113年中崙高中教甄-數學詳解

臺北市立中崙高級中學 113 學年度第 1 次數學科正式教師甄選

壹、填充題(每題 5 分)

解答:

$$\overline{BC}=2 \Rightarrow \overline{AD}=\sqrt 3 \Rightarrow \overline{AE}={\sqrt 3\over 3} \\ \cos \angle DAB= \cos 30^\circ ={\sqrt 3\over 2}= {\overline{AE}^2 +\overline{AB}^2 - \overline{BE}^2 \over 2\cdot \overline{AE} \cdot \overline{AB}} ={1/3+4-\overline{BE}^2 \over 4\sqrt 3/3} \\ \Rightarrow \overline{BE}^2={7\over 3} \Rightarrow \cos \angle DEB={{4\over 3}+{7\over 3}-1\over {4\over 3}\sqrt 7} ={2\over \sqrt 7} \Rightarrow \sin \angle DEB={3\over \sqrt 7} \\ \Rightarrow \overline{BF}=\overline{BD} \cos \angle DBE= \overline{BD} \sin \angle EDB = 1\cdot {3\over \sqrt 7}= \bbox[red, 2pt]{\sqrt{21}\over 7}$$

解答: $$56\times 5\times 4\times(24+24)+ 56\times 5\times(24+48)= \bbox[red, 2pt]{73920}$$

解答: $$可能n值之和=138+140+141+142+143+145= \bbox[red, 2pt]{849}$$

解答: $$x^2+y^2=1 \Rightarrow x^2=1-y^2 代回曲線\Rightarrow (1-y^2)+2xy+2y^2-2x-2y=0 \\ \Rightarrow y^2-2y+1+2xy-2x=0 \Rightarrow (y-1)^2+2x(y-1)=0 \Rightarrow (y-1)(y-1+2x)=0 \\ \Rightarrow \cases{y=1 \Rightarrow x=0\\ y=1-2x \Rightarrow x(5x-4)=0 } \Rightarrow 交點(x,y)=(0,1), \bbox[red, 2pt]{ ({4\over 5},-{3\over 5})}$$

解答: $$z+w={2\sqrt 3\over 3}i \Rightarrow 取\cases{z=a+({\sqrt 3\over 3}+b)i\\ w=-a+ ({\sqrt 3\over 3}-b)i} \Rightarrow z^2+w^2=2a^2-{2\over 3}-2b^2+4abi={2\sqrt 3\over 3}i \\ \Rightarrow \cases{a^2-b^2={1\over 3}\\ ab={\sqrt 3\over 6}} \Rightarrow a^2-({\sqrt 3\over 6a})^2 ={1\over 3} \Rightarrow 12a^4-4a^2-1=0 \Rightarrow (6a^2+1)(2a^2-1)=0 \\ \Rightarrow a^2={1\over 2} \Rightarrow a= \bbox[red, 2pt]{\sqrt 2\over 2}$$

解答: $$(3^x)^2-2(m+1)\cdot 3^x-(m-1)=0 \Rightarrow 3^x=(m+1)\pm \sqrt{m^2+3m} \\由於\cases{兩相異實根 \Rightarrow m^2+3m\gt 0 \Rightarrow m\gt 0或m\lt -3\\ 3^x \gt 0 \Rightarrow m+1\gt \sqrt{m^2+3m} \Rightarrow m\lt 1} \Rightarrow \bbox[red, 2pt]{0\lt m\lt 1}$$

解答:

$$圓C:(x-2)^2+(y-2)^2=8-k \Rightarrow \cases{圓心M(2,2)\\ 圓半徑r=\sqrt{8-k}}\\ A(-3,3)對稱於x軸的對稱點A'(-3,-3) \Rightarrow L=\overleftrightarrow{A'M}:x=y \Rightarrow L為\angle PA'Q的角平分線\\ \Rightarrow L\cap x軸=O(0,0) \Rightarrow \cases{\overline{PO} =a \\ \overline{QO}=7/4-a},其中\cases{ P(-a,0), \\ Q(-a+{7\over 4},0)} \\ \Rightarrow {\overline{A'P} \over \overline{A'Q} }={\overline{PO} \over \overline{QO}} \Rightarrow {\sqrt{(-3+a)^2+9} \over \sqrt{(a-19/4)^2+9}} ={a\over 7/4-a} \Rightarrow 4a^2-31a+21=0 \Rightarrow a=3/4(a=7不合)\\ \Rightarrow P(-{3\over 4},0) \Rightarrow L_2=\overleftrightarrow{A'P}:  3y=4x+3 \Rightarrow r=d(M,L_2)= 1=\sqrt{8-k} \Rightarrow k=\bbox[red, 2pt]7$$

解答: $$\cases{A在L_1:x+2y=0上 \Rightarrow A(-2a,a)\\ B在L_2:x+y=0上 \Rightarrow B(b,-b)} \Rightarrow 交點為重心G(0,0) \Rightarrow C(2a-b,-a+b) \\ \Rightarrow \cases{\overrightarrow{CA}= (-4a+b,2a-b)\\ \overrightarrow{CB}= (-2a+2b,a-2b)} \Rightarrow \overrightarrow{CA} \cdot \overrightarrow{CB}=0 \Rightarrow 10a^2-15ab+4b^2=0 \cdots(1)\\ 又\overline{AB}=36 \Rightarrow (-2a-b)^2+(a+b)^2=36^2 \Rightarrow 5a^2+6ab+2b^2=36^2 \cdots(2)\\ 由(1)及(2)可得ab={2\cdot 36^2\over 27} \Rightarrow  \triangle ABC面積={1\over 2}\begin{Vmatrix} -2a & a& 1\\ b& -b & 1\\ 2a-b& -a+b& 1 \end{Vmatrix}={3\over 2}ab ={3\over 2} \cdot {2\cdot 36^2\over 27}\\ = \bbox[red, 2pt]{144}$$

解答: $$\cos \theta= \left(1+\cos 2\theta \over 2\right)^{1/2} \Rightarrow a_0=\cos {\pi\over 3}={1\over 2} \Rightarrow a_1=\cos{\pi\over 3\cdot 2} = \left(1+(1/2) \over 2\right)^{1/2} ={\sqrt 3\over 2} \\ \Rightarrow a_2=\cos{\pi\over 3\cdot 2^2} \Rightarrow \cdots\Rightarrow  a_n= \cos {\pi \over 3\cdot 2^n} =1-{({\pi \over 3\cdot 2^n})^2\over 2!} +{({\pi \over 3\cdot 2^n})^4 \over 4!} -\cdots \text{ (Taylor series）} \\ \Rightarrow \lim_{n\to \infty} 4^n\cdot (1-a_n) =\lim_{n\to \infty} 4^n\cdot \left( {({\pi \over 3\cdot 2^n})^2\over 2!} -{({\pi \over 3\cdot 2^n})^4 \over 4!} +\cdots \right) =\lim_{n\to \infty} 4^n\cdot {({\pi \over 3\cdot 2^n})^2\over 2!} \\={\pi^2 \over 9}\cdot {1\over 2!}= \bbox[red, 2pt] {\pi^2\over 18}$$

解答: $$3\sin 2x+4\cos 2x=1 \Rightarrow (4\cos 2x)^2= (1-3\sin 2x)^2 \Rightarrow 16\cos^2 2x=1-6\sin 2x+9\sin^2 2x \\ \Rightarrow 16(1-\sin^2 2x)=1-6\sin 2x+9\sin^2 2x \Rightarrow 25\sin^2 2x-6\sin 2x-15=0 \\\Rightarrow \sin 2\alpha \sin 2\beta=-{3\over 5} \cdots(1)\\ 同理,(3\sin 2x)^2=(1-4\cos 2x)^2 \Rightarrow 9\sin^2 2x= 9(1-\cos^2 2x)=1-8\cos 2x+ 16\cos^2 2x \\ \Rightarrow 25\cos^2 2x-8\cos 2x-8=0 \Rightarrow \cos 2\alpha \cos 2\beta=-{8\over 25} \cdots(2)\\ 由(1)及(2)可得 \cos 2(\alpha+\beta)= \cos 2\alpha \cos 2\beta-\sin 2\alpha \sin 2\beta={7\over 25}  \\ \Rightarrow \sin 2(\alpha+\beta)= \sqrt{1-\cos^2 2(\alpha+\beta)} =\sqrt{1-{49\over 625}} = \bbox[red, 2pt]{24\over 25}$$

解答: $$X=7 \Rightarrow 111111排列數=1\\ X=6 \Rightarrow 211111排列數={6!\over 5!}=6\\ X=5\Rightarrow \cases{11113排列數=5!/4!=5\\ 11122排列數= 5!/(3!2!)= 10} \Rightarrow 合計15\\ X=4 \Rightarrow \cases{1114排列數=4!/3!=4\\ 1123排列數=4!/2!=12\\ 1222排列數=4} \Rightarrow 合計20 \\ X=3 \Rightarrow \cases{115排列數=3!/2!=3\\ 124排列數=6\\ 133排列數=3\\ 223排列數=3} \Rightarrow 合計15 \\X=2 \Rightarrow \cases{16 排列數=2\\ 25排列數=2\\ 34排列數=2} \Rightarrow 合計6\\  X=1 \Rightarrow 7排列數=1\\ \Rightarrow E(X)={1\cdot 1+2\cdot 6+ 3 \cdot 15+ 4\cdot 20+ 5\cdot 15+6\cdot 6+ 7\cdot 1\over 1+6+15+20+15+6+1} ={256\over 64} =\bbox[red, 2pt]4$$

解答: $$f(x)-2f({1\over x})-{3\over x}=0 \Rightarrow f({1\over x})={1\over 2}\left( f(x)-{3\over x}\right) \cdots(1)\\又f({1\over x})-2f(x)-3x=0 \Rightarrow f({1\over x})=2f(x)+3x \cdots(2)\\ 由(1)及(2)可得f({1\over x})={1\over 2}\left( f(x)-{3\over x}\right) =2f(x)+3x \Rightarrow f(x)=-2x-{1\over x} \\ 令g(x)=(f(x))^2 =4x^2+{1\over x^2}+4 \Rightarrow g'(x)=0 \Rightarrow x^2={1\over 2} \Rightarrow g(x)=4\cdot {1\over 2}+2+4= \bbox[red, 2pt]8$$

貳、計算證明題(每小題 10 分)

解答: $$x^2+x+1=0的兩根為\omega,\omega^2 且\omega^3=1\\ 由於\cases{(\omega+2)^2=\omega^2+4\omega+4=3\omega+3=3(\omega+1)\\ (\omega+1)^2=\omega^2+2\omega +1=\omega}, 因此f(\omega)=(\omega+2)^{24}-(\omega+1)^{24} \\=(3(\omega+1))^{12} -(\omega+1)^{24} =3^{12}(\omega+1)^{12}-(\omega+1)^{24}=3^{12}\omega^6-\omega^{12}= \bbox[red, 2pt]{ 3^{12}-1}$$

解答: $$a_1=2^{20}-24(偶數) \Rightarrow a_4=2^{17}-3(奇數) \Rightarrow a_5=2^{17}-4 \Rightarrow a_7=2^{15}-1 \\ \Rightarrow a_8=2^{15}-2 \Rightarrow a_9=2^{14}-1 \Rightarrow a_{11}=2^{13}-1 \Rightarrow a_{13}= 2^{12}-1 \Rightarrow a_{33}=2^2-1\\ \Rightarrow a_{35}=2-1=1 \Rightarrow M=\bbox[red, 2pt]{35}$$

解答:

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