國立臺南第二高級中學 113 學年度第一次教師甄選
一、填充題(每題 5 分,共 60 分)
解答:G是重心⇒→AG=13→AB+13→AC,又{¯BC=42¯GI=2¯GI∥¯BC⇒→GI=121→BC因此→AI=→AG+→GI=13→AB+13→AC+121→BC=13→AB+13→AC+121(−→AB+→AC)=621→AC+821→AC⇒{x=6/21=2/7y=8/21⇒(x,y)=(27,821)解答:
假設{α=log2x≥0β=log2y≥0⇒α2+β2=log2(4x4)+log2(8x8)=2+4log2x+3+8log2x=5+12α⇒(α−6)2+β2=41為一橢圓,但僅限第一象限(如圖)⇒{α=√41cosθ+6β=√41sinθ⇒log2(xy)=log2x+log2y=α+β=√41(cosθ+sinθ)+6=√82sin(θ+π/4)+6⇒θ=π2時有極大值6+√82當α=0時,β=√5⇒α+β=√5為極小值⇒M+m=6+√82+√5
解答:|z|=1⇒z=cosθ+isinθ=eiθ⇒z3=e3iθ⇒¯z3=cos3θ−isin3θ⇒|−1+√3i−z|=|(−1+√3i)22−¯z3|⇒|(−1−cosθ)+(√3−sinθ)i|=|(−1−cos3θ)+(sin3θ−√3)i|⇒(−1−cosθ)2+(√3−sinθ)2=(−1−cos3θ)2+(sin3θ−√3)2⇒2cosθ−2√3sinθ+5=2cos3θ−2√3sin3θ+5⇒cos3θ−cosθ=√3(sin3θ−sinθ)⇒−2sin2θsinθ=2√3cos2θsinθ⇒sinθ(√3cos2θ+sin2θ)=0⇒{sinθ=0⇒θ=0⇒z=1tan2θ=−√3⇒θ=π/3⇒z=12+√32i⇒z=1,12+√32i
解答:{f(a,b)=√4a+13+√6b+15g(a,b)=a+b−2⇒{fa=λgafb=λgbg=0⇒{2√4a+13=λ3√6b+15=λ⇒2√4a+13=3√6b+15⇒44a+13=96b+15⇒36a−24b+115=0⇒36a−24(2−a)+115=0⇒a=229300⇒b=371300⇒f(229300,371300)=√22975+13+√37150+15=15(13+12)√762=√7626
解答:f(x)=∫x0g(t)dt+1⇒f(0)=1g(x)=12x2−6x+∫10[f(t)+g′(t)]dt⇒g(0)=∫10[f(t)+g′(t)]dt為一常數⇒f(x)=∫x0[12t2−6t+g(0)]dt+1=4x3−3x2+xg(0)+1⇒g(x)=12x2−6x+∫10f(t)dt+∫10g′(t)dt=12x2−6x+∫10(4t3−3t2+tg(0)+1)dt+g(1)−g(0)=12x2−6x+[t4−t3+12g(0)t2+t]|10+g(1)−g(0)=12x2−6x+1+g(1)−12g(0)⇒g(x)=12x2−6x+1+g(1)−12g(0)⇒g(1)=7+g(1)−12g(0)⇒7−12g(0)=0⇒g(0)=14
解答:lim
解答:丟一次骰子點數為偶數的機率=2/6=1/3,奇數機率為2/3\\ 假設a_n為擲骰子n次點數和為偶數的機率,則a_n=前n-1次點數和為偶數且第n次也為偶數或\\前n-1次點數和為奇數且第n次也為奇數.因此a_n={1\over 3}a_{n-1} +{2\over 3}(1-a_{n-1}),n\ge 2\\ \Rightarrow a_n={2\over 3}-{1\over 3}a_{n-1} \Rightarrow a_n-{1\over 2}=-{1\over 3}(a_{n-1}-{1\over 2})\\ 取b_n=a_n-{1\over 2} \Rightarrow b_n=-{1\over 3}b_{n-1},b_1=a_1-{1\over 2}={1\over 3}-{1\over 2}=-{1\over 6} \\ \Rightarrow b_n=b_1(-{1\over 3})^{n-1} =-{1\over 6}(-{1\over 3})^{n-1} \Rightarrow a_n={1\over 2}-{1\over 6}(-{1\over 3})^{n-1} ={1\over 2}(1+(-{1\over 3})\cdot (-{1\over 3})^{n-1} \\ \Rightarrow a_n=\bbox[red, 2pt]{{1\over 2}\left(1+(-{1\over 3})^n \right)}
解答:假設\cases{f(x)=x^{2024}-2x^{427}+ 3x^{113}-4x \\ p(x)= x^3+2x^2+2x+1} \\ p(x)= x^3+x^2+ x^2+2x+1 =x^2(x+1)+(x+1)^2=(x+1)(x^2+x+1)\\ 假設\omega為x^2+x+1=0的根,則\omega^3=1 \Rightarrow f(\omega) = (\omega^3)^{674}\cdot \omega^2-2(\omega^3)^{142} \cdot \omega+3(\omega^3)^{37}\cdot \omega^2-4\omega \\ \Rightarrow f(\omega)=4\omega^2-6\omega =4(-\omega-1)-6\omega=-10\omega-4 \\ \Rightarrow f(x)=p(x)Q(x)+ a(x^2+x+1)-10x-4 \\ \Rightarrow f(-1)=1+2-3+4=a(1-1+1)+10-4 \Rightarrow 4=a+6 \Rightarrow a=-2 \\\Rightarrow 餘式: -2(x^2+x+1)-10x-4= \bbox[red, 2pt]{-2x^2-12x-6}
解答:
解答:|z|=1⇒z=cosθ+isinθ=eiθ⇒z3=e3iθ⇒¯z3=cos3θ−isin3θ⇒|−1+√3i−z|=|(−1+√3i)22−¯z3|⇒|(−1−cosθ)+(√3−sinθ)i|=|(−1−cos3θ)+(sin3θ−√3)i|⇒(−1−cosθ)2+(√3−sinθ)2=(−1−cos3θ)2+(sin3θ−√3)2⇒2cosθ−2√3sinθ+5=2cos3θ−2√3sin3θ+5⇒cos3θ−cosθ=√3(sin3θ−sinθ)⇒−2sin2θsinθ=2√3cos2θsinθ⇒sinθ(√3cos2θ+sin2θ)=0⇒{sinθ=0⇒θ=0⇒z=1tan2θ=−√3⇒θ=π/3⇒z=12+√32i⇒z=1,12+√32i
解答:{f(a,b)=√4a+13+√6b+15g(a,b)=a+b−2⇒{fa=λgafb=λgbg=0⇒{2√4a+13=λ3√6b+15=λ⇒2√4a+13=3√6b+15⇒44a+13=96b+15⇒36a−24b+115=0⇒36a−24(2−a)+115=0⇒a=229300⇒b=371300⇒f(229300,371300)=√22975+13+√37150+15=15(13+12)√762=√7626
解答:f(x)=∫x0g(t)dt+1⇒f(0)=1g(x)=12x2−6x+∫10[f(t)+g′(t)]dt⇒g(0)=∫10[f(t)+g′(t)]dt為一常數⇒f(x)=∫x0[12t2−6t+g(0)]dt+1=4x3−3x2+xg(0)+1⇒g(x)=12x2−6x+∫10f(t)dt+∫10g′(t)dt=12x2−6x+∫10(4t3−3t2+tg(0)+1)dt+g(1)−g(0)=12x2−6x+[t4−t3+12g(0)t2+t]|10+g(1)−g(0)=12x2−6x+1+g(1)−12g(0)⇒g(x)=12x2−6x+1+g(1)−12g(0)⇒g(1)=7+g(1)−12g(0)⇒7−12g(0)=0⇒g(0)=14
解答:lim
解答:丟一次骰子點數為偶數的機率=2/6=1/3,奇數機率為2/3\\ 假設a_n為擲骰子n次點數和為偶數的機率,則a_n=前n-1次點數和為偶數且第n次也為偶數或\\前n-1次點數和為奇數且第n次也為奇數.因此a_n={1\over 3}a_{n-1} +{2\over 3}(1-a_{n-1}),n\ge 2\\ \Rightarrow a_n={2\over 3}-{1\over 3}a_{n-1} \Rightarrow a_n-{1\over 2}=-{1\over 3}(a_{n-1}-{1\over 2})\\ 取b_n=a_n-{1\over 2} \Rightarrow b_n=-{1\over 3}b_{n-1},b_1=a_1-{1\over 2}={1\over 3}-{1\over 2}=-{1\over 6} \\ \Rightarrow b_n=b_1(-{1\over 3})^{n-1} =-{1\over 6}(-{1\over 3})^{n-1} \Rightarrow a_n={1\over 2}-{1\over 6}(-{1\over 3})^{n-1} ={1\over 2}(1+(-{1\over 3})\cdot (-{1\over 3})^{n-1} \\ \Rightarrow a_n=\bbox[red, 2pt]{{1\over 2}\left(1+(-{1\over 3})^n \right)}
解答:假設\cases{f(x)=x^{2024}-2x^{427}+ 3x^{113}-4x \\ p(x)= x^3+2x^2+2x+1} \\ p(x)= x^3+x^2+ x^2+2x+1 =x^2(x+1)+(x+1)^2=(x+1)(x^2+x+1)\\ 假設\omega為x^2+x+1=0的根,則\omega^3=1 \Rightarrow f(\omega) = (\omega^3)^{674}\cdot \omega^2-2(\omega^3)^{142} \cdot \omega+3(\omega^3)^{37}\cdot \omega^2-4\omega \\ \Rightarrow f(\omega)=4\omega^2-6\omega =4(-\omega-1)-6\omega=-10\omega-4 \\ \Rightarrow f(x)=p(x)Q(x)+ a(x^2+x+1)-10x-4 \\ \Rightarrow f(-1)=1+2-3+4=a(1-1+1)+10-4 \Rightarrow 4=a+6 \Rightarrow a=-2 \\\Rightarrow 餘式: -2(x^2+x+1)-10x-4= \bbox[red, 2pt]{-2x^2-12x-6}
解答:
\cases{\sqrt{x^4-4x^2-12x+25} =\sqrt{(x^2-4)^2+(2x-3)^2} =\overline{PA}\\ \sqrt{x^4+2x^2+1} =\sqrt{(x^2-1)^2+(2x-0)^2} =\overline{PB}},其中\cases{P(2x,x^2)\\ A(3,4)\\ B(0,1)} \\ 而P\in \Gamma:y=({x\over 2})^2,\Gamma 為一拋物線,B剛好為焦點,準線L:y=-1 \\ 因此\overline{PB}=d(P,L) \Rightarrow 當\overleftrightarrow{PA}成一垂直線時, \overline{PA} +\overline{PB}=d(A,L)=\bbox[red, 2pt]5為最小值
解答:假設C為原點,則\cases{A(0,2\sqrt 3,2\sqrt 3)\\ B(0,4\sqrt 3,0)\\ C(0,0,0)\\ D(4,0,0)} \Rightarrow \cases{\overrightarrow{AD}=(4,-2\sqrt 3,-2\sqrt 3)\\ \overrightarrow{BC} =(0,-4\sqrt 3,0)} \\\Rightarrow \cases{\vec n= \overrightarrow{AD} \times \overrightarrow{BD}=(-24,0,-16\sqrt 3) \\ \overrightarrow{CD}=(4,0,0)} \Rightarrow d( \overleftrightarrow{AD},\overleftrightarrow{BC}) = \overrightarrow{CD}在\vec n上的投影\\ ={|\vec n\cdot \overrightarrow{CD}| \over |\vec n|} ={96\over 8\sqrt{21}}= \bbox[red, 2pt] {{4\over 7}\sqrt{21}}
解答:f(p)=(p+q)^{10}=\sum_{k=0}^{10}p^kq^{10-k}C^{10}_k \Rightarrow f'(p)=10(p+q)^9= \sum_{k=1}^{10}p^{k-1}q^{10-k}kC^{10}_k \\ \Rightarrow g(p)=pf'(p)=10p(p+q)^9 = \sum_{k= 1}^{10}p^{k}q^{10 -k}kC^{10}_k \\\Rightarrow g'(p)=10(10p+q) (p+q)^8 = \sum_{k= 1}^{10}p^{k-1}q^{10 -k}k^2C^{10}_k\\ \Rightarrow pg'(p)= 10p(10p+q)(p+q)^8 =\sum_{k= 1}^{10}p^{k}q^{10 -k}k^2C^{10}_k \\ \sum_{k=1}^{10}3^{k-1}\cdot 2^{9-k}\cdot k^2C^{10}_k = {1\over 6}\sum_{k=1}^{10}3^{k}\cdot 2^{10-k}\cdot k^2C^{10}_k = {1\over 6}\cdot 10\cdot 3\cdot 32\cdot 5^8 \\ = 10\cdot 2^4\cdot 5^8 =10\cdot (2\cdot 5)^4\cdot 5^4=\bbox[red, 2pt]{625 00000}
解答:{a+b\over a}={\sin B\over \sin B-\sin A}={b\over b-a} \Rightarrow b^2-a^2=ab \Rightarrow a^2+ab-b^2=0 \\ \Rightarrow a={-b+b\sqrt5\over 2} =b\left( {\sqrt 5-1\over 2}\right) \cdots(1) \\ \cos(A-B)+\cos C=1-\cos 2C \Rightarrow \cos(A-B)-\cos (A+B)=2\sin^2 C \\ \Rightarrow 2\sin A\sin B=2\sin^2 C\Rightarrow ab=c^2 \Rightarrow c=\sqrt{ab} =\sqrt{b^2\cdot {\sqrt 5-1 \over 2}} =b \sqrt{\sqrt 5-1\over 2}\cdots(2) \\\text{By (1) and (2), we have } {a+c\over b}= \bbox[red, 2pt]{ \left( {\sqrt 5-1\over 2}\right)+\sqrt{\sqrt 5-1\over 2}}
解答:\cases{(a-b)^2=a^2-2ab +b^2\ge 0\\-1\le \sin x\le 1} \Rightarrow a^2-2ab\sin x+b^2\ge 0, \forall x \\ 取x=\theta+30^\circ \Rightarrow a^2-2ab(\theta+30^\circ)+b^2\ge 0 \Rightarrow a^2-\sqrt 3ab\sin \theta -ab \cos \theta+b^2\ge 0 \\ \Rightarrow a^2-ab\cos \theta+b^2 \ge \sqrt 3ab\sin \theta=2\sqrt 3 \cdot {1\over 2}ab\sin \theta =2\sqrt 3\triangle \\ \Rightarrow 2a^2-2ab\cos \theta+2 b^2\ge 4\sqrt 3\triangle \Rightarrow a^2+b^2+(a^2-2ab\cos \theta+b^2)\ge 4\sqrt 3\triangle \\ \Rightarrow a^2+b^2+c^2 \ge 4\sqrt 3\triangle \Rightarrow {a^2+b^2 +c^2\over \triangle }\ge 4\sqrt 3. \;\bbox[red, 2pt]{QED.}
解答:
解答:f(p)=(p+q)^{10}=\sum_{k=0}^{10}p^kq^{10-k}C^{10}_k \Rightarrow f'(p)=10(p+q)^9= \sum_{k=1}^{10}p^{k-1}q^{10-k}kC^{10}_k \\ \Rightarrow g(p)=pf'(p)=10p(p+q)^9 = \sum_{k= 1}^{10}p^{k}q^{10 -k}kC^{10}_k \\\Rightarrow g'(p)=10(10p+q) (p+q)^8 = \sum_{k= 1}^{10}p^{k-1}q^{10 -k}k^2C^{10}_k\\ \Rightarrow pg'(p)= 10p(10p+q)(p+q)^8 =\sum_{k= 1}^{10}p^{k}q^{10 -k}k^2C^{10}_k \\ \sum_{k=1}^{10}3^{k-1}\cdot 2^{9-k}\cdot k^2C^{10}_k = {1\over 6}\sum_{k=1}^{10}3^{k}\cdot 2^{10-k}\cdot k^2C^{10}_k = {1\over 6}\cdot 10\cdot 3\cdot 32\cdot 5^8 \\ = 10\cdot 2^4\cdot 5^8 =10\cdot (2\cdot 5)^4\cdot 5^4=\bbox[red, 2pt]{625 00000}
解答:{a+b\over a}={\sin B\over \sin B-\sin A}={b\over b-a} \Rightarrow b^2-a^2=ab \Rightarrow a^2+ab-b^2=0 \\ \Rightarrow a={-b+b\sqrt5\over 2} =b\left( {\sqrt 5-1\over 2}\right) \cdots(1) \\ \cos(A-B)+\cos C=1-\cos 2C \Rightarrow \cos(A-B)-\cos (A+B)=2\sin^2 C \\ \Rightarrow 2\sin A\sin B=2\sin^2 C\Rightarrow ab=c^2 \Rightarrow c=\sqrt{ab} =\sqrt{b^2\cdot {\sqrt 5-1 \over 2}} =b \sqrt{\sqrt 5-1\over 2}\cdots(2) \\\text{By (1) and (2), we have } {a+c\over b}= \bbox[red, 2pt]{ \left( {\sqrt 5-1\over 2}\right)+\sqrt{\sqrt 5-1\over 2}}
二、計算證明題(每題10 分,共40 分)(需將演算過程寫在答案卷上,並註明題號)
解答:A=\begin{bmatrix} -1 & 4 & 2 \\ -1 & 3 & 1 \\-1 & 2 & 2 \end{bmatrix} = \begin{bmatrix}2 & 1 & 2 \\ 1 & 0 & 1 \\0 & 1 & 1 \end{bmatrix} \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\0 & 0 & 2 \end{bmatrix} \begin{bmatrix}1 & -1 & -1 \\ 1 & -2 & 0 \\-1 & 2 & 1 \end{bmatrix} \\ \Rightarrow A^n = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\0 & 0 & 2^n \end{bmatrix} \begin{bmatrix}1 & -1 & -1 \\ 1 & -2 & 0 \\-1 & 2 & 1 \end{bmatrix} = \bbox[red, 2pt]{ \begin{bmatrix}3- 2^{n+1} & 2^{n+2}-4 & 2^{n+1}-2 \\-2^n+1 & 2^{n+1}-1 & 2^n-1 \\1-2^n & 2^{n+1}-2 & 2^n \end{bmatrix}}解答:\cases{(a-b)^2=a^2-2ab +b^2\ge 0\\-1\le \sin x\le 1} \Rightarrow a^2-2ab\sin x+b^2\ge 0, \forall x \\ 取x=\theta+30^\circ \Rightarrow a^2-2ab(\theta+30^\circ)+b^2\ge 0 \Rightarrow a^2-\sqrt 3ab\sin \theta -ab \cos \theta+b^2\ge 0 \\ \Rightarrow a^2-ab\cos \theta+b^2 \ge \sqrt 3ab\sin \theta=2\sqrt 3 \cdot {1\over 2}ab\sin \theta =2\sqrt 3\triangle \\ \Rightarrow 2a^2-2ab\cos \theta+2 b^2\ge 4\sqrt 3\triangle \Rightarrow a^2+b^2+(a^2-2ab\cos \theta+b^2)\ge 4\sqrt 3\triangle \\ \Rightarrow a^2+b^2+c^2 \ge 4\sqrt 3\triangle \Rightarrow {a^2+b^2 +c^2\over \triangle }\ge 4\sqrt 3. \;\bbox[red, 2pt]{QED.}
解答:
P(x,y) \Rightarrow \overrightarrow{OP}=(x,y)=t^2\overrightarrow{OA}+ t\overrightarrow{OB} =(\sqrt 3t^2-t,t^2+\sqrt 3 t) \\ \Rightarrow \begin{bmatrix}x \\y \end{bmatrix}= \begin{bmatrix} \sqrt 3 & -1 \\1 & \sqrt 3 \end{bmatrix} \begin{bmatrix}t^2 \\t \end{bmatrix} = \begin{bmatrix} \sqrt 3/2 & -1/2 \\1/2 & \sqrt 3/2 \end{bmatrix} \begin{bmatrix}t^2/2 \\t/2 \end{bmatrix} =\begin{bmatrix}\cos 30^\circ & -\sin 30^\circ \\\sin 30^\circ & \cos 30^\circ \end{bmatrix} \begin{bmatrix}x' \\y' \end{bmatrix}\\ \qquad 其中\;\Gamma: y'^2=2x' 為一拋物線\\ 因此P點軌跡就是圖形\Gamma 逆時針旋轉30^\circ,而直線L: y=-{1\over \sqrt 3}x逆時針旋轉30^\circ 就是x軸\\ 欲求之圖形面積就上圖紅色區域,而藍色區域就是旋轉前的區域,兩者面積相等\\而藍色區域較易計算,藍色區域為兩圖形\cases{x=-\sqrt 3y\\ x=y^2/2}所夾區域,交點為\cases{O(0,0)\\ A(6,-2\sqrt 3)} \\ \Rightarrow 面積=\int_{-2\sqrt 3}^y \left(-\sqrt 3y-{1\over 2}y^2 \right)\,dy =\bbox[red, 2pt]{2\sqrt 3}
解答:$$這是有名的歐拉定理,證明過程可參考維基百科$$
解答:$$這是有名的歐拉定理,證明過程可參考維基百科$$
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解題僅供參考,其他歷年試題及詳解
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