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2020年10月25日 星期日

108年新北市板橋高中教甄-數學詳解

新北市立板橋高級中學108學年度第一次教師甄選
數學科試題卷

本題

A,BΓ:y=mx21A,BL1:x+y=0;A,BL2:y=x+k,L1L2;ΓL2(AB),mx21=x+kmx2x1k=0(1)a,b(1){A(a,a+k)B(b,b+k)A,BC(a+b2,a+b2+k)L1a+b2+a+b2+k=0k=(a+b)=1/m((1)=a+b=1/m)(1)>01+4m(k+1)>01+4m(1m+1)>014+4m>0m>34



asinA=bsinB=csinC=k{a=ksinAb=ksinBc=ksinC¯AB2¯AC2=¯AB¯BCc2b2=ack2sin2Ck2sin2B=k2sinAsinCsinAsinC=sin2Csin2B=(sinC+sinB)(sinCsinB)=2sinC+B2cosCB22cosC+B2sinCB2=sin(C+B)sin(CB)=sinAsin(CB)sinC=sin(CB)180C=CBC=90+B/2A+B+C=30+B+90+B/2=18032B=60B=40


4.有甲、乙兩個袋子,甲袋內裝有兩顆 1 號球,乙袋內裝有兩顆 2 號球,每一顆球被取到的機會都相同,若每次從各袋中取一顆球交換,則交換 5 次後甲袋內兩顆球的和為偶數的機率為?

{S1:(1,1),(2,2)S2:(1,2),(1,2)S3:(2,2,),(1,1){P(S1S1)=0,P(S1S2)=1,P(S1S3)=0P(S2S1)=1/4,P(S2S2)=1/2,P(S2S3)=1/4P(S3S3)=0,P(S3S2)=1,P(S3S3)=0T=[0101/41/21/4010]T5=[5/3211/165/32]P(S1)+P(S3)=532+532=516



本題




{f(1)=3f(2)=6f(3)=9g(x)=f(x)+3x,1,2,3g(x)=0g(x)=a(x1)(x+2)(x3)f(5)=153g(5)=a472=56a=f(5)+35=168a=3g(0)=f(0)+30=a(1)2(3)=18f(0)=18



{y=logx5y=log5x}:{A(5,1)E(1/5,1){y=logx15y=log1/5x}:{B(5,1)D(1/5,1){y=log5xy=log1/5x}:C(1,0)5




α,βx2x1=0{α+β=1αβ=1(αnβn)(α+β)=αn+1+αnβαβnβn+1=αn+1βn+1+αβ(αn1βn1)αnβn=αn+1βn+1(αn1βn1)αn+1βn+1=(αnβn)+(αn1βn1)an+1=an+an1,an=αnβna2019=a2018+a2017=(a2017+a2016)+(a2016+a2015)=a2017+2a2016+a2015=a2017+2(a2018a2017)+a2015=2a2018a2017+a2015=2a2018(a2019a2018)+a2015=3m+na20192a2019=3m+na2019=3m+n2






{A(6,13)B(12,11){A,BC(9,12)AB=(6,2)¯ABL:6(x9)2(y12)=03xy=15LXC(5,0);OLO(t,3t15)r=¯OA=¯OB¯OC2=¯OA2+¯AC2(t5)2+(3t15)2=(t6)2+(3t28)2+(65)2+132100t+250=180t+990t=37/4=¯OA2π=((37/46)2+(111/428)2)π=((13/4)2+(1/4)2)π=858π



6^{108}+ 8^{108} =(7-1)^{108} +(7+1)^{108} =\sum_{k=0}^{108}C^{108}_k 7^k(-1)^{108-k} +\sum_{k=0}^{108}C^{108}_k 7^k =2 \sum_{k=0}^{54}C^{108}_{2k} 7^{2k} \\ 又343=7^3 \Rightarrow (6^{108}+ 8^{108}) \mod{343} =2 \sum_{k=0}^{54}C^{108}_{2k} 7^{2k} \mod{7^3} =2(1+C^{54}_2 7^2) \mod{7^3} \\ =(2+108\times 107\times 49) \mod{7^3} = (2+ (2\cdot 7^2+10)(2\cdot 7^2+9)7^2) \mod{7^3} \\= (2+4\cdot 7^6 +38\cdot 7^4+90\cdot 7^2) \mod{7^3} = (2+90\cdot 7^2) \mod{7^3}=(2+ (6+12\cdot 7)7^2) \mod{7^3} \\= (2+6\cdot 7^2) \mod{7^3} = \bbox[red,2pt]{296}



\cases{\overline{IC} 為\angle C的角平分線\\ \overline{DE}\parallel \overline{BC}} \Rightarrow \cases{\angle ECI= \angle ICB \\ \angle ICB=\angle EIC} \Rightarrow \angle ECI=\angle EIC \Rightarrow \overline{EC} =\overline{EI}=a;\\ 同理\overline{DI} =\overline{DB}=b; \\ \overline{AF} 為\angle A的角平分線 \Rightarrow {\overline{AB} \over \overline{AC}} ={\overline{BF} \over \overline{FC}} \Rightarrow \cases{\overline{BF} =7/2\\ \overline{FC}=9/2}\\ \cases{\triangle AIE\sim \triangle AFC \\ \triangle ADI \sim \triangle ABF} \Rightarrow \cases{{\overline{AE} \over \overline{AC}} ={\overline{IE} \over \overline{FC}} \Rightarrow {9-a\over 9}={a\over 9/2}\\ {\overline{AD} \over \overline{AB}} ={\overline{DI} \over \overline{BF}} \Rightarrow {7-b\over 7}= {b\over 7/2}} \Rightarrow \cases{a=3\\ b=7/3} \Rightarrow \overline{DE}=a+b=\bbox[red,2pt]{16\over 3}




(x+y+z)^2 =x^2+y^2+z^2 +2(xy+yz +zx) \Rightarrow 4^2=10+2(xy +yz+zx) \Rightarrow xy +yz+zx=3\\ 又(x+y+z)^3 =x^3+y^3+z^3 +3(x+y+z)( xy+yz+zx)- 3xyz\\ \Rightarrow 4^3 =22 +3 \times 4\times 3-3xyz \Rightarrow 64=58-3xyz \Rightarrow xyz=\bbox[red, 2pt]{-2}



由上圖可知: A→B→C→D→E→F→G→H→I→J→A,不含頭尾的A,共有B-J,\bbox[red,2pt]{9} 個反射點。




{3\over 7} < {q\over p} < {9\over 19} \Rightarrow {19\over 9}< {p\over q} < {7\over 3} \Rightarrow {19\over 9}+{1\over q}< {p+1\over q} < {7\over 3}+{1\over q} \\ \Rightarrow \begin{array}{}q& p & {p+1\over q} \\\hline 1 & 2.1< p< 2.3 非整數& \\2 & 4.2 < p < 4.6非整數& \\3 & 6.3 < p < 7 非整數& \\4 & 8.4 < p< 9.3 \Rightarrow p=9& {5\over 2} \\ 5 & 10.5 < p < 11.6 \Rightarrow p=11 & {12\over 5} < {5\over 2}\\\hline\end{array}\\ \Rightarrow 最大值為\bbox[red,2pt]{5\over 2}




令\langle a_n \rangle = \left \langle {n(n+1)\over 2^n}\right \rangle,則S_n= \sum_{k=1}^n a_n \Rightarrow S=\lim_{n\to \infty} S_n\\ S_n={1\cdot 2\over 2} + {2\cdot 3\over 2^2} +\cdots +{(n-1)n\over 2^{n-1}} + {n(n+1)\over 2^{n}} \\ \Rightarrow {1\over 2}S_n={1\cdot 2\over 2^2} + {2\cdot 3\over 2^3} +\cdots +{(n-1)n\over 2^{n}} + {n(n+1)\over 2^{n+1}}\\ \Rightarrow S_n-{1\over 2}S_n={1\cdot 2\over 2}+{2\cdot 2\over 2^2} + {3\cdot 2\over 2^3} +\cdots +{n\cdot 2\over 2^{n}} - {n(n+1)\over 2^{n+1}} =S' - {n(n+1)\over 2^{n+1}}\\ \Rightarrow {1\over 2}S_n=S' - {n(n+1)\over 2^{n+1}}\\ 而S'={1\cdot 2\over 2}+{2\cdot 2\over 2^2} + {3\cdot 2\over 2^3} +\cdots +{n\cdot 2\over 2^{n}} \Rightarrow {1\over 2}S'={1\over 2}+{2\over 2^2} + {3\over 2^3} +\cdots +{n\over 2^{n}} \\ \Rightarrow {1\over 4}S' ={1\over 2^2}+{2\over 2^3} + {3\over 2^4} +\cdots +{n\over 2^{n+1}} \Rightarrow {1\over 2}S'-{1\over 4}S'= {1\over 2}+ {1\over 2^2} + \cdots +{1\over 2^n}-{n\over 2^{n+1}} \\ \Rightarrow {1\over 4}S'=1-{1\over 2^n}-{n\over 2^{n+1}}\Rightarrow S'=4-{1\over 2^{n-2}}-{n\over 2^{n-1}} \\ \Rightarrow {1\over 2}S_n=S' - {n(n+1)\over 2^{n+1}} =4-{1\over 2^{n-2}}-{n\over 2^{n-1}} - {n(n+1)\over 2^{n+1}} \\ \Rightarrow S_n=8-{1\over 2^{n-3}}-{n\over 2^{n-2}} - {n(n+1)\over 2^{n}} \Rightarrow \lim_{n\to \infty }S_n = 8-0-0-0 =\bbox[red,2pt]{8}




\omega =\cos{2\pi \over 7}+i\sin{2\pi\over 7} \Rightarrow \cases{\omega^7=1\\ |\omega|=1} ,另1,\omega,\omega^2,..,\omega^6 可視為內接單位圓上的正七邊形的頂點,如上圖;\\因此\cases{\overline{AB}=|1-\omega|=a\\ \overline{AC}=|1-\omega^2|=b \\ \overline{AD}= |1-\omega^3|=c\\ \overline{AD}=\overline{AE}\\ \overline{AF}= \overline{AC} \\ \overline{AG}= \overline{AB}}; \\又 圓內接四邊形ABCD \Rightarrow \overline{AC}\times \overline{BD} =\overline{BC} \times \overline{AD}+ \overline{AB}\times \overline{CD} \Rightarrow b^2=ac+a^2\\同理,圓內接四邊形ACDF \Rightarrow \overline{AD}\times \overline{CF} = \overline{CD}\times \overline{AF}+ \overline{AC}\times \overline{DF}   \Rightarrow c^2=ab+b^2\\ f(k)=|1-\omega^k||\omega^{k+1}-\omega^{2k+1}| + |1-\omega^{2k+1}||\omega^{k}-\omega^{k+1}|  \\ \Rightarrow \cases{f(0)=0+|1-\omega||1-\omega| = |1-\omega|^2 \\f(1)=|1-\omega||\omega^{2}-\omega^{3}| + |1-\omega^{3}||\omega -\omega^{2}| =|\omega^2||1-\omega|^2 +|\omega||1-\omega^{3}||1-\omega|\\ f(2)= |1-\omega^2||\omega^{3}-\omega^{5}| + |1-\omega^{5}||\omega^{2}-\omega^{3}|= |\omega^3||1-\omega^2|^2  +|\omega^2| |1-\omega^{5}||1-\omega |} \\ \Rightarrow \cases{f(0)=|1-\omega|^2 =a^2\\ f(1)=|1-\omega|^2+|1-\omega^3||1-\omega| =a^2+ca \\f(2) =|1-\omega^2|^2 +|1-\omega^5||1-\omega| =b^2+ba} \\ \Rightarrow f(0) \times f(1) \times f(2) =a^2(a^2+ac)(b^2+ab) =a^2b^2c^2 = |1-\omega|^2|1-\omega^2|^2|1-\omega^3|^2 \\ =|1-\omega||1-\omega^6||1-\omega^2||1-\omega^5||1-\omega^3||1-\omega^4| =|(1-\omega)(1-\omega^2) (1-\omega^4)(1-\omega^5) (1-\omega^6)|\\ 由於\omega,\omega^2,...,\omega^6 為g(x)=x^6+x^5+x^4 +x^3+x^2+x+1=0 之六根\\ \Rightarrow g(x)=(x-\omega)(x-\omega^2)\cdots (x-\omega^6) \Rightarrow g(1)=1+\dots+1=7 = (1-\omega)(1-\omega^2)\cdots (1-\omega^6) \\ \Rightarrow f(0) \times f(1) \times f(2) = |7| =\bbox[red,2pt]{7}



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