99)√10−√1+√10−√2+⋯+√10−√99=√2(√10−√100−1)+√2(√10−√100−2)+⋯+√2(√10−√100−99)√10−√1+√10−√2+⋯+√10−√99=√2(√10−√99+√10−√98+⋯+√10−√1)√10−√1+√10−√2+⋯+√10−√99=√2⇒L=√2+1
解答:等比數列:a1=a,a2=ar,a3=ar2⇒r=(ar+2a)rar+2a=a3+2a2a2+2a1⇒r=−4a+4√2+log82023+2(2a−2√2+log42023)2a−2√2+log42023+2(−a+√2+log22023)=log82023+2log42023log42023+2log22023=13log22023+log2202312log22023+2log22023=4352=815
解答:sin(37∘)=sin(30∘+7∘)=sin30∘cos7∘+sin7∘cos30∘=12cos7∘+√32sin7∘原式:sin237∘+cos27∘−sin37∘×cos7∘=(12cos7∘+√32sin7∘)2+cos27∘−(12cos7∘+√32sin7∘)cos7∘=14cos27∘+√32sin7∘cos7∘+34sin27∘+cos27∘−12cos27∘−√32sin7∘cos7∘=34cos27∘+34sin27∘=34
解答:取g(x)=f(x)−x,則1,2,3為g(x)=0的三根⇒g(x)=(x−1)(x−2)(x−3)(x−k)⇒f(x)=g(x)+x=(x−1)(x−2)(x−3)(x−k)+x由於f(x)=0的四根之積=d=1⋅2⋅3⋅k=6k⇒k=d6⇒f(x)=(x−1)(x−2)(x−3)(x−d6)+x因此{f(0)=df(4)=3⋅2⋅1⋅(4−d6)+4=28−d⇒14(f(0)+f(4))=14⋅28=7
解答:原式=limn→∞n∑k=114n2√4n2−k2=limn→∞n∑k=114n√4−(kn)2=∫1014√4−x2dx令x=2sinθ⇒dx=2cosθdθ,則原式=∫π/6014√4cos2θ⋅2cosθdθ=∫π/60cos2θdθ=∫π/6012(cos2θ+1)dθ=[14sin2θ+12θ]|π/60=√38+π12
解答:
解答:{|z1|=2|z2|=3|z1−z2|=√5⇒{¯OA=2¯OB=3¯AB=√5⇒¯OB2=¯OA2+¯AB2⇒∠OAB=90∘⇒cos∠AOB=cosθ=22+32−52⋅2⋅3=23令{A′(iz1)A″
解答:
解答:等比數列:a1=a,a2=ar,a3=ar2⇒r=(ar+2a)rar+2a=a3+2a2a2+2a1⇒r=−4a+4√2+log82023+2(2a−2√2+log42023)2a−2√2+log42023+2(−a+√2+log22023)=log82023+2log42023log42023+2log22023=13log22023+log2202312log22023+2log22023=4352=815
解答:sin(37∘)=sin(30∘+7∘)=sin30∘cos7∘+sin7∘cos30∘=12cos7∘+√32sin7∘原式:sin237∘+cos27∘−sin37∘×cos7∘=(12cos7∘+√32sin7∘)2+cos27∘−(12cos7∘+√32sin7∘)cos7∘=14cos27∘+√32sin7∘cos7∘+34sin27∘+cos27∘−12cos27∘−√32sin7∘cos7∘=34cos27∘+34sin27∘=34
解答:取g(x)=f(x)−x,則1,2,3為g(x)=0的三根⇒g(x)=(x−1)(x−2)(x−3)(x−k)⇒f(x)=g(x)+x=(x−1)(x−2)(x−3)(x−k)+x由於f(x)=0的四根之積=d=1⋅2⋅3⋅k=6k⇒k=d6⇒f(x)=(x−1)(x−2)(x−3)(x−d6)+x因此{f(0)=df(4)=3⋅2⋅1⋅(4−d6)+4=28−d⇒14(f(0)+f(4))=14⋅28=7
解答:原式=limn→∞n∑k=114n2√4n2−k2=limn→∞n∑k=114n√4−(kn)2=∫1014√4−x2dx令x=2sinθ⇒dx=2cosθdθ,則原式=∫π/6014√4cos2θ⋅2cosθdθ=∫π/60cos2θdθ=∫π/6012(cos2θ+1)dθ=[14sin2θ+12θ]|π/60=√38+π12
解答:
解答:{|z1|=2|z2|=3|z1−z2|=√5⇒{¯OA=2¯OB=3¯AB=√5⇒¯OB2=¯OA2+¯AB2⇒∠OAB=90∘⇒cos∠AOB=cosθ=22+32−52⋅2⋅3=23令{A′(iz1)A″
解答:
先求過A(1,8)的圓切線,假設圓切點P(a,\sqrt{16-a^2}) \Rightarrow \overrightarrow{PA} \cdot \overrightarrow{OP}=0 \\ \Rightarrow (1-a,8-\sqrt{16-a^2}) \cdot(a, \sqrt{16-a^2})=0 \Rightarrow 65a^2-32a-768=0\\ \Rightarrow (13a-48)(5a+16)=0 \Rightarrow a={48\over 13},-{16\over 5} \Rightarrow 兩切點\cases{P(-16/5, 12/5)\\ Q(48/13,20/13)}\\ 又x^2+y^2=16 \Rightarrow 2x+2yy'=0 \Rightarrow y'=-{x\over y} \Rightarrow 兩切線斜率\cases{m_1=4/3\\ m_2=-12/5} \\ \Rightarrow 兩切線\cases{L_1= \overleftrightarrow{AP} : 4x-3y+20=0 \\ L_2 =\overleftrightarrow{AQ}: 12x+5y=52} \Rightarrow \cases{B=L_1\cap (y=-4) =(-8,-4)\\ C=L_2 \cap (y=-4)= (6,-4)}\\ \cases{過A垂直x軸的垂線L_3:x=1\\ 與L_2垂直的垂線L_4斜率為5/12 \Rightarrow L_4: 5x-12y=8} \Rightarrow H=L_3\cap L_4 =\bbox[red,2pt]{(1,-{1\over 4})}
解答:A^2+4AB=0 \Rightarrow A(A+4B)=0 \Rightarrow A=-4B \Rightarrow A^{-1}=-{1\over 4}B^{-1} \\ \Rightarrow A^{-1}+B^{-1}={3\over 4}B^{-1}= \begin{bmatrix} 3 & 6\\ 9 & 12\end{bmatrix} \Rightarrow B^{-1}=\begin{bmatrix} 4 & 8\\12 & 16\end{bmatrix} \Rightarrow B=\begin{bmatrix} -1/2 & 1/4\\ 3/8 & -1/8\end{bmatrix} =\begin{bmatrix} a & b\\ c & d\end{bmatrix} \\ \Rightarrow a=\bbox[red,2pt]{-{1\over 2}}
解答:令\overrightarrow{PD} =x\overrightarrow{PA}+ y\overline{PB} , 由於x+y=1 \Rightarrow D\in \overline{BC}\\ 因此\overrightarrow{PD} +z\overrightarrow{PC}=0, 若\cases{z=0 \Rightarrow P=D\\ z=1 \Rightarrow P為\overline{CD}中點\\ z=2 \Rightarrow \overline{PC}: \overline{CD}=1:3\\ z=3 \Rightarrow \overline{PC}: \overline{CD} = 1:4} \\ \Rightarrow {S\over \triangle ABC}={1-1/16}=\bbox[red,2pt]{15\over 16}
解答:5A,4B,3C串連數至少是3,因此只要考慮增加一個串連的情形:\\ A拆成2個串連:\cases{A+4A:4B與3C插入兩個A串連之間,共有12個\\ 2A+3A:4B與3C插入兩個A串連之間,共有12個}\Rightarrow 合計24種\\ B拆成2個串連:\cases{B+3B:5A與3C插入兩個B串連之間,共有12個\\ 2B+2B: 5A與3C插入兩個B串連之間,共有6個},合計18個\\ C拆成1C+2C:5A與4B插入兩個B串連之間,共有12個\\ 以上共有24+18+12=\bbox[red,2pt]{54}種
解答:由於a^{527}+b^{527} =(a+b)(a^{526}-a^{525}b+ a^{524} b^2-\cdots-ab^{526}+ b^{527}) \Rightarrow a+b可以整除a^{527}+b^{527}\\ 而\sum_{k=1}^{2022}k^{527} =(1^{527}+2022^{527})+(2^{527}+2021^{527}) +\cdots+ (1011^{527}+ 1012^{527})\\ 上式每個括號內都是1+2022=2023的倍數,因此2023可以整除\sum_{k=1}^{2022}k^{527},\bbox[red,2pt]{故得證}
解答:\cases{ax+by+cz=0\\ bx+cy+az=0\\ cx+ay+bz=0}至少有一解(0,0,0)\\ 依題意有無限多解 \Rightarrow {a\over b}={b\over c}={c\over a}=k \Rightarrow \cases{a=bk\\ b=ck\\ c=ak},k\ne 0 \\ \Rightarrow a=bk=ck^2=ak^3 \Rightarrow k=1 \Rightarrow a=b=c, \bbox[red, 2pt]{故得證}
解答:\cases{{4^n+3^n\over 8^n+3^n} \le {4^n+4^n\over 8^n }={2\cdot 4^n\over 8^n} =2({1\over 2})^n \\ {4^n+3^n\over 8^n+3^n} \ge{4^n \over 8^n+8^n} ={1 \over 2}\cdot {4^n\over 8^n}} \Rightarrow {1 \over 2}\cdot {4^n\over 8^n}\le {4^n+3^n\over 8^n+3^n} \le 2({1\over 2})^n \\ \Rightarrow \left( {1\over 2} \right)^{1/n}\cdot {1\over 2} \le \left({4^n+3^n\over 8^n+3^n} \right)^{1/n} \le 2^{1/n}\cdot {1\over 2} \\ \Rightarrow \lim_{n\to \infty}\left( {1\over 2} \right)^{1/n}\cdot {1\over 2} \le \lim_{n\to \infty} \left({4^n+ 3^n\over 8^n+3^n} \right)^{1/n} \le \lim_{n\to \infty}2^{1/n}\cdot {1\over 2} \\ \Rightarrow {1\over 2}\le \lim_{n\to \infty} \left({4^n+ 3^n\over 8^n+3^n} \right)^{1/n} \le {1\over 2} \Rightarrow \lim_{n\to \infty} \left({4^n+ 3^n\over 8^n+3^n} \right)^{1/n} = \bbox[red, 2pt]{1\over 2}
解答:A^2+4AB=0 \Rightarrow A(A+4B)=0 \Rightarrow A=-4B \Rightarrow A^{-1}=-{1\over 4}B^{-1} \\ \Rightarrow A^{-1}+B^{-1}={3\over 4}B^{-1}= \begin{bmatrix} 3 & 6\\ 9 & 12\end{bmatrix} \Rightarrow B^{-1}=\begin{bmatrix} 4 & 8\\12 & 16\end{bmatrix} \Rightarrow B=\begin{bmatrix} -1/2 & 1/4\\ 3/8 & -1/8\end{bmatrix} =\begin{bmatrix} a & b\\ c & d\end{bmatrix} \\ \Rightarrow a=\bbox[red,2pt]{-{1\over 2}}
解答:令\overrightarrow{PD} =x\overrightarrow{PA}+ y\overline{PB} , 由於x+y=1 \Rightarrow D\in \overline{BC}\\ 因此\overrightarrow{PD} +z\overrightarrow{PC}=0, 若\cases{z=0 \Rightarrow P=D\\ z=1 \Rightarrow P為\overline{CD}中點\\ z=2 \Rightarrow \overline{PC}: \overline{CD}=1:3\\ z=3 \Rightarrow \overline{PC}: \overline{CD} = 1:4} \\ \Rightarrow {S\over \triangle ABC}={1-1/16}=\bbox[red,2pt]{15\over 16}
解答:5A,4B,3C串連數至少是3,因此只要考慮增加一個串連的情形:\\ A拆成2個串連:\cases{A+4A:4B與3C插入兩個A串連之間,共有12個\\ 2A+3A:4B與3C插入兩個A串連之間,共有12個}\Rightarrow 合計24種\\ B拆成2個串連:\cases{B+3B:5A與3C插入兩個B串連之間,共有12個\\ 2B+2B: 5A與3C插入兩個B串連之間,共有6個},合計18個\\ C拆成1C+2C:5A與4B插入兩個B串連之間,共有12個\\ 以上共有24+18+12=\bbox[red,2pt]{54}種
解答:由於a^{527}+b^{527} =(a+b)(a^{526}-a^{525}b+ a^{524} b^2-\cdots-ab^{526}+ b^{527}) \Rightarrow a+b可以整除a^{527}+b^{527}\\ 而\sum_{k=1}^{2022}k^{527} =(1^{527}+2022^{527})+(2^{527}+2021^{527}) +\cdots+ (1011^{527}+ 1012^{527})\\ 上式每個括號內都是1+2022=2023的倍數,因此2023可以整除\sum_{k=1}^{2022}k^{527},\bbox[red,2pt]{故得證}
解答:\cases{ax+by+cz=0\\ bx+cy+az=0\\ cx+ay+bz=0}至少有一解(0,0,0)\\ 依題意有無限多解 \Rightarrow {a\over b}={b\over c}={c\over a}=k \Rightarrow \cases{a=bk\\ b=ck\\ c=ak},k\ne 0 \\ \Rightarrow a=bk=ck^2=ak^3 \Rightarrow k=1 \Rightarrow a=b=c, \bbox[red, 2pt]{故得證}
解答:\cases{{4^n+3^n\over 8^n+3^n} \le {4^n+4^n\over 8^n }={2\cdot 4^n\over 8^n} =2({1\over 2})^n \\ {4^n+3^n\over 8^n+3^n} \ge{4^n \over 8^n+8^n} ={1 \over 2}\cdot {4^n\over 8^n}} \Rightarrow {1 \over 2}\cdot {4^n\over 8^n}\le {4^n+3^n\over 8^n+3^n} \le 2({1\over 2})^n \\ \Rightarrow \left( {1\over 2} \right)^{1/n}\cdot {1\over 2} \le \left({4^n+3^n\over 8^n+3^n} \right)^{1/n} \le 2^{1/n}\cdot {1\over 2} \\ \Rightarrow \lim_{n\to \infty}\left( {1\over 2} \right)^{1/n}\cdot {1\over 2} \le \lim_{n\to \infty} \left({4^n+ 3^n\over 8^n+3^n} \right)^{1/n} \le \lim_{n\to \infty}2^{1/n}\cdot {1\over 2} \\ \Rightarrow {1\over 2}\le \lim_{n\to \infty} \left({4^n+ 3^n\over 8^n+3^n} \right)^{1/n} \le {1\over 2} \Rightarrow \lim_{n\to \infty} \left({4^n+ 3^n\over 8^n+3^n} \right)^{1/n} = \bbox[red, 2pt]{1\over 2}
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