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2024年8月18日 星期日

113年彰化女中第2次教甄-數學詳解

 113 學年度 國立彰化女中 第二次教師甄選初試


解答:
ACD=θ{ADC=120+θB=60+θADC:1sinθ=3sin(120+θ)tanθ=37{sinθ=3/213cosθ=7/213BCD:¯BCsin(60θ)=2sin60=43¯BC=43sin(60θ)=613cosA=9+93613299=1113

解答:

|z1(3+3i)|=2P(z1)C1:(x3)2+(y3)2=22z2=a+biiz21=(1b)+ai|iz21|=1Q(z2)C2:x2+(y+1)2=1|z1z2|=¯PQ==32+42(2+1)=2


解答:{a=logx3b=logy2{2logx81=24logx3=24a=k3logy16=34logy2=34b=k{4a=log2k=logk/log24b=log3k=logk/log3{a=logk/4log2b=logk/4log3log3x+log2y=1a+1b=4log2logk+4log3logk=log(2434)logk=1k=2434=1296

解答:limn(1n2+2n+1n2+4n++1n2+2n2)=limnnk=11n2+2kn=limnnk=11n1+2kn=1011+2xdx=[1+2x]|10=31
解答:a2a=a(a1)=×=1000k=(125×8)ka=125p=8q±1{p=1,2,p=3a=125×3=375a+1=8×47=376a=376376×375=1000×141p=4,p=5a=125×5=625a1=624=8×78a=625625×624=1000×390p=6,7,8,a=376,625

解答:f(t)=(t1)2+(2t1)2+(2t3)2+(t2)2+(2t)2+(2t+1)2=9t218t+11+9t2+5f(t)=09t99t218t+11+9t9t2+5=03t210t+5=0t=5±103f(5103)<f(5+103)t=5103
解答:x2+y22x6y+8=0(x1)2+(y3)2=2y=32(x1)2(0,0)(2,2),=π20(32(x1)2)2dxπ20x2dx=π(4633π)83π=383π3π2

解答:

{()O1,r=1()O2,RC=¯O1Q¯AB¯BC=12¯AB=123{CO1B=60O1BC=30¯CQ=¯O1P¯PQ¯O1C=332tanQBC=¯CQ¯BC=23{QBC=15CQB=75¯O2B=¯O2Q=RBO2C=18075×2=30R=¯BC×2=3{AQBO2=16(3)2π=π2ABO2=1233sin60=334AQB=π2334,{APBO1=13πABO1=12sin120=3/4APB=π334=(π334)(π2334)=π6+32滿=(π6π+32)/π=32π16
解答:limn(13n2+13n2++13n2)<<limn(13n2+2n+13n2+2n++13n2+2n)limn(2n3n2)<<limn(2n3n2+2n)23<<23=233
解答:{C(0,0)A(1,1)B(1,1)P(12cosθ,12sinθ){AB=(2,0)BC=(1,1)AP=(12cosθ+1,12sinθ+1)AP=αAB+βBC(12cosθ+1,12sinθ+1)=(2αβ,β){α=1+14(sinθ+cosθ)β=12sinθ+14αβ=3+12sinθ+cosθ=3+52sin(θ+δ)=3+52
解答:a34a2=22034a2da=433

解答:{A(0,8)L1B(9,2)L4C(6,0)L2D(5,4)L3,{L1L2L3L4L1L3{L1:y=mx+a1a1=8y=mx+8L2:y=mx+a20=6m+a2y=mx6mL3:x+my=b15+4m=b1x+my=4m5L4:x+my=b29+2m=b2x+my=2m+9d(L1,L2)=d(L3,L4)|6m+8|m2+1=|2m14|m2+1(6m+8)2=(2m14)2{m=3/4d(L1,L2)=10m=11/2d(L1,L2)=25=102=100
解答:{(x1)20x2+2x+3=(x+1)2+2>0,(xk)(x6)0{k<6k<x<610x=5,4,,45k<4k>66<x<k9(x=1,10)x=7,8,,1515<k16{a=15b=16c=5d=4a+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:



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}
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解題僅供參考,教甄歷年試題及詳解






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