臺中第一高級中等學校111學年度第1次教師甄選
一、填充題(甲)(每題 4 分,共 24 分。 )
解答:當[x]=1或2或3時,[x]⋅{x}≱3;考慮[x]=4時,[x]⋅{x}=4⋅{x}=3⇒{x}=34符合條件;因此x=434=194為符合條件之最小正數
解答:
作{¯MN∥¯AB¯ST∥¯BC¯UV∥¯AC,見上圖;則△PUM、△PSV、及△PTN皆為正△,且1,2,3分別為正△的高;因此{¯PU=¯PM=¯UM=2/√3¯PV=¯PS=¯VS=4/√3¯PN=¯PT=¯NT=6/√3⇒{¯PA2=22+(6/√3+2/√3)2=76/3¯PB2=12+(4/√3+1/√3)2=28/3¯PC2=32+(2/√3+3/√3)2=52/3⇒¯PA2:¯PB2:¯PC2=19:7:13

解答:an+3SnSn−1=Sn−Sn−1+3SnSn−1=0⇒1Sn−1−1Sn+3=0⇒bn=bn−1+3,bn=1Sn⇒b2022=b2021+3=b2020+2⋅3=⋯=b1+2021⋅3=4+2021⋅3=6067
解答:依題意{A(0,16)B(5,0)C(0,0)D(x,y),並取E(0,1),則滿足14¯AD=¯DE,即116(x2+(y−16)2)=x2+(y−1)2⇒x2+y2=16,也就是D的軌跡剛好就是圓Γ;因此14¯AD+¯BD=¯DE+¯BD≥¯BE=√26⇒最小值為√26
解答:{f(x+3)≤f(x)+3f(x)+2≤f(x+2)⇒f(x+3)−3≤f(x)≤f(x+2)−2⇒f(x+3)≤f(x+2)+1⇒f(2)≤f(1)+1=4⇒f(3)≤f(2)+1=5⋯(1)又f(x)≤f(x+2)−2⇒f(1)≤f(3)−2⇒5≤f(3)⋯(2)由(1)及(2)可得f(3)=5;同理{f(2)≤f(1)+1f(3)≤f(2)+1⇒4≤f(2)≤4⇒f(2)=4因此f(n)=n+2⇒f(2022)=2024
解答:
解答:依題意{A(0,16)B(5,0)C(0,0)D(x,y),並取E(0,1),則滿足14¯AD=¯DE,即116(x2+(y−16)2)=x2+(y−1)2⇒x2+y2=16,也就是D的軌跡剛好就是圓Γ;因此14¯AD+¯BD=¯DE+¯BD≥¯BE=√26⇒最小值為√26
解答:{f(x+3)≤f(x)+3f(x)+2≤f(x+2)⇒f(x+3)−3≤f(x)≤f(x+2)−2⇒f(x+3)≤f(x+2)+1⇒f(2)≤f(1)+1=4⇒f(3)≤f(2)+1=5⋯(1)又f(x)≤f(x+2)−2⇒f(1)≤f(3)−2⇒5≤f(3)⋯(2)由(1)及(2)可得f(3)=5;同理{f(2)≤f(1)+1f(3)≤f(2)+1⇒4≤f(2)≤4⇒f(2)=4因此f(n)=n+2⇒f(2022)=2024
解答:令{u=lnxdv=x2dx⇒{du=1xdxv=13x3⇒∫x2lnxdx=13x3lnx−13∫x2dx=13x3lnx−19x3+C,其中C為常數
二、填充題(乙)(每題 6 分,共 54 分。 )
解答:在複數平面上,Ai為z111=1的根,i=1−111⇒z111−1=(z−A1)(z−A2)⋯(z−A111)⇒|z111−1|=|cos111θ+isin111θ−1|=¯PA1ׯPA2×⋯ׯPA111而|cos111θ+isin111θ−1|≤2,因此最大值為2,此時θ=π111解答:
不失一般性,假設{D(0,0)B(−5,0)C(5,0)∠ACB=θ⇒¯BN=45ׯBC=8⇒¯CN=6⇒{sinθ=4/5cosθ=3/5⇒N=(¯CD−¯CNcosθ,¯CNsinθ)=(7/5,24/5)又△ABC∼△BNC(AAA)⇒¯AB¯BC=¯BN¯CN⇒¯AB=10×8÷6=40/3⇒A(−5,403)⇒{L1=↔AD:8x+3y=0L2=↔BN:4y=3x+15⇒M=L1∩L2=(−4541,12041)⇒¯BM=20041⇒¯MN=8−¯BM=12841⇒¯BM¯MN=200128=2516
解答:k⋅2k(k+1)(k+2)=ak+1+bk+2⇒a(k+2)+b(k+1)=(a+b)k+2a+b=k⋅2k⇒{a+b=2k2a+b=0⇒{a=−2kb=2⋅2k=2k+1⇒k⋅2k(k+1)(k+2)=2k+1k+2−2kk+1⇒An=n∑k=1(2k+1k+2−2kk+1)=(223−1)+(234−223)+⋯+(2n+1n+2−2nn+1)=2n+1n+2−1;又Bn=n∑k=12k=2n+1−2,因此|(n+2)An−Bn|=|2n+1−(n+2)−2n+1+2|=|−n|>2022⇒n=2023
解答:(an+1)2+(an)2+1=2(an+1⋅an+an+1+an)⇒(an+1)2+(an)2+1−2an+1⋅an+2an−2an+1=4an⇒(an+1−an−1)2=4an⇒an+1−an−1=2√an⇒an+1=an+2√an+1=(√an+1)2⇒√an+1=√an+1⇒√an=√an−1+1=√an−2+2=√a1+(n−1)=n⇒an=n2⇒Sn=n∑k=1ak=n∑k=1k2=16n(n+1)(2n+1)⇒lim
解答:由\cases{x\gt 0\\ x,y,z\in \mathbb{R}}可假設\cases{x=\tan A\\ y=\tan B \\ z=\tan C},其中\cases{ 0\lt A\lt \pi/2 \\ -\pi/2 \lt B,C\lt \pi/2} \\ 因此\cases{5\left(x+{1\over x}\right) = 12\left(y+{1\over y}\right) =13\left(z+{1\over z}\right) \\ xy +yz +zx=1} \Rightarrow \cases{5\cdot {2\over \sin 2A} =12\cdot {2\over \sin 2B} =13\cdot {2\over \sin 2C} \\ \tan A\tan B+ \tan B\tan C+ \tan C\tan A=1} \\ \Rightarrow \cases{{5\over \sin 2A} ={12\over \sin 2B} ={13\over \sin 2C} \\ A+B+C = 90^\circ(或\pm 180^\circ)} \Rightarrow \cases{\tan 2A= 5/12\\ \tan 2B=12/5\\ \tan 2C=\infty} \Rightarrow \cases{\tan A=1/5 \\ \tan B= 2/3\\\tan C=1} \\ \Rightarrow (x,y,z)= \bbox[red, 2pt]{\left({1\over 5},{2\over 3},1\right)}
解答:\left((4\sqrt 6)^2-(2\sqrt 6)^2\right)\cdot {\sqrt 3\over 4}\cdot 4= \bbox[red, 2pt]{72\sqrt 3}\\ \href{https://m.tgjsw.com/k12/exam/detail/2747654.html}{這裡有圖及詳解}
解答:令\cases{a+ 2b+c = x\\ a+b+2c = y\\ a+b+3c= z} \Rightarrow \cases{a= -x+5y-3z\\ b=x-2y+z \\ c=-y+z} \Rightarrow \cases{a+3c=-x+2y \\ 4b= 4x-8y+4z\\ 8c = -8y+8z} \\ \Rightarrow {a+3c \over a+2b+c} +{4b\over a+b+2c}-{8c\over a+b+3c}+17 ={-x+2y\over x} +{4x-8y+4z\over y}-{-8y+8z\over z}+17 \\=-1+{2y\over x} -8+{4(x+z)\over y}-8+{8y\over z}+17 = \left( {2y\over x}+{4x\over y} \right)+ \left({4z\over y}+{8y\over z} \right) \\\ge 2\sqrt{{2y\over x}\cdot {4x\over y}} +2\sqrt{{4z\over y} \cdot {8y\over z}} =4\sqrt 2+8\sqrt 2=12\sqrt 2 \Rightarrow 最小值為\bbox[red, 2pt]{12\sqrt 2}
解答:
解答:(an+1)2+(an)2+1=2(an+1⋅an+an+1+an)⇒(an+1)2+(an)2+1−2an+1⋅an+2an−2an+1=4an⇒(an+1−an−1)2=4an⇒an+1−an−1=2√an⇒an+1=an+2√an+1=(√an+1)2⇒√an+1=√an+1⇒√an=√an−1+1=√an−2+2=√a1+(n−1)=n⇒an=n2⇒Sn=n∑k=1ak=n∑k=1k2=16n(n+1)(2n+1)⇒lim
解答:由\cases{x\gt 0\\ x,y,z\in \mathbb{R}}可假設\cases{x=\tan A\\ y=\tan B \\ z=\tan C},其中\cases{ 0\lt A\lt \pi/2 \\ -\pi/2 \lt B,C\lt \pi/2} \\ 因此\cases{5\left(x+{1\over x}\right) = 12\left(y+{1\over y}\right) =13\left(z+{1\over z}\right) \\ xy +yz +zx=1} \Rightarrow \cases{5\cdot {2\over \sin 2A} =12\cdot {2\over \sin 2B} =13\cdot {2\over \sin 2C} \\ \tan A\tan B+ \tan B\tan C+ \tan C\tan A=1} \\ \Rightarrow \cases{{5\over \sin 2A} ={12\over \sin 2B} ={13\over \sin 2C} \\ A+B+C = 90^\circ(或\pm 180^\circ)} \Rightarrow \cases{\tan 2A= 5/12\\ \tan 2B=12/5\\ \tan 2C=\infty} \Rightarrow \cases{\tan A=1/5 \\ \tan B= 2/3\\\tan C=1} \\ \Rightarrow (x,y,z)= \bbox[red, 2pt]{\left({1\over 5},{2\over 3},1\right)}
解答:\left((4\sqrt 6)^2-(2\sqrt 6)^2\right)\cdot {\sqrt 3\over 4}\cdot 4= \bbox[red, 2pt]{72\sqrt 3}\\ \href{https://m.tgjsw.com/k12/exam/detail/2747654.html}{這裡有圖及詳解}
解答:令\cases{a+ 2b+c = x\\ a+b+2c = y\\ a+b+3c= z} \Rightarrow \cases{a= -x+5y-3z\\ b=x-2y+z \\ c=-y+z} \Rightarrow \cases{a+3c=-x+2y \\ 4b= 4x-8y+4z\\ 8c = -8y+8z} \\ \Rightarrow {a+3c \over a+2b+c} +{4b\over a+b+2c}-{8c\over a+b+3c}+17 ={-x+2y\over x} +{4x-8y+4z\over y}-{-8y+8z\over z}+17 \\=-1+{2y\over x} -8+{4(x+z)\over y}-8+{8y\over z}+17 = \left( {2y\over x}+{4x\over y} \right)+ \left({4z\over y}+{8y\over z} \right) \\\ge 2\sqrt{{2y\over x}\cdot {4x\over y}} +2\sqrt{{4z\over y} \cdot {8y\over z}} =4\sqrt 2+8\sqrt 2=12\sqrt 2 \Rightarrow 最小值為\bbox[red, 2pt]{12\sqrt 2}
解答:
\cases{直角\triangle OAB =\sqrt 3/2\\ 扇形OBC = \pi/6} \Rightarrow 4\times{\sqrt 3\over 2}+2 \times {\pi \over 6} = \bbox[red, 2pt]{2\sqrt 3+{\pi \over 3}}
解答:看英文比較容易懂,\href{https://math.stackexchange.com/questions/2052209/least-positive-integer-n-such-that-any-set-of-n-pairwise-relatively-prime}{說明在此},該題不含1,所以n=15,本題含1,故n=\bbox[red,2pt]{16}\\第i個質數a_i及其平方數:\\ \begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c}i&1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15\\ \hline a_i &2 & 3 & 5 & 7 & 11 & 13 & 17 & 19 & 23 & 29 & 31 & 37 & 41 & 43 & 47\\ \hline a_i^2 &4 & 9 & 25 & 49 & 121 & 169 & 289 & 361 & 529 & 841 & 961 & 1369 & 1681 & 1849 & 2209\end{array} \\ 從上表中,我們可以找到15個數:1,a_1^2, a_2^2,\dots, a_{14}^2,滿足兩兩互質且小於等於2022;\\若再加上a_{15}^2,就有16個數滿足兩兩互質,但a_{15}^2 \gt 2022,因此符合題意的n值可能是16;\\現在要來證明n=16,也就是任取16個兩兩互質的數,其中至少有一個質數。\\若此16個數依序為1,x_1,x_2,\dots , x_{15}\le 2022,滿足兩兩互質且都不是質數,\\因為x_1\sim x_{15}都不是質數,因此每一個x_i = p_{i} \times k_i,其中p_{i } 是質數,k_i\ge 2;\\將15個不同的質數依序排列p_1\lt p_2\lt \cdots \lt p_{15},則p_{15}\ge 47,最大的x_{15}\ge 47^2 \gt 2022,矛盾
三、 計算與證明題 (請先標示題號,再詳列計算/證明過程,未有過程僅有答案者不給分。 共 22 分。 )
解答:
(1)\cases{A(0,0,0)\\ B(2,0,0)\\ P(0,0,2)\\ C(a,b,0)} \Rightarrow \cases{F=(P+B)\div 2=(1,0,1) \\ \overrightarrow{CA}=(-a,-b,0)\\ \overrightarrow{CB} =(2-a,-b,0) \\ \overleftrightarrow{PC}:{x\over a} ={y\over b}= {z-2\over -2}}\Rightarrow \cases{ \overrightarrow{CA} \cdot \overrightarrow{CB} =0 \Rightarrow a^2+b^2 =2a\\ E=(at,bt,-2t+2,t\in \mathbb{R}}\\ \Rightarrow \overrightarrow{EA} =(-at,-bt,2t-2) \Rightarrow \overrightarrow{EA} \cdot \overrightarrow{PC} = -t(a^2+b^2)-4t+4 =0 \Rightarrow -t(2a)=4t-4\\ \Rightarrow a=-2+{2\over t} =-2+2m(取m=1/t)\Rightarrow b^2=2a-a^2 = -4(m^2-3m+2)\\ \Rightarrow b= 2\sqrt{-m^2+3m-2} (1\lt m\lt 2) \Rightarrow E=(-2t+2,2\sqrt{-1+3t-2t^2},-2t+2)\\ \Rightarrow \cases{\overrightarrow{AE}= (-2t+2,2\sqrt{-1+3t-2t^2},-2t+2)\\ \overrightarrow{AF}=(1,0,1)} \Rightarrow f(t)= |\overrightarrow{AE}|^2|\overrightarrow{AF}|^2-(\overrightarrow{AE}\cdot \overrightarrow{AF})^2 \\ =-16t^2+24t-8 \Rightarrow t={24\over 32}={3\over 4} 有極大值f(3/4)=1 \Rightarrow \triangle AEF面積={1\over 2}f(2/4)= \bbox[red,2pt]{1\over 2}\\ (2)t=3/4 \Rightarrow E=({1\over 2}, {1\over \sqrt 2},{1\over 2}) \Rightarrow \cases{\overrightarrow{FP} =(-1,0,1)\\ \overrightarrow{FE}=(-1/2,\sqrt 2/2,-1/2)} \Rightarrow \overrightarrow{FP} \cdot \overrightarrow{FE}=0 \\ \Rightarrow \angle PFE=90^\circ \Rightarrow \tan \theta =\overline{FE}/\overline{FP} ={1\over \sqrt 2} =\bbox[red,2pt]{\sqrt 2\over 2}
(1)假設\cases{\Gamma: y^2=6x \\ L=\overleftrightarrow{AB}: y=mx+b},將L代入\Gamma可得(mx+b)^2=6x \Rightarrow m^2x^2+ (2mb-6)x +b^2=0 \\ \Rightarrow x_1+x_2 = {6-2mb \over m^2} =4 \Rightarrow b= \cfrac{4m^2-6}{-2m}={3\over m}-2m \Rightarrow L: y=mx +{3\over m}-2m \\ \Rightarrow \overline{AB}中點M(2,{3\over m}) \Rightarrow \overline{AB}中垂線L',其斜率為(-{1\over m})且通過點D \Rightarrow L': y=-{1\over m}(x-2)+{3\over m} \\ \Rightarrow L'與x軸交於\bbox[red, 2pt]{C(5,0)}(2)由於\Gamma 對稱x軸且 0\le x_1,x_2,因此取2個端點,當\cases{A=(0,0) \Rightarrow B=(4,\pm 2\sqrt 6) \Rightarrow M=(2,\pm \sqrt 6)\\ A=B=(2,\pm 2\sqrt 3) \Rightarrow M=(2,\pm 2\sqrt 3)} \\ \Rightarrow \bbox[red, 2pt]{-2\sqrt 3\lt y_0\lt 2\sqrt 3}(因為A、B相異,沒有等號)(3)
L:y=m x+b ,由(1)知:b={3\over m}-2m\Rightarrow x={y-b\over m} 代回\Gamma \Rightarrow my^2-6y+6b=0 \\ \Rightarrow \cases{兩根之和:y_1+y_2= 6/m\\ 兩根之積:y_1y_2= 6b/m} \Rightarrow (y_1-y_2)^2 =(y_1+y_2)^2 -4y_1y_2 = 36/m^2-24b/m \\ ={48m^2-36\over m^2} \Rightarrow y_1-y_2= \sqrt{48m^2-36\over m^2}\\ 令D=L\cap x軸=(-{b\over m},0) \Rightarrow \overline{CD} =5+{b\over m} =3+{3\over m^2} \\ 因此\triangle ABC = {1\over 2}\overline{CD}\times |y_1-y_1|={1\over 2}(3+{3\over m^2}) \sqrt{ 48m^2-36\over m^2} =3(1+{1\over m^2})\sqrt{12-{9\over m^2}}\\ 令a=\sqrt{12-{9\over m^2}},則\triangle ABC=g(a)= 7a-{1\over 3}a^3 \Rightarrow g'(a)=0 \Rightarrow a^2=7 \\\Rightarrow g(\sqrt 7)= 7\sqrt 7-{7\over 3}\sqrt 7 = \bbox[red, 2pt]{{14\over 3}\sqrt{7}}為面積最大值
======================== END ==============================
沒有留言:
張貼留言