國立蘭陽女子高級中學 115 學年度第 1 次教師甄選
解答:$$f(x)=a\sin x+b\cos x+c =\sqrt{a^2+b^2} \sin (x+\theta)+c, 其中\cases{\sin \theta=b/\sqrt{a^2+b^2} \\ \cos \theta=a/\sqrt{a^2+b^2}} \\ \Rightarrow \cases{最大值=\sqrt{a^2+b^2}+c=3\\ 最小值=-\sqrt{a^2+b^2} +c=-1 \\{11\pi\over 6}+\theta=\pi/2} \Rightarrow \cases{\sqrt{a^2+b^2}=2\\c=1 \\ \theta=-8\pi/6} \Rightarrow \cases{\sin \theta= \sqrt 3/2 \\ \cos \theta=-1/2} \Rightarrow \cases{a=-1\\ b=\sqrt 3} \\ \Rightarrow (a+bi)^7=(-1+\sqrt 3i)^7 =(2(\cos {2\pi\over 3}+i\sin{2\pi\over 3}))^7 =2^7 (\cos {140\pi\over 3}+ i\sin{140\pi\over 3}) \\=2^7(-{1\over 2}+{\sqrt 3i\over 2})= \bbox[red, 2pt]{-63+64\sqrt 3i}$$
解答:$$n=1 \Rightarrow p_1={1\over 6} \\n=2 \Rightarrow 第1次是2-5 \Rightarrow p_2={4\over 6}\cdot {1\over 6} \\ n=3\Rightarrow 第1次是2-5,第2次是2-5 \Rightarrow p_3=({4\over 6})^2\cdot {1\over 6} \\ n=k \Rightarrow 前k-1次都是2-5\Rightarrow p_k= ({4\over 6})^{k-1} \cdot {1\over 6} \\ \Rightarrow \lim_{n\to \infty}\sum_{k=1}^n p_k ={1\over 6} \left(1+ {4\over 6}+({4\over 6}^2) +\cdots\right) ={1\over 6} \left( {1\over 1-4/6} \right) =\bbox[red, 2pt]{1\over 2}$$
解答:$$\Gamma_1:y^2=4x \Rightarrow 2yy'=4 \Rightarrow y'={2\over y} \Rightarrow 切點(t^2,2t) \in \Gamma_1 \Rightarrow 切線L_1:y={1\over t}(x-t^2)+2t \\ \Gamma_2: x^2=2y-3 \Rightarrow 2x=2y' \Rightarrow y'=x \Rightarrow 切點(s,{s^2+3\over 2}) \in \Gamma_2 \Rightarrow 切線L_2: y=s(x-s)+{s^2+3\over 2} \\ L_1=L_2 \Rightarrow \cases{斜率:{1\over t}=s\\ 截距:t={-s^2+3\over 2}} \Rightarrow s^3-3s+2=0 \Rightarrow (s-1)^2(s+2)=0 \\ \Rightarrow \cases{s=1 \Rightarrow t=1 \Rightarrow L_1:y=x-1+2\\ s=-2\Rightarrow t=-1/2 \Rightarrow L_2:y=-2(x+2)+7/2} \Rightarrow 公切線: \bbox[red, 2pt]{y=x+1,y=-2x-{1\over 2}}$$
解答:$$f(x)=\sin x(1+\cos x) =\sin x+ {1\over 2}\sin 2x \Rightarrow f'(x)=\cos x+\cos(2x) =\cos x+2\cos^2x-1 \\=(2\cos x-1)(\cos x+1) =0 \Rightarrow \cos x=\cases{1/2 \Rightarrow \cases{x=\pi/3 \Rightarrow \sin x=\sqrt 3/2 \Rightarrow f(x)= 3\sqrt 3/4\\ x=5\pi/3 \Rightarrow \sin x=-\sqrt 3/2 \Rightarrow f(x)=-3\sqrt 3/4} \\ -1 \Rightarrow x=\pi \Rightarrow \sin x=0 \Rightarrow f(x) =0} \\ \Rightarrow (M,m) = \bbox[red, 2pt]{\left( {3\sqrt 3\over 4},-{3\sqrt 3\over 4} \right)}$$
解答:$$3\tan 10^\circ+ 4\sqrt 3 \sin 10^\circ = 3{\sin 10^\circ \over \cos 10^\circ}+ 4\sqrt 3 \sin 10^\circ ={3\sin 10^\circ +4\sqrt 3\sin 10^\circ \cos 10^\circ \over \cos 10^\circ} \\={3\sin 10^\circ +2\sqrt 3\sin 20^\circ \over \cos 10^\circ} ={\sqrt 3(\sqrt3\sin 10^\circ +2 \sin (30^\circ-10^\circ)) \over \cos 10^\circ} \\={\sqrt 3(\sqrt 3\sin 10^\circ+ \cos 10^\circ-\sqrt 3 \sin 10^\circ) \over \cos 10^\circ} ={\sqrt 3\cos 10^\circ\over \cos 10^\circ} = \bbox[red, 2pt]{\sqrt 3}$$
解答:$$\cases{A= \begin{bmatrix}3&-1\\ a&-2 \end{bmatrix} \\[1ex] B= \begin{bmatrix}-2& b\\ -6& 3 \end{bmatrix} } \Rightarrow \cases{AB= \begin{bmatrix}0&3b-3\\ -2a+12& ab-6 \end{bmatrix} \\[1ex] BA= \begin{bmatrix}-6+ab& 2-2b\\ -18+3a& 0 \end{bmatrix}} \Rightarrow AB=BA \Rightarrow \cases{a=6\\ b=1} \\ \Rightarrow \cases{A= \begin{bmatrix}3&-1\\ 6&-2 \end{bmatrix} \\[1ex] B= \begin{bmatrix}-2& 1\\ -6& 3 \end{bmatrix} } \Rightarrow C=A-B = \begin{bmatrix}5& -2\\ 12& -5 \end{bmatrix} \Rightarrow \det(C-\lambda I)= (\lambda-1)(\lambda +1)=0 \Rightarrow \lambda=\pm 1 \\ \Rightarrow C= P \begin{bmatrix}1&0\\0& -1 \end{bmatrix}P^{-1} \Rightarrow C^n =P \begin{bmatrix}1&0\\0& (-1)^n \end{bmatrix}P^{-1} =\cases{PIP^{-1}=I, n是正偶數\\ C, n是正奇數} \\ \Rightarrow \bbox[red, 2pt]{(A-B)^n = \begin{cases} \begin{bmatrix}1&0\\0& 1 \end{bmatrix},& n是正偶數\\ \begin{bmatrix}5&-2\\12&-5 \end{bmatrix}, &n是正奇數\end{cases}}$$
解答:$$\left( x-\sqrt{x-{1\over x}} \right)^2 = \left( \sqrt{1-{1\over x}} \right)^2 \Rightarrow x^2-2x \sqrt{x-{1\over x}}+x-{1\over x}=1-{1\over x} \\ \Rightarrow x^2+x-1=2x\sqrt{x-{1\over x}} \Rightarrow x+1-{1\over x}=2\sqrt{x-{1\over x}} \Rightarrow \left( x-{1\over x} \right)-2\sqrt{x-{1\over x}} +1=0 \\ \Rightarrow \left( \sqrt{x-{1\over x}} -1\right)^2=0 \Rightarrow \sqrt{x-{1\over x}}=1 \Rightarrow x-{1\over x}=1 \Rightarrow x^2-x-1=0 \Rightarrow x={1\pm \sqrt 5\over 2}\\ 由於1-{1\over x} \ge 0 \Rightarrow x\ge 1 \Rightarrow x= \bbox[red, 2pt]{1+\sqrt 5\over 2}$$
解答:
$$\cases{\angle A=90^\circ\\ \overline{AB}=3\\ \overline{AC}=4} \Rightarrow \overline{BC}=5, 又\triangle CPR及\triangle BPQ皆為等腰直角\triangle \Rightarrow \angle CPR=\angle BPQ=45^\circ \\ \angle RPQ = \theta \Rightarrow \tan \theta ={2\over \sqrt 5} \Rightarrow \sin \theta={2\over 3}, 又 \overline{PB}:\overline{PC}=5:3 \Rightarrow \cases{\overline{PC} =3k \\ \overline{PB}=5k}\\ \Rightarrow \cos \angle BPC =\cos (90^\circ+\theta)= {9k^2+25k^2-5^2\over 30k^2} \Rightarrow -\sin \theta=-{2\over 3} ={34k^2-25\over 30k^2} \\ \Rightarrow k^2={75\over 162} \Rightarrow k={5\over 3\sqrt{6}} \Rightarrow \overline{PC}=3k={5\over \sqrt 6} = \bbox[red, 2pt]{5\sqrt 6\over 6}$$
解答:
$$假設\cases{\overline{AD}=a\\ \angle BDO =\angle ODF=\alpha\\ \angle CEO= \angle FEO=\beta} \Rightarrow \cases{\overline{DB}=\overline{DF}=4\\ \overline{EC} =\overline{EF}=3} \Rightarrow \cases{\tan \alpha=5/4\\ \tan \beta=5/3} \\ \Rightarrow 2\beta=\angle A+ \angle ADE =\angle A+180^\circ-2\alpha \Rightarrow \angle A= 2\alpha+ 2\beta -180^\circ \Rightarrow \tan {\angle A\over 2} = \tan (\alpha+ \beta-90^\circ) \\=-\cot(\alpha+ \beta) ={\tan \alpha \cdot \tan \beta-1\over \tan \alpha+\tan \beta} ={(5/4)\cdot (5/3)-1 \over (5/4) +(5/3)} ={13\over 35} ={\overline{OB} \over \overline{AB}} ={5\over \overline{AB}} \Rightarrow \overline{AB}= {175\over 13} \\ \Rightarrow 4+a={175\over 13} \Rightarrow a={123\over 13} \\ 四邊形ABOC面積= 2\triangle ABO= \triangle ADE+ 2\triangle DBO+ 2\triangle EOF \Rightarrow 5(a+4)= \triangle ADE+20+15 \\ \Rightarrow \triangle ADE=5(a-3) =5 \left( {123\over 13}-3 \right) =\bbox[red, 2pt]{240\over 13}$$
解答:$$f(2026)=-{1\over 2025} \Rightarrow f(f(2026)) =f(-{1\over 2025})={2025\over 2026} \Rightarrow f(f(f(2026))) =f({2025\over 2026}) =2026 \\ \Rightarrow \underbrace{f(f(f(f(2026))))}_{4個f} =f(2026) \Rightarrow 循環數3 \Rightarrow 2026=2\times 675+1\\ \Rightarrow \underbrace{f(f( \cdots(f(2026))))}_{2026個f} =f(2026)= \bbox[red, 2pt]{-{1\over 2025}}$$
解答:$$令\cases{u=a+\sqrt{2a-1} \\v=a-\sqrt{2a-1}} \Rightarrow \cases{u+v=2a\\ uv=a^2-2a+1=(a-1)^2} \\ 令k=\sqrt{u}+\sqrt v \Rightarrow k^2=u+v+2 \sqrt{uv} =2a+2(1-a) =2 \Rightarrow k=\sqrt 2\\ \Rightarrow \log_8k= \log_8 \sqrt 2 ={\log_2 \sqrt 2\over \log_2 8} ={1/2\over 3}= \bbox[red, 2pt]{1\over 6}$$
解答:$$x=\sqrt[3]9-\sqrt[3]3 \Rightarrow x^3=9-3-3\cdot \sqrt[3]9\cdot \sqrt[3]3(\sqrt[3]9-\sqrt[3]3) =6-9x \\ \Rightarrow x^5= x^2\cdot x^3=x^2(6-9x )=6x^2-9x^3 \\ \Rightarrow 原式:x^5+8x^3-6x^2-9x+8 =(6x^2-9x^3)+8x^3-6x^2-9x+8 =-x^3-9x+8\\=-(6-9x)-9x+8= \bbox[red, 2pt]2$$
解答:$$假設欲求之平面E，由於\cases{A(1,0,0) \\B(0,2,0)} 皆在xy平面(z=0)上，因此E與xy平面的交線為\overleftrightarrow{AB} \\ \Rightarrow L=\overleftrightarrow{AB} : \cases{2x+y-2=0\\z=0} \Rightarrow E: 2x+y-2+kz=0 \Rightarrow \cases{xy平面法向量\vec n_1=(0,0,1)\\ 平面E法向量\vec n_2=(2,1,k)} \\ \Rightarrow \cos 60^\circ ={\vec n_1\cdot \vec n_2\over |\vec n_1||\vec n_2|} \Rightarrow {1\over 2}={k\over \sqrt{5+k^2}} \Rightarrow {1\over 4}={k^2\over 5+k^2} \Rightarrow 3k^2=5 \Rightarrow k=\pm {\sqrt{15}\over 3} \\ \Rightarrow E:2x+y\pm {\sqrt {15}\over 3}z=2 \Rightarrow \bbox[red, 2pt]{6x+3y \pm \sqrt{15}z=6}$$
解答:$$a=106b \Rightarrow (106b)^b=b^{106b} \Rightarrow 106b=b^{106} \Rightarrow106=b^{105} \Rightarrow b=106^{1/105} \\\Rightarrow a=106\cdot 106^{1/105} =106^{106/105} \Rightarrow ab= 106^{106/105+1/105} =106^{107/105} \Rightarrow \log_{106}(ab)= \bbox[red, 2pt]{107\over 105}$$
解答:
$$假設\cases{D為原點\\ \overleftrightarrow{BC}為x軸} \Rightarrow \cases{A(0,2) \\B(-2,0) \\C(2,0)\\D(0,0) \\E(a,b)} \Rightarrow G=E順時針旋轉90^\circ=(b,-a) \Rightarrow F=E+G=(a+b,b-a)\\ F\in \overleftrightarrow{AC}: x+y=2 \Rightarrow (a+b) +(b-a)=2 \Rightarrow b=1 \Rightarrow \cases{E(a,1)\\G(1,-a)} \\ \overline{BE}= \sqrt 3\cdot \overline{CG} \Rightarrow (-2-a)^2+1=3\cdot (1+(-a)^2) \Rightarrow a^2-2a-1=0 \Rightarrow a=1\pm \sqrt 2 \\E在原點D的左側\Rightarrow a\lt 0 \Rightarrow a=1-\sqrt 2 \Rightarrow 正方形面積=\overline{DE}^2 =(1-\sqrt 2)^2+1^2= \bbox[red, 2pt]{4-2\sqrt 2}$$
$$1-\sqrt{4-y^2}\le x\le 0 \Rightarrow (x-1)^2+y^2 \le 4 且x\le 0 \\ x=0代入圓:(x-1)^2+y^2=4 \Rightarrow y^2=3 \Rightarrow y=\pm \sqrt 3 \Rightarrow 圓與y軸交於\cases{A(0,\sqrt 3) \\B(0,-\sqrt 3)} \\ f(x,y)=3x-y \Rightarrow \cases{f(0,\sqrt 3) =-\sqrt 3\\f(0,-\sqrt 3)=\sqrt 3}, 接下來要找圓與直線L:3x-y=k的切點\\ (x-1)^2+y^2=4 \Rightarrow 2(x-1)+2yy'=0 \Rightarrow y'={1-x\over y} =3 (L斜率為3) \Rightarrow x=1-3y \\ \Rightarrow (-3y)^2+y^2=4 \Rightarrow y^2= {2\over 5} \Rightarrow y=\sqrt{2\over 5} \Rightarrow x=1-3y=1-3\sqrt{2\over 5} \Rightarrow 切點P(1-3\sqrt{2\over 5},\sqrt{2\over 5})\\ \Rightarrow f(1-3\sqrt{2\over 5},\sqrt{2 \over 5}) =3-2\sqrt{10} \Rightarrow (M,m) =\bbox[red, 2pt] {\left( \sqrt 3,3-2\sqrt{10} \right)}$$
解答:$$假設\cases{\vec a= \overrightarrow{OA} \\\vec b= \overrightarrow{OB} \\\vec c= \overrightarrow{OC} } \Rightarrow \cases{|\vec a| =|\vec b|= |\vec c|=R (外接圓半徑)\\ \overrightarrow{BC} =\vec c-\vec b\\ \overrightarrow{CA} = \vec a-\vec c\\ \overrightarrow{AB} =\vec b-\vec a} \\ \Rightarrow 5\overrightarrow{AO} \cdot \overrightarrow{BC} +3\overrightarrow{BO} \cdot \overrightarrow{CA}+8 \overrightarrow{CO} \cdot \overrightarrow{AB} = 5(-\vec a) \cdot (\vec c-\vec b)+ 3(-\vec b)\cdot (\vec a-\vec c)+8(-\vec c)\cdot (\vec b-\vec a) =0 \\ \Rightarrow 2\vec a\cdot \vec b-5\vec b\cdot \vec c+3\vec c\cdot \vec a \Rightarrow 5\vec b\cdot \vec c=3\vec c\cdot \vec a +2\vec a\cdot \vec b =\vec a \cdot (2\vec b+3\vec c) \Rightarrow (\vec a \cdot (2\vec b+3\vec c))^2 \le |\vec a|^2 |2\vec b+3\vec c|^2 \\ \Rightarrow 25(\vec b\cdot \vec c)^2 \le R^2(13R^2+12(\vec b\cdot \vec c)) \\取t=\vec b\cdot \vec c, 則25t^2 \le13R^4+12R^2 t \Rightarrow 25t^2-12R^2t+13R^4\le 0 \Rightarrow (25t+13R^2) (t-R^2) \le 0 \\ \Rightarrow -{13\over 25}R^2 \le t= |\vec b||\vec c| \cos 2A\le R^2 \Rightarrow -{13\over 25}R^2 \le R^2\cos 2A \le R^2 \\ \Rightarrow \cos 2A=2\cos^2A-1 \ge -{13\over 25}\Rightarrow \cos^2A \ge {6\over 25} \Rightarrow \cos A\ge \bbox[red, 2pt]{\sqrt 6\over 5}$$
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