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

解：

本題$\bbox[red,2pt]{送分}$

解：

$$假設A,B在拋物線\Gamma:y=mx^2-1上，且A,B對稱於直線L_1:x+y=0;\\因此A,B同在直線L_2:y=x+k,其中L_1\bot L_2;\\ 先求\Gamma與L_2的交點(即A與B),即mx^2-1=x+k \Rightarrow mx^2-x-1-k=0\cdots(1)\\ 令a,b為式(1)的解，即\cases{A(a,a+k)\\ B(b,b+k)} \Rightarrow A,B的中點C({a+b\over 2},{a+b\over 2}+k)在L_1上\\\Rightarrow {a+b\over 2}+{a+b\over 2}+k=0 \Rightarrow k=-(a+b)=-1/m(\because 式(1)兩根之和=a+b=1/m)\\ 再由式(1)有兩相異實根\Rightarrow 判別式>0 \Rightarrow 1+4m(k+1) > 0 \Rightarrow 1+4m(-{1\over m}+1) > 0 \\ \Rightarrow 1-4+4m >0 \Rightarrow \bbox[red,2pt]{m > {3\over 4}}$$

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解： $${a\over \sin A} ={b\over \sin B} ={c\over \sin C} =k \Rightarrow \cases{a=k\sin A\\ b=k\sin B\\ c=k\sin C} \Rightarrow \overline{AB}^2- \overline{AC}^2 =\overline{AB}\cdot \overline{BC} \equiv c^2-b^2=ac \\ \Rightarrow k^2\sin^2 C-k^2\sin ^2B=k^2\sin A\sin C \Rightarrow \sin A\sin C=\sin^2C-\sin^2B =(\sin C +\sin B)(\sin C-\sin B) \\ =2\sin{C+B\over 2}\cos {C-B\over 2}\cdot 2\cos{C+B\over 2}\sin{C-B\over 2} =\sin(C+B)\sin(C-B) = \sin A\sin(C-B) \\ \Rightarrow \sin C=\sin (C-B) \Rightarrow 180^\circ-C=C-B \Rightarrow C=90^\circ+B/2 \\\Rightarrow A+B+C =30^\circ +B+90^\circ +B/2=180^\circ \Rightarrow {3\over 2}B=60^\circ \Rightarrow B=\bbox[red,2pt]{40^\circ}$$

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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： $$令\cases{S1:甲(1,1),乙(2,2)\\ S2:甲(1,2),乙(1,2)\\ S3:甲(2,2,),乙(1,1)} \Rightarrow \cases{P(S_1\to S_1)=0,P(S_1\to S_2)=1, P(S_1\to S_3)=0\\ P(S_2\to S_1)=1/4, P(S_2\to S_2)=1/2,P(S_2\to S_3)=1/4\\ P(S_3\to S_3)=0, P(S_3 \to S_2)=1, P(S_3\to S_3)=0 }\\ \Rightarrow T=\begin{bmatrix}0 & 1& 0 \\ 1/4& 1/2 & 1/4 \\ 0 & 1 & 0\end{bmatrix} \Rightarrow T^5=\begin{bmatrix}5/32 & 11/16& 5/32 \\ &無所謂 \\ &無所謂\end{bmatrix} \\ \Rightarrow P(S_1)+P(S_3) = {5\over 32} +{5\over 32} =\bbox[red,2pt]{5\over 16}$$

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解：

本題$\bbox[red,2pt]{送分}$

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解： $$\cases{f(1)=-3\\ f(-2)=6 \\ f(3)=-9} \Rightarrow 令g(x)=f(x)+3x,則1,-2,3為g(x)=0之三根 \Rightarrow g(x)=a(x-1)(x+2)(x-3)\\ 又f(5)=153 \Rightarrow g(5)=a\cdot 4\cdot 7\cdot 2=56a=f(5)+3\cdot 5=168 \Rightarrow a=3\\ \Rightarrow g(0)=f(0)+3\cdot 0=a\cdot (-1)\cdot 2\cdot (-3)=18 \Rightarrow f(0)=\bbox[red,2pt]{18}$$

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解：

$$\left\{\matrix{y=\log_x 5\\ y=\log_5 x }\right\} 交點:\cases{A(5,1)\\ E(1/5,-1)} \\\left\{\matrix{y=\log_x {1\over 5}\\ y=\log_{1/5} x }\right\} 交點:\cases{B(5,-1)\\ D(1/5,1)} \\ \left\{\matrix{y=\log_5 x\\ y=\log_{1/5} x }\right\} 交點:C(1,0)\\ \Rightarrow 共\bbox[red,2pt]{5}個交點$$

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解： $$\alpha,\beta 為x^2-x-1=0之二根\Rightarrow \cases{\alpha+\beta =1 \\ \alpha\beta=-1}\Rightarrow  (\alpha^n-\beta^n)(\alpha+\beta) =\alpha^{n+1} +\alpha^n\beta- \alpha\beta^n -\beta^{n+1}\\ =\alpha^{n+1}-\beta^{n+1} +\alpha\beta(\alpha^{n-1}- \beta^{n-1})  \Rightarrow  \alpha^n-\beta^n=\alpha^{n+1}-\beta^{n+1}-(\alpha^{n-1}- \beta^{n-1}) \\ \Rightarrow \alpha^{n+1}-\beta^{n+1}= (\alpha^n-\beta^n) +(\alpha^{n-1}- \beta^{n-1}) \equiv a_{n+1}=a_n+a_{n-1}, 其中a_n=\alpha^n-\beta^n\\ 因此a_{2019}=a_{2018}+a_{2017}=(a_{2017}+a_{2016})+ (a_{2016}+a_{2015}) =a_{2017}+2a_{2016}+a_{2015}\\ =a_{2017}+2(a_{2018}-a_{2017} )+a_{2015} =2a_{2018}-a_{2017}+a_{2015} =2a_{2018}-(a_{2019}-a_{2018})+a_{2015} \\ =3m+n-a_{2019}\Rightarrow 2a_{2019}=3m+n \Rightarrow a_{2019}= \bbox[red,2pt]{3m+n\over 2}$$

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解：

$$\cases{A(6,13)\\ B(12,11)} \Rightarrow \cases{A,B的中點C(9,12)\\ \overrightarrow{AB}=(6,-2)}\Rightarrow \overline{AB}的中垂線L: 6(x-9)-2(y-12)=0 \Rightarrow 3x-y=15\\ \Rightarrow L交X軸於C(5,0);又圓心O在L上\Rightarrow O(t,3t-15) \Rightarrow 圓半徑r=\overline{OA} =\overline{OB} \Rightarrow \overline{OC}^2= \overline{OA}^2+ \overline{AC}^2\\ \Rightarrow (t-5)^2+(3t-15)^2 = (t-6)^2+(3t-28)^2 +(6-5)^2+13^2 \\ \Rightarrow -100t+250=-180t+990 \Rightarrow t=37/4 \Rightarrow 圓面積=\overline{OA}^2\pi = ((37/4-6)^2+(111/4-28)^2)\pi \\ =((13/4)^2+(1/4)^2)\pi=\bbox[red,2pt]{{85\over 8}\pi}$$

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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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解：

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

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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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解：

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。

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

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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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解：

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

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