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
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解:
假設A,B在拋物線Γ:y=mx2−1上,且A,B對稱於直線L1:x+y=0;因此A,B同在直線L2:y=x+k,其中L1⊥L2;先求Γ與L2的交點(即A與B),即mx2−1=x+k⇒mx2−x−1−k=0⋯(1)令a,b為式(1)的解,即{A(a,a+k)B(b,b+k)⇒A,B的中點C(a+b2,a+b2+k)在L1上⇒a+b2+a+b2+k=0⇒k=−(a+b)=−1/m(∵式(1)兩根之和=a+b=1/m)再由式(1)有兩相異實根⇒判別式>0⇒1+4m(k+1)>0⇒1+4m(−1m+1)>0⇒1−4+4m>0⇒m>34
解:asinA=bsinB=csinC=k⇒{a=ksinAb=ksinBc=ksinC⇒¯AB2−¯AC2=¯AB⋅¯BC≡c2−b2=ac⇒k2sin2C−k2sin2B=k2sinAsinC⇒sinAsinC=sin2C−sin2B=(sinC+sinB)(sinC−sinB)=2sinC+B2cosC−B2⋅2cosC+B2sinC−B2=sin(C+B)sin(C−B)=sinAsin(C−B)⇒sinC=sin(C−B)⇒180∘−C=C−B⇒C=90∘+B/2⇒A+B+C=30∘+B+90∘+B/2=180∘⇒32B=60∘⇒B=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?
解:令{S1:甲(1,1),乙(2,2)S2:甲(1,2),乙(1,2)S3:甲(2,2,),乙(1,1)⇒{P(S1→S1)=0,P(S1→S2)=1,P(S1→S3)=0P(S2→S1)=1/4,P(S2→S2)=1/2,P(S2→S3)=1/4P(S3→S3)=0,P(S3→S2)=1,P(S3→S3)=0⇒T=[0101/41/21/4010]⇒T5=[5/3211/165/32無所謂無所謂]⇒P(S1)+P(S3)=532+532=516
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本題送分
解:{f(1)=−3f(−2)=6f(3)=−9⇒令g(x)=f(x)+3x,則1,−2,3為g(x)=0之三根⇒g(x)=a(x−1)(x+2)(x−3)又f(5)=153⇒g(5)=a⋅4⋅7⋅2=56a=f(5)+3⋅5=168⇒a=3⇒g(0)=f(0)+3⋅0=a⋅(−1)⋅2⋅(−3)=18⇒f(0)=18
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{y=logx5y=log5x}交點:{A(5,1)E(1/5,−1){y=logx15y=log1/5x}交點:{B(5,−1)D(1/5,1){y=log5xy=log1/5x}交點:C(1,0)⇒共5個交點解:α,β為x2−x−1=0之二根⇒{α+β=1αβ=−1⇒(αn−βn)(α+β)=αn+1+αnβ−αβn−βn+1=αn+1−βn+1+αβ(αn−1−βn−1)⇒αn−βn=αn+1−βn+1−(αn−1−βn−1)⇒αn+1−βn+1=(αn−βn)+(αn−1−βn−1)≡an+1=an+an−1,其中an=αn−βn因此a2019=a2018+a2017=(a2017+a2016)+(a2016+a2015)=a2017+2a2016+a2015=a2017+2(a2018−a2017)+a2015=2a2018−a2017+a2015=2a2018−(a2019−a2018)+a2015=3m+n−a2019⇒2a2019=3m+n⇒a2019=3m+n2
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{A(6,13)B(12,11)⇒{A,B的中點C(9,12)→AB=(6,−2)⇒¯AB的中垂線L:6(x−9)−2(y−12)=0⇒3x−y=15⇒L交X軸於C(5,0);又圓心O在L上⇒O(t,3t−15)⇒圓半徑r=¯OA=¯OB⇒¯OC2=¯OA2+¯AC2⇒(t−5)2+(3t−15)2=(t−6)2+(3t−28)2+(6−5)2+132⇒−100t+250=−180t+990⇒t=37/4⇒圓面積=¯OA2π=((37/4−6)2+(111/4−28)2)π=((13/4)2+(1/4)2)π=858π
解: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}
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\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}解:(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}
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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。解:{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}
解: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}
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\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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