113 學年度 國立彰化女中 第二次教師甄選初試
解答:
假設∠ACD=θ⇒{∠ADC=120∘+θ∠B=60∘+θ△ADC:1sinθ=3sin(120∘+θ)⇒tanθ=√37⇒{sinθ=√3/2√13cosθ=7/2√13△BCD:¯BCsin(60∘−θ)=2sin60∘=4√3⇒¯BC=4√3sin(60∘−θ)=6√13因此cosA=9+9−36132⋅9⋅9=1113

解答:

解答:
|z1−(3+3i)|=2⇒P(z1)在圓C1:(x−3)2+(y−3)2=22上z2=a+bi⇒iz2−1=(−1−b)+ai⇒|iz2−1|=1⇒Q(z2)在圓C2:x2+(y+1)2=1上⇒|z1−z2|=¯PQ的最小值=兩圓心距離−兩圓半徑和=√32+42−(2+1)=2

解答:令{a=logx3b=logy2⇒{2logx81=24logx3=24a=k3logy16=34logy2=34b=k⇒{4a=log2k=logk/log24b=log3k=logk/log3⇒{a=logk/4log2b=logk/4log3⇒log3x+log2y=1a+1b=4log2logk+4log3logk=log(24⋅34)logk=1⇒k=24⋅34=1296

解答:


解答:令{a=logx3b=logy2⇒{2logx81=24logx3=24a=k3logy16=34logy2=34b=k⇒{4a=log2k=logk/log24b=log3k=logk/log3⇒{a=logk/4log2b=logk/4log3⇒log3x+log2y=1a+1b=4log2logk+4log3logk=log(24⋅34)logk=1⇒k=24⋅34=1296

解答:limn→∞(1√n2+2n+1√n2+4n+⋯+1√n2+2n2)=limn→∞n∑k=11√n2+2kn=limn→∞n∑k=11n√1+2kn=∫101√1+2xdx=[√1+2x]|10=√3−1
解答:a2−a=a(a−1)=奇數×偶數=1000k=(125×8)k⇒a=125p=8q±1⇒{p=1,2,不合p=3⇒a=125×3=375⇒a+1=8×47=376⇒取a=376⇒376×375=1000×141p=4,不合p=5⇒a=125×5=625⇒a−1=624=8×78⇒取a=625⇒625×624=1000×390p=6,7,8,不合⇒a=376,625

解答:a2−a=a(a−1)=奇數×偶數=1000k=(125×8)k⇒a=125p=8q±1⇒{p=1,2,不合p=3⇒a=125×3=375⇒a+1=8×47=376⇒取a=376⇒376×375=1000×141p=4,不合p=5⇒a=125×5=625⇒a−1=624=8×78⇒取a=625⇒625×624=1000×390p=6,7,8,不合⇒a=376,625

解答:f(t)=√(t−1)2+(2t−1)2+(2t−3)2+√(t−2)2+(2t)2+(2t+1)2=√9t2−18t+11+√9t2+5⇒f′(t)=0⇒9t−9√9t2−18t+11+9t√9t2+5=0⇒3t2−10t+5=0⇒t=5±√103⇒f(5−√103)<f(5+√103)⇒t=5−√103有最小值
解答:x2+y2−2x−6y+8=0⇒(x−1)2+(y−3)2=2⇒y=3−√2−(x−1)2兩圖形交於(0,0)及(2,2),因此旋轉體積=π∫20(3−√2−(x−1)2)2dx−π∫20x2dx=π(463−3π)−83π=383π−3π2解答:

假設{小圓(月亮)圓心O1,半徑r=1大圓(太陽)圓心O2,半徑RC=¯O1Q與¯AB的交點⇒¯BC=12¯AB=12√3⇒{∠CO1B=60∘∠O1BC=30∘又¯CQ=¯O1P−¯PQ−¯O1C=√3−32⇒tan∠QBC=¯CQ¯BC=2−√3⇒{∠QBC=15∘∠CQB=75∘由於¯O2B=¯O2Q=R⇒∠BO2C=180∘−75∘×2=30∘⇒R=¯BC×2=√3⇒{扇形AQBO2面積=16(√3)2π=π2△ABO2面積=12√3⋅√3sin60∘=3√34⇒弓形AQB面積=π2−3√34同理,{扇形APBO1面積=13π△ABO1面積=12sin120∘=√3/4⇒弓形APB面積=π3−√34⇒月偏食亮面面積=(π3−√34)−(π2−3√34)=−π6+√32⇒月偏食亮面面積滿月圓面積=(−π6π+√32)/π=√32π−16
解答:limn→∞(1√3n2+1√3n2+⋯+1√3n2)<原式<limn→∞(1√3n2+2n+1√3n2+2n+⋯+1√3n2+2n)⇒limn→∞(2n√3n2)<原式<limn→∞(2n√3n2+2n)⇒2√3<原式<2√3⇒原式=2√33
解答:依題意取{C(0,0)A(−1,−1)B(1,−1)P(12cosθ,12sinθ)⇒{→AB=(2,0)→BC=(−1,1)→AP=(12cosθ+1,12sinθ+1)→AP=α→AB+β→BC⇒(12cosθ+1,12sinθ+1)=(2α−β,β)⇒{α=1+14(sinθ+cosθ)β=12sinθ+1⇒4α−β=3+12sinθ+cosθ=3+√52sin(θ+δ)最大值=3+√52
解答:假設正△的邊長為a,其面積為√34a2⇒體積=2∫20√34a2da=4√33
解答:假設{A(0,8)∈L1B(9,2)∈L4C(6,0)∈L2D(−5,4)∈L3,其中{L1∥L2L3∥L4L1⊥L3⇒{L1:y=mx+a1⇒a1=8⇒y=mx+8L2:y=mx+a2⇒0=6m+a2⇒y=mx−6mL3:x+my=b1⇒−5+4m=b1⇒x+my=4m−5L4:x+my=b2⇒9+2m=b2⇒x+my=2m+9又d(L1,L2)=d(L3,L4)⇒|6m+8|√m2+1=|2m−14|√m2+1⇒(6m+8)2=(2m−14)2⇒{m=3/4⇒d(L1,L2)=10m=−11/2⇒d(L1,L2)=2√5⇒最大面積=102=100
解答:由於{(x−1)2≥0x2+2x+3=(x+1)2+2>0,因此僅需考慮(x−k)(x−6)≤0若{k<6⇒k<x<6有10個整數解⇒x=5,4,…,−4⇒−5≤k<−4k>6⇒6<x<k有9個整數解(還有x=1,共10個)⇒x=7,8,…,15⇒15<k≤16⇒{a=15b=16c=−5d=−4⇒a+b+c+d=22
解答:\cases{O(-1,2) \\ I(2,2)\\ A(2,8)} \Rightarrow \cases{\triangle ABC外接圓\Gamma_1: (x+1)^2+ (y-2)^2= 45 \\ L=\overleftrightarrow{IA}: x=2} \Rightarrow D= \Gamma \cap L=(2,-4) \\ 以D為圓心,\overline{DI}=6為半徑的圓\Gamma_2:(x-2)^2+ (y+4)^2=36\\ \Gamma_1\cap \Gamma_2 \Rightarrow (x+1)^2+ (y-2)^2- 45=(x-2)^2+ (y+4)^2-36 \\ \Rightarrow \bbox[red, 2pt]{x-2y=4}\\ 依雞爪定理:\Gamma_1\cap \Gamma_2 的兩點即為B,C\; \href{http://w.mathsgreat.com/SC/geom_th/geom_th_123.pdf}{參考資料}
解答:
解答:limn→∞(1√3n2+1√3n2+⋯+1√3n2)<原式<limn→∞(1√3n2+2n+1√3n2+2n+⋯+1√3n2+2n)⇒limn→∞(2n√3n2)<原式<limn→∞(2n√3n2+2n)⇒2√3<原式<2√3⇒原式=2√33
解答:依題意取{C(0,0)A(−1,−1)B(1,−1)P(12cosθ,12sinθ)⇒{→AB=(2,0)→BC=(−1,1)→AP=(12cosθ+1,12sinθ+1)→AP=α→AB+β→BC⇒(12cosθ+1,12sinθ+1)=(2α−β,β)⇒{α=1+14(sinθ+cosθ)β=12sinθ+1⇒4α−β=3+12sinθ+cosθ=3+√52sin(θ+δ)最大值=3+√52
解答:假設正△的邊長為a,其面積為√34a2⇒體積=2∫20√34a2da=4√33
解答:由於{(x−1)2≥0x2+2x+3=(x+1)2+2>0,因此僅需考慮(x−k)(x−6)≤0若{k<6⇒k<x<6有10個整數解⇒x=5,4,…,−4⇒−5≤k<−4k>6⇒6<x<k有9個整數解(還有x=1,共10個)⇒x=7,8,…,15⇒15<k≤16⇒{a=15b=16c=−5d=−4⇒a+b+c+d=22
解答:\cases{O(-1,2) \\ I(2,2)\\ A(2,8)} \Rightarrow \cases{\triangle ABC外接圓\Gamma_1: (x+1)^2+ (y-2)^2= 45 \\ L=\overleftrightarrow{IA}: x=2} \Rightarrow D= \Gamma \cap L=(2,-4) \\ 以D為圓心,\overline{DI}=6為半徑的圓\Gamma_2:(x-2)^2+ (y+4)^2=36\\ \Gamma_1\cap \Gamma_2 \Rightarrow (x+1)^2+ (y-2)^2- 45=(x-2)^2+ (y+4)^2-36 \\ \Rightarrow \bbox[red, 2pt]{x-2y=4}\\ 依雞爪定理:\Gamma_1\cap \Gamma_2 的兩點即為B,C\; \href{http://w.mathsgreat.com/SC/geom_th/geom_th_123.pdf}{參考資料}
解答:

y=x^2+bx+c= (x-\alpha)^2-\beta^2\; (\beta\gt 0)\\ \Rightarrow \cases{A(\alpha-\beta,0)\\ B(\alpha+\beta,0)\\ C(0,\alpha^2-\beta^2\lt 0) \Rightarrow \alpha\lt \beta\\ M(\alpha,-\beta^2)} \Rightarrow S_{\triangle ABM} ={1\over 2}\cdot 2\beta\cdot \beta^2 =\beta^3 \\ 由阿基米德某性質(\href{https://www.cnblogs.com/james-wangx/p/16111454.html}{參考資料})可知:S_{\triangle ACM}={1\over 8} S_{\triangle ABM}\\ 現在 S_{\triangle ABM}+S_{\triangle ACM} =9 \Rightarrow S_{\triangle ABM}= \beta^3=8 \Rightarrow \beta=2 \\又 通過(-2,5) \Rightarrow 5=(-2-\alpha)^2-4 \Rightarrow \alpha=1 \Rightarrow y=(x-1)^2-4\\ \Rightarrow y=x^2-2x-3 =x^2+bx+c \Rightarrow (b,c)=\bbox[red, 2pt]{(-2,-3)}
解答:令\cases{P(x,y)\\ Q(0,a)},且\overline{PQ}={2\over 3}\overline{PB} \Rightarrow x^2+(y-a)^2={4\over 9}\left( x^2+ (y-6)^2\right) \\ \Rightarrow 5x^2+5y^2-18ay+48y+9a^2-144=0 \Rightarrow 5\cdot 16-6y(3a-8)+9a^2-144=0 \\ \Rightarrow 9a^2-64-6y(3a-8)=0 \Rightarrow (3a-8)(3a+8-6y)=0 \Rightarrow a={8\over 3} \Rightarrow Q(0,{8\over 3}) \\ 因此3\overline{PA}+ 2\overline{PB}=3\overline{PA}+ 2\cdot {3\over 2}\overline{PQ}= 3(\overline{PA}+ \overline{PQ})最小值出現在Q,P,A在一直線\\ \Rightarrow 最小值 =3\overline{AQ} =3\cdot \sqrt{8^2+(8/3)^2} = 3\cdot \sqrt {640\over 9} = \bbox[red, 2pt]{8\sqrt{10}}
解答:將\cases{k=6\\ n=7} 代入公式 {(k-1)\left[ (k-1)^{n-1}+(-1)^n\right] \over k} ={5(5^6-1)\over 6} = \bbox[red, 2pt]{13020}
解答:令\cases{\angle EAF=\theta \\ \overline{AB}=\overline{AE}=a}\Rightarrow \cos \theta ={a^2-3\over 2a} \Rightarrow \sin \theta={\sqrt{-a^4+10a^2-9} \over 2a} \\ \Rightarrow S_{\triangle ABC} ={1\over 2}\overline{AB} \cdot \overline{AC}\sin \angle BAC ={1\over 2}a\sin (240^\circ-\theta) =-{1\over 2}a\sin(60^\circ-\theta) =-{1\over 4}(\sqrt 3\cos \theta-\sin \theta) \\=f(a)=-{1\over 8}(\sqrt 3a^2-3\sqrt 3-\sqrt{-a^4+10a^2-9}) \Rightarrow f'(a)=0 \Rightarrow 2\sqrt 3a ={-2a^3+10a \over \sqrt{-a^4+10a^2-9}} \\ \Rightarrow 16a^6-160a^4+208a^2=0 \Rightarrow 16a^2(a^4-10a^2+10) =0 \Rightarrow a^2=5-2\sqrt 3 \\ \Rightarrow f(a^2=5-2\sqrt 3) =-{1\over 8}(\sqrt 3(5-2\sqrt 3)-3\sqrt 3-\sqrt{-(5-2\sqrt 3)^2+10(5-2\sqrt 3)-9} \\= -{1\over 8}(2\sqrt 3-8) = \bbox[red, 2pt]{1-{\sqrt 3\over 4}}
解答:\cases{\cos^2 A+\cos^2 B+ 2\sin A\sin B \cos C={15\over 8} \\ \cos^2 B+ \cos^2 C+ 2\sin B\sin C \cos A={14\over 9}}\\兩式相加 \Rightarrow \cos^2 A+ 2\cos^2B +\cos^2 C+2\sin B(\sin A\cos C+ \sin C \cos A) ={15\over 8} +{14\over 9} \\ \Rightarrow \cos^2 A+ 2\cos^2B +\cos^2 C+2\sin B\sin(A+C) =\cos^2 A+ 2\cos^2B +\cos^2 C+2\sin^2 B ={247\over 72}\\\Rightarrow \cos^2A+ \cos^2C ={247\over 72}-2 \Rightarrow \cos^2A+ \cos^2C={103\over 72} \cdots(1)\\ 兩式相減\Rightarrow \cos^2 A-\cos^2C +2\sin B(\sin A\cos C-\sin C\cos A) ={15\over 8}-{14\over 9} \\ \Rightarrow \cos^2A-\cos^2 C+2 \sin (A+C)\sin (A-C) =\cos^2A-\cos^2 C+ \cos(2C)-\cos (2A) ={23\over 72} \\ \Rightarrow \cos^2 A-\cos^2C+ (2\cos^2 C-1)-(2\cos^2A -1) ={23\over 72} \Rightarrow -\cos^2 A+ \cos^2 C={23\over 72} \cdots(2) \\ 由(1)及(2)可得\cases{\cos^2 A=5/9\\ \cos^2 C=7/8} \\因此\cos^2C+ \cos^2 A+2\sin C\sin A\cos B={103\over 72}+ 2\sin C\sin A(-\cos(A+C)) \\={103\over 72}+ 2\sin C\sin A(\sin A\sin C- \cos A\cos C) \\={103\over 72}+2\sin^2 A\sin ^2 C-2\sin C\cos C\sin A\cos A \\={103\over 72}+2\cdot {4\over 9}\cdot {1\over 8}\pm 2\cdot {1\over \sqrt 8} \cdot {\sqrt 7\over \sqrt 8}\cdot {2\over 3}\cdot {\sqrt 5\over 3} = \bbox[red, 2pt]{{111\over 72}\pm {4\sqrt{35}\over 72}}
解答:
解答:令\cases{P(x,y)\\ Q(0,a)},且\overline{PQ}={2\over 3}\overline{PB} \Rightarrow x^2+(y-a)^2={4\over 9}\left( x^2+ (y-6)^2\right) \\ \Rightarrow 5x^2+5y^2-18ay+48y+9a^2-144=0 \Rightarrow 5\cdot 16-6y(3a-8)+9a^2-144=0 \\ \Rightarrow 9a^2-64-6y(3a-8)=0 \Rightarrow (3a-8)(3a+8-6y)=0 \Rightarrow a={8\over 3} \Rightarrow Q(0,{8\over 3}) \\ 因此3\overline{PA}+ 2\overline{PB}=3\overline{PA}+ 2\cdot {3\over 2}\overline{PQ}= 3(\overline{PA}+ \overline{PQ})最小值出現在Q,P,A在一直線\\ \Rightarrow 最小值 =3\overline{AQ} =3\cdot \sqrt{8^2+(8/3)^2} = 3\cdot \sqrt {640\over 9} = \bbox[red, 2pt]{8\sqrt{10}}
解答:將\cases{k=6\\ n=7} 代入公式 {(k-1)\left[ (k-1)^{n-1}+(-1)^n\right] \over k} ={5(5^6-1)\over 6} = \bbox[red, 2pt]{13020}
解答:令\cases{\angle EAF=\theta \\ \overline{AB}=\overline{AE}=a}\Rightarrow \cos \theta ={a^2-3\over 2a} \Rightarrow \sin \theta={\sqrt{-a^4+10a^2-9} \over 2a} \\ \Rightarrow S_{\triangle ABC} ={1\over 2}\overline{AB} \cdot \overline{AC}\sin \angle BAC ={1\over 2}a\sin (240^\circ-\theta) =-{1\over 2}a\sin(60^\circ-\theta) =-{1\over 4}(\sqrt 3\cos \theta-\sin \theta) \\=f(a)=-{1\over 8}(\sqrt 3a^2-3\sqrt 3-\sqrt{-a^4+10a^2-9}) \Rightarrow f'(a)=0 \Rightarrow 2\sqrt 3a ={-2a^3+10a \over \sqrt{-a^4+10a^2-9}} \\ \Rightarrow 16a^6-160a^4+208a^2=0 \Rightarrow 16a^2(a^4-10a^2+10) =0 \Rightarrow a^2=5-2\sqrt 3 \\ \Rightarrow f(a^2=5-2\sqrt 3) =-{1\over 8}(\sqrt 3(5-2\sqrt 3)-3\sqrt 3-\sqrt{-(5-2\sqrt 3)^2+10(5-2\sqrt 3)-9} \\= -{1\over 8}(2\sqrt 3-8) = \bbox[red, 2pt]{1-{\sqrt 3\over 4}}
解答:\cases{\cos^2 A+\cos^2 B+ 2\sin A\sin B \cos C={15\over 8} \\ \cos^2 B+ \cos^2 C+ 2\sin B\sin C \cos A={14\over 9}}\\兩式相加 \Rightarrow \cos^2 A+ 2\cos^2B +\cos^2 C+2\sin B(\sin A\cos C+ \sin C \cos A) ={15\over 8} +{14\over 9} \\ \Rightarrow \cos^2 A+ 2\cos^2B +\cos^2 C+2\sin B\sin(A+C) =\cos^2 A+ 2\cos^2B +\cos^2 C+2\sin^2 B ={247\over 72}\\\Rightarrow \cos^2A+ \cos^2C ={247\over 72}-2 \Rightarrow \cos^2A+ \cos^2C={103\over 72} \cdots(1)\\ 兩式相減\Rightarrow \cos^2 A-\cos^2C +2\sin B(\sin A\cos C-\sin C\cos A) ={15\over 8}-{14\over 9} \\ \Rightarrow \cos^2A-\cos^2 C+2 \sin (A+C)\sin (A-C) =\cos^2A-\cos^2 C+ \cos(2C)-\cos (2A) ={23\over 72} \\ \Rightarrow \cos^2 A-\cos^2C+ (2\cos^2 C-1)-(2\cos^2A -1) ={23\over 72} \Rightarrow -\cos^2 A+ \cos^2 C={23\over 72} \cdots(2) \\ 由(1)及(2)可得\cases{\cos^2 A=5/9\\ \cos^2 C=7/8} \\因此\cos^2C+ \cos^2 A+2\sin C\sin A\cos B={103\over 72}+ 2\sin C\sin A(-\cos(A+C)) \\={103\over 72}+ 2\sin C\sin A(\sin A\sin C- \cos A\cos C) \\={103\over 72}+2\sin^2 A\sin ^2 C-2\sin C\cos C\sin A\cos A \\={103\over 72}+2\cdot {4\over 9}\cdot {1\over 8}\pm 2\cdot {1\over \sqrt 8} \cdot {\sqrt 7\over \sqrt 8}\cdot {2\over 3}\cdot {\sqrt 5\over 3} = \bbox[red, 2pt]{{111\over 72}\pm {4\sqrt{35}\over 72}}
解答:
無法連成一直線的有28種,如上圖(其中右邊三圖是左圖的旋轉)\\ 因此欲求之機率=1-{28\over C^9_5} =1-{2\over 9}=\bbox[red, 2pt]{7\over 9}
======================== END ===========================
解題僅供參考,教甄歷年試題及詳解
沒有留言:
張貼留言