113年景美女中教甄-數學詳解

臺北市立景美女子高級中學113學年度第1次教師甄選數學科試題

一、 填充題： (每題6 分，共72 分)

解答: $$(A)\times: g'(0)不存在\\(B) \bigcirc: 依均值定理\\ (C)\times: f(x)=x^4 \Rightarrow \cases{f'(0)=0\\ f(0)為極小值}, 但f''(x)= 12x^2 \Rightarrow f''(0)=0 \not \gt 0\\ (D)\bigcirc: f(x)\ge 0 \Rightarrow 圖形凹向上\Rightarrow f(x)為偶數次\\ (E)\times: f(x)=x^3為嚴格遞增,但f'(x)=3x^2 \Rightarrow f'(0)=0 \not \gt 0\\故選\bbox[red, 2pt]{(BD)}$$

解答: $$L_1:{x\over 1}={y\over 2}= {z\over 2} \Rightarrow L_1的方向量\vec u=(1,2,2)\\若P\in L_1 \Rightarrow P(t,2t,2t) \Rightarrow \overrightarrow{AP}=(t-8,2t+11,2t+2) \Rightarrow \overrightarrow{AP} \cdot \vec n=0 \Rightarrow t=-2\\ \Rightarrow A在L_1上的垂足P(-2,-4,-4)； 同理可求A在L_2上的垂足Q(1,-1,0) \\ 因此\cases{A對稱L_1的對稱點A'(-12,3,-6) \in \overleftrightarrow{BC} \\A對稱L_2的對稱點A''(-6,9,2) \in \overleftrightarrow{BC}} \Rightarrow \overleftrightarrow{BC} =\overleftrightarrow{A'A''} :{x+12\over 6}={y-3\over 6}={z+6\over 8} \\ \Rightarrow \overleftrightarrow{BC} : \bbox[red, 2pt]{{x+12\over 3}={y-3\over 3}={z+6\over 4}}$$

解答:

$$假設\cases{P在平面E的投影點Q，對稱點P' \Rightarrow \overline{PQ} =\overline{QP'}=3 \\P在平面F的投影點R，對稱點P'' \Rightarrow \overline{PR} =\overline{RP''}=4 } \\ 又\angle QOR=60^\circ \Rightarrow \angle QPR=120^\circ \Rightarrow \cos 120^\circ={6^2+8^2-\overline{P'P''}^2\over 2\cdot 6\cdot 8} \\ \Rightarrow \overline{P'P''}= \bbox[red, 2pt]{2\sqrt{37}}, 此時\cases{ \overline{PA}= \overline{P'A}\\ \overline{PB}= \overline{P''B}} \Rightarrow \triangle PAB周長=\overline{P'P''}$$

解答:

$$f(x)=\sqrt{x^4-x^2-6x+10} -\sqrt{x^4-3x^2+4} \\\qquad =\sqrt{(x^2-1)^2 +(x-3)^2} -\sqrt{(x^2-2)^2+(x-0)^2} =\overline{PA}-\overline{PB},其中\cases{P(x,x^2)\\ A(3,1)\\ B(0,2)}\\ 由於 \overline{PA}-\overline{PB} \le \overline{AB} =\sqrt{3^2+1^2}=\sqrt{10} \Rightarrow 最大值 = \bbox[red, 2pt]{\sqrt{10}}$$

解答: $$\textbf{Case I }k=0: -x+1=0 \Rightarrow x=1\in \mathbb Z\\ \textbf{Case II }k\ne 0:k(x^2+x+1)=x-1 \Rightarrow k={x-1\over x^2+x+1} \cdots(1)\\\qquad 假設\alpha, \beta為kx^2+(k-1)x+(k+1)=0的兩根\Rightarrow \alpha+\beta ={1-k\over k} ={1\over k}-1 \in \mathbb Z\\\qquad 將(1)代入上式\Rightarrow \alpha+\beta ={x^2+x+1\over x-1}-1 ={x^2+2\over x-1} \in \mathbb Z \Rightarrow \cases{x-1\mid x^2+2 \\x-1\mid x^2-1} \\ \qquad \Rightarrow x-1\mid 3 \Rightarrow \cases{x=2,4 \Rightarrow k=1/7\\x=0,-2 \Rightarrow k=-1}\\ \Rightarrow k=\bbox[red, 2pt]{-1,0,{1\over 7}}$$

解答: $$(x^3y^4)^3=x^9y^{12}=(xy^2)^5\cdot (x^2y)^2 \ge 81\cdot 243=3^{30} \Rightarrow x^3y^4 \ge 3^{10} \\ \Rightarrow x^3y^4的最小值為3^{10},此時\cases{xy^2=81\\ x^2y=243} \Rightarrow {y\over x}={1 \over 3} \Rightarrow x=3y \Rightarrow 3y^3=81\\ \Rightarrow y=3 \Rightarrow x=9 \Rightarrow \cases{p=9\\ q=3\\ m=3^{10}} \Rightarrow {m\over pq}={3^{10} \over 3^2\cdot 3} =3^7= \bbox[red, 2pt]{2187}$$

解答: $$O為重心\Rightarrow \overrightarrow{OA} +\overrightarrow{OB} +\overrightarrow{OC}=0 \Rightarrow (\overrightarrow{OA} +\overrightarrow{OB} +\overrightarrow{OC}) \cdot (\overrightarrow{OA} +\overrightarrow{OB} +\overrightarrow{OC})=0\\ \Rightarrow 2+5+9 +2(\overrightarrow{OA} \cdot \overrightarrow{OB} +\overrightarrow{OB}\cdot \overrightarrow{OC}+ \overrightarrow{OC}\cdot \overrightarrow{OA}) =0\\ \Rightarrow 16+2(\overrightarrow{OA} (\overrightarrow{OB} +\overrightarrow{OC})+ \overrightarrow{OB}\cdot \overrightarrow{OC})=16+2(-(\sqrt 2)^2 +\overrightarrow{OB} \cdot \overrightarrow{OC}) =0 \\ \Rightarrow \overrightarrow{OB}\cdot \overrightarrow{OC}=-6\\ 若\cases{z_2=a+bi\\ z_3=c+di} \Rightarrow \overline{Z_2}Z_3 =(a-bi)(c+di)=ac+bd+(ad-bc)i \\ \Rightarrow Re(\overline{Z_2}Z_3)= ac+bd =\overrightarrow{OB}\cdot \overrightarrow{OC} =\bbox[red, 2pt]{-6}$$

解答: $$只需考慮次數小於等於4的情形:\\(1+2x+3x^2+4x^3+ 5x^4+ \cdots)^4=(\cdots+ 35x^4+20x^3+10x^2+4x+1)^2 \\ \Rightarrow x^4係數=35 \cdot 1+20\cdot 4+10\cdot 10+4\cdot 20+1\cdot 35 =\bbox[red, 2pt]{330}$$

解答: $$\cases{C(1,1)\\ A(1+3\cos \theta, 1+4\sin \theta)\\ B(1-3\cos \theta, 1+4\sin \theta)} \Rightarrow S_{\triangle ABC} ={1\over 2} \begin{Vmatrix}1& 1& 1\\ 1+3\cos \theta & 1+ 4\sin \theta& 1\\ 1-3\cos \theta& 1+4\sin \theta& 1 \end{Vmatrix} =6\sin 2\theta \le \bbox[red, 2pt]{6}$$

解答: $$假設四位數字分別為a,b,c,d，而千位數字a不能為0，至少為1，因此有H^4_{18}=1330\\ 需扣除\cases{a\ge 10:H^4_9=220\\ b\ge 10: H^4_8=165 \\c\ge 10: H^4_8=165 \\d\ge 10: H^4_8=165 } \Rightarrow 1330-220-165\times 3= \bbox[red, 2pt]{615}$$

解答 : $$x^4-3x^3-6x^2+ax-24=(x-\alpha)^3(x-\beta) \Rightarrow \cases{3\alpha+\beta=3 \cdots(1)\\ 3\alpha\beta+3\alpha^2= -6 \cdots(2) \\ \alpha^3\beta=-24  \cdots(3)} \\(1)\Rightarrow \beta=3-3\alpha代入(2) \Rightarrow 2\alpha^2-3\alpha-2=0 \Rightarrow (2\alpha+1)(\alpha-2)=0 \\ \Rightarrow \cases{\alpha=2 \Rightarrow \beta=-3 \Rightarrow \alpha^3\beta= -24符合(3)\\ \alpha=-1/2 \Rightarrow \beta=9/2 \Rightarrow \alpha^3\beta=-9/16 \ne -24} \\ \Rightarrow a=-(\alpha^3+ 3\alpha^2 \beta)=-(8-36)= \bbox[red, 2pt]{28}$$

解答: $$\cases{d\gt 0 \Rightarrow 相交二點或四點\\ d=2 \Rightarrow 相交一點或三點,其中一點為(0,2)\\ d\lt 2 \Rightarrow 最多相交二點} \Rightarrow d=2\\y=cx^2+2代入橢圓方程式 \Rightarrow 4x^2+9(cx^2+2)^2=36 \\ \Rightarrow 9c^2x^4+(4+36c)x^2 =0 \Rightarrow x^2(9c^2x^2+36c+4)=0 \\ \Rightarrow 9c^2x^2+36c+4=0 \Rightarrow x^2=-{4(9c+1)\over 9c^2} \gt 0 \Rightarrow 9c+1\lt 0 \Rightarrow \bbox[red, 2pt]{c\lt -{1\over 9}}$$

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解題僅供參考，其他歷年試題及詳解