108年新北市立高中教甄聯招-數學詳解

新北市立高級中等學校108學年度
教師聯合甄選數學科試題

一： 共 40 題，總分 100 分，每題 2.5 分

解：

$$A=旋轉120^\circ 矩陣 = \left[\matrix{\cos 120^\circ & -\sin 120^\circ \\ \sin 120^\circ & \cos 120^\circ} \right] = \left[\matrix{-{1\over 2} & -{\sqrt 3\over 2} \\ {\sqrt 3\over 2} & -{1\over 2}} \right] \\ \Rightarrow A\left[\matrix{2 \\ 8\sqrt 3} \right] =\left[\matrix{-1-12 \\ \sqrt 3-4\sqrt 3} \right] =\left[\matrix{-13 \\ -3\sqrt 3 } \right] \Rightarrow A\left[\matrix{-13 \\ -3\sqrt 3 } \right]  =\left[\matrix{{13\over 2}+{9\over 2} \\ -{13\sqrt 3\over 2}+{3\sqrt 3\over 2} } \right] =\left[\matrix{11\\ -5\sqrt 3 } \right] \\ \Rightarrow 另兩腿所對應的複數為\bbox[red, 2pt]{\cases{(-13,-3\sqrt 3 i)\\ (11,-5\sqrt 3 i}}$$

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解：

$$假設\cases{a:被乘數 \\b:乘數(個位數字是d) \\c:積}，又\cases{8\times a是三位數\\  d\times a是四位數} \Rightarrow a\ge 112;\\但c只能是四位數，因此a=112 \Rightarrow c=\bbox[red, 2pt]{9968}$$

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解：

$$令f(x)=ax^2+bx+c,由y截距為1 \Rightarrow c=1,即f(x)=ax^2+bx+1;\\又f(x-2)=f(-x-2) \Rightarrow a(x-2)^2+b(x-2)+1 =a(-x-2)^2 +b(-x-2)+1 \\ \Rightarrow ax^2-4ax+4a+bx-2b+1=ax^2+4ax+4a-bx-2b+1\\ \Rightarrow 2bx=8ax \Rightarrow b=4a \Rightarrow f(x)=ax^2+4ax+1\\ x軸截線長為2\sqrt 2,由此可假設f(x)=0的兩根為\alpha,\alpha+2\sqrt 2 \\\Rightarrow 兩根之和=-{4a\over a}=-4 =2\alpha +2\sqrt 2 \Rightarrow \alpha = -2-\sqrt 2 \Rightarrow 另一根為-2+\sqrt 2\\ \Rightarrow 兩根之積={1\over a} =(-2+\sqrt 2) (-2-\sqrt 2)=2 \Rightarrow a={1\over 2} \Rightarrow \bbox[red, 2pt]{f(x)={1\over 2}x^2+2x+1}$$

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解：

$$\cases{f(x,y,z)=\sin x\sin y\sin z\\ g(x,y,z)=\tan x\tan y\tan z-1} ，利用\text{Lagrange }算子\Rightarrow \cases{{\partial \over \partial x}(f+\lambda g)=0 \\ {\partial \over \partial y}(f+\lambda g)=0 \\ {\partial \over \partial z}(f+\lambda g)=0 \\g=0 } \\ \Rightarrow \cases{\cos x\sin y\sin z+\lambda \sec^2x\tan y\tan z=0 \\ \sin x\cos y\sin z+\lambda \tan x\sec^2 y\tan z=0 \\ \sin x\sin y\cos z+\lambda \tan x\tan y\sec^2 z=0 \\ \tan x\tan y\tan z=1\cdots(1)} \Rightarrow \cases{\cos x+{\lambda \over \cos^2x\cos y\cos z}=0 \\ \cos y+{\lambda \over \cos x\cos^2 y\cos z}=0 \\ \cos z+{\lambda \over \cos x\cos y\cos^2 z}=0 }\\ \Rightarrow \cases{\cos^3 x\cos y\cos z=-\lambda \\ \cos x\cos^3 y\cos z=-\lambda \\\cos x\cos y\cos^3 z=-\lambda } \Rightarrow \cases{{\cos^ 2 x\over \cos ^2y}=1 \\ {\cos^ 2 y\over \cos ^2z}=1 \\{\cos^ 2 z\over \cos ^2x}=1 } \Rightarrow \cos^2 x = \cos^2 y= \cos^2 z \\ \Rightarrow 1-\cos^2 x = 1-\cos^2 y= 1-\cos^2 z\Rightarrow \sin^2 x = \sin^2 y= \sin^2 z \\ 由(1) \Rightarrow \tan^2 x\tan^2 y\tan^2 z=1 \Rightarrow \tan^2 x=\tan^2 y=\tan^2 z = a \\ \Rightarrow \tan^2 x\tan^2 y\tan^2 z=a^3=1 \Rightarrow a=1 \Rightarrow \tan x=\tan y=\tan z = \pm 1 \\\Rightarrow x=y=z=45^\circ 時f有最大值{\sqrt 2\over 2} \times{\sqrt 2\over 2} \times {\sqrt 2\over 2} = \bbox[red,2pt]{\sqrt 2\over 4}$$

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解：

$$由上表可知，7的次方再除以100取餘數可得:(7、49、43、1)四個數字循環\\，2019=4\times 506+3 \Rightarrow 第3個循環數為\bbox[red, 2pt]{43}$$

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解：

此題相當於求正五邊形扣除中間交集區域(R)的面積，區域R是由一個小的正五邊形與五個弓形所組成的。因此我們先求$P,Q$兩點的坐標，就可以知道小正五邊形的邊長，即$\overline{PQ}$長度。

$$\cases{\overline{AP}=1 \\ \overline{BP}=1 \\ \overline{AB}=1 } \Rightarrow \triangle PAB為正\triangle \Rightarrow \angle PAB=60^\circ；同理，\angle QAE=60^\circ \\\Rightarrow \angle PAQ = \angle QAE +\angle PAB- \angle A =60^\circ +60^\circ-108^\circ  = 12^\circ \Rightarrow \angle QAB =60^\circ-12^\circ=48^\circ\\\triangle APQ\Rightarrow  \cos 12^\circ= {1+1-\overline{PQ}^2 \over 2} \Rightarrow \overline{PQ}^2 =2-2\cos 12^\circ\\\cases{ \triangle APQ面積={1\over 2}\times \overline{AP}\times \overline{AQ}\times \sin \angle PAQ= {1\over 2}\sin 12^\circ \\ 扇形APQ面積= \overline{AP}^2\pi \times {12\over 360}= {\pi\over 30}} \Rightarrow 弓形= {\pi\over 30}- {1\over 2}\sin 12^\circ \\ \Rightarrow R面積= 小正五邊形+5個弓形 ={5\over 4}\overline{PQ}^2\tan 54^\circ +5({\pi\over 30}- {1\over 2}\sin 12^\circ) \\ = {5\over 4}\left(2-2\cos 12^\circ\right)\tan 54^\circ +{\pi \over 6} -{5\over 2}\sin 12^\circ\\ 依題意所求面積=正五邊形-R面積 ={5\over 4}\tan 54^\circ-{5\over 4}\left(2-2\cos 12^\circ\right)\tan 54^\circ -{\pi \over 6} +{5\over 2}\sin 12^\circ \\ ={5\over 4}\tan 54^\circ(2\cos 12^\circ-1)+{5\over 2}\sin 12^\circ -{\pi \over 6} = {5\over 4}\cot 36^\circ(2\cos 12^\circ-1)+{5\over 2}\sin 12^\circ -{\pi \over 6}\\={5\over 4}{\cos 36^\circ \over \sin 36^\circ}(2\cos 12^\circ-1)+{5\over 2}\sin 12^\circ -{\pi \over 6} ={5\over 4} \times {2\cos 24^\circ-\cos 36^\circ \over \sin 36^\circ}-{\pi \over 6} \\ = \bbox[red, 2pt]{{5\sqrt 3\over 4}-{\pi \over 6}}$$

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解：

$$假設長寬高分別為a,b,c,則各頂點坐標如上圖;\\\cases{\overrightarrow{EH}=(0,b,0) \\ \overrightarrow{DF}=(a,-b,c)\\ \overrightarrow{DH}=(0,0,c)} \Rightarrow \vec n_1 =  \overrightarrow{EH}\times \overrightarrow{DF}=(-c,0,a) \\\Rightarrow 兩歪斜線\overline{DF}與\overline{EH}的距離 = {\overrightarrow{DH} \cdot \vec n_1\over |\vec n_1|} ={ac\over \sqrt {a^2+c^2}} \\ 同理,兩歪斜線\overline{DF}與\overline{AB}的距離 = {bc\over \sqrt {b^2+c^2}},\overline{DF}與\overline{CG}的距離 = {ab\over \sqrt {a^2+b^2}}; \\因此由題意知:\cases{{ab\over \sqrt{a^2+b^2}}=2\sqrt 5 \\{bc\over \sqrt{b^2+c^2}}={30\over \sqrt{13}}\\ {ac\over \sqrt{a^2+c^2}} ={15\over \sqrt {10}} } \Rightarrow \cases{{a^2b^2 \over a^2+b^2}=20 \cdots(1)\\{b^2c^2 \over b^2+c^2}={900\over 13} \cdots(2) \\ {a^2c^2\over  a^2+c^2} ={45\over 2} \cdots(3)} ,由(1)及(2)可得 \cases{a^2={20b^2\over b^2-20} \\ c^2={900b^2 \over 13b^2-900}} 帶入(3)\\ \Rightarrow {{18000b^4 \over (b^2-20)(13b^2-900)}\over {20b^2\over b^2-20}+{900b^2 \over 13b^2-900}}={18000b^4 \over 1160b^4-36000b^2}={45\over 2} \Rightarrow 16200b^4-1620000b^2=0 \Rightarrow b^2(b^2-100)=0 \\\Rightarrow b=10 \Rightarrow \cases{a^2={20b^2\over b^2-20} ={2000\over 80}=25\\ c^2={900b^2 \over 13b^2-900} ={90000 \over 400}=225} \Rightarrow \cases{a=5 \\b=15} \Rightarrow 體積=abc= \bbox[red, 2pt]{750}$$

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解：

$$將坐標化為球坐標，即\cases{x=r\sin \theta \cos \varphi\\ y= r\sin\theta \sin \varphi\\ z= r\cos \theta} 代入x^2+y^2-2z^2=0 \Rightarrow r^2\sin^2 \theta=2r^2\cos^2 \theta \\ \Rightarrow \tan^2 \theta =2 \Rightarrow \tan \theta =\sqrt 2 \Rightarrow \cases{\sin \theta=\sqrt 2/3\\ \cos \theta = 1/3} \Rightarrow \cases{x={\sqrt 2\over 3}r \cos \varphi\\ y= {\sqrt 2\over 3}r \sin \varphi\\ z= {1\over 3}r } 代入x+y-3z=5 \\ \Rightarrow {\sqrt 2\over 3}r(\cos \varphi+\sin \varphi)-r=5 \Rightarrow r={5\over {\sqrt 2\over 3}(\cos \varphi+\sin \varphi)-1} \Rightarrow 要使|r|最大,\varphi =\pi/4\\ \Rightarrow r={5\over {\sqrt 2\over 3}(\sqrt 2/2+\sqrt 2/2)-1} =-15 \Rightarrow \cases{x={\sqrt 2\over 3}\cdot (-15) \cdot{\sqrt 2\over 2}= -5\\ y= {\sqrt 2\over 3}\cdot (-15) \cdot{\sqrt 2\over 2} =-5\\ z= {1\over 3}\cdot (-15)=-5 } \\\Rightarrow 最遠的點\bbox[red, 2pt]{(-5,-5,-5)}$$

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解：

$$x^3+3x^2+px-q=0的三根a-d,a,a+d成等差\\\Rightarrow \cases{(a-d)+a+(a+d)=-3\cdots(1) \\ (a-d)a(a+d)=q \cdots(2) \\ a(a-d)+a(a+d)+(a-d)(a+d)=p \cdots (3)}\\由(1) \Rightarrow 3a=-3 \Rightarrow a=-1 代入(2)及(3) \Rightarrow \cases{(1+d)(d-1)=q \\ 3-d^2=p} \\\Rightarrow d^2=q+1=3-p\Rightarrow p+q=2 \cdots(7)\\ 又x^3+(2-p)x^2-(q+3)x-8=0的三根\alpha,\alpha r,\alpha r^2成等比 \\\Rightarrow \cases{\alpha\cdot \alpha r\cdot \alpha r^2 =8 \\ \alpha+\alpha r+\alpha r^2=p-2 \\ \alpha^2 r+ \alpha^2 r^3 + \alpha^2 r^2=-q-3} \Rightarrow \cases{\alpha^3 r^3 \Rightarrow \alpha r=2 \cdots(4)\\ \alpha(1+r+r^2)=p-2 \cdots(5)\\ \alpha^2r (1+r+r^2)=-q-3 \cdots(6)} \\ \Rightarrow {(5)\over (6)} = {1\over \alpha r} ={p-2 \over -q-3} ={1\over 2} \Rightarrow 2p+q=1 \cdots(8) \\ 由(7)及(8) \Rightarrow \cases{p+q=2 \\ 2p+q=1} \Rightarrow (p,q)=\bbox[red, 2pt]{(-1,3)}$$

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解：

$${1\over \cos^2 10^\circ} = {4\sin^2 10^\circ\over (2\cos 10^\circ \sin 10^\circ)(2\cos 10^\circ \sin 10^\circ)} ={4\sin^2 10^\circ \over \sin^2 20^\circ} = {16\sin^2 10^\circ \cos^2 20^\circ \over (2\sin 20^\circ \cos 20^\circ)(2\sin 20^\circ \cos 20^\circ)}\\\qquad = {16\sin^2 10^\circ \cos^2 20^\circ \over \sin^2 40^\circ } \\ {1\over \sin^2 20^\circ} = {4\cos^2 20^\circ\over (2\sin 20^\circ \cos 20^\circ)(2\sin 20^\circ \cos 20^\circ)} = {4\cos^2 20^\circ\over \sin^2 40^\circ  }\\ 原式:{1\over \cos^2 10^\circ} +{1\over \sin^2 20^\circ} +{1\over \sin^2 40^\circ} = {16\sin^2 10^\circ \cos^2 20^\circ +4\cos^2 20^\circ+1\over \sin^2 40^\circ }\\ ={4(1-\cos 20^\circ)(1+ \cos 40^\circ) +2(1+\cos 40^\circ)+1\over \sin^2 40^\circ }  ={7-4\cos 20^\circ+6\cos 40^\circ -4\cos 20^\circ \cos 40^\circ\over \sin^2 40^\circ } \\= {7-4\cos 20^\circ+6\cos 40^\circ -2(\cos 20^\circ +\cos 60^\circ)\over \sin^2 40^\circ }= {7-4\cos 20^\circ+6\cos 40^\circ -2\cos 20^\circ -1\over \sin^2 40^\circ } \\ = {6-6(\cos 20^\circ-\cos 40^\circ) \over \sin^2 40^\circ } = {6-12\sin 10^\circ\sin 30^\circ \over \sin^2 40^\circ }={6-12\cos 80^\circ\cdot ({1\over 2}) \over \sin^2 40^\circ } ={6-6\cos 80^\circ  \over \sin^2 40^\circ } \\={6(1-\cos 80^\circ)  \over {1\over 2}(1-\cos 80^\circ) }=\bbox[red, 2pt]{12}\\  本題使用的公式:\cases{2\cos^2 \theta = 1+\cos 2\theta\\ 2\sin^2 \theta =1-\cos 2\theta \\ 2\cos \alpha \cos \beta = \cos(\alpha+\beta)+ \cos (\alpha-\beta) \\\cos \alpha -\cos \beta= -2\sin {\alpha +\beta \over 2}\sin {\alpha -\beta \over 2} }$$

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解：

$$X=2: (甲不進、乙不進、甲投進)，(甲不進、乙不進、甲不進、乙投進)\\，機率為{2\over 5} \times {3\over 5} \times {3\over 5} +{2\over 5} \times {3\over 5} \times {2\over 5} \times{2\over 5} = {18\over 125}+ {24\over 625} = \bbox[red, 2pt]{114\over 625}$$

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解：

$$若各色球皆不同，就有\bbox[red,2pt]{{1\over 2}C^{6n}_{3n}}分法$$

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解：

$$，故選\bbox[red, 2pt]{( )}$$

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解：

$$令\cases{\overline{BC}=w\\ \overline{AB}=h \\ \overline{BE}=a \\\overline{CF}=b}，則各頂點坐標如上圖；因此\cases{R=wh \\a_1={1\over 2}ah\\ a_2={1\over 2}w(h-b)\\ \triangle CEF={1\over 2}b(w-a)} \\ \Rightarrow a_3= R-a_1-a_2-\triangle CEF = wh- {1\over 2}(ah+ w(h-b)+b(w-a)) = wh- {1\over 2}(ah+wh-ab) \\ \Rightarrow R^2-2a_3R-4a_1a_2 =w^2h^2-2wh(wh- {1\over 2}(ah+wh-ab)) -ahw(h-b)\\ = w^2h^2-2w^2h^2+wh(ah+wh-ab)-ah^2w+abhw \\ =w^2h^2-2w^2h^2+ah^2w+w^2h^2-abhw-ah^2w+abhw =0 \\ \Rightarrow R^2-2a_3R-4a_1a_2 =0,故得證$$

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