115 學年度臺北市立明倫高中第一次教師甄試
一、填空題： (12 格，每格 5 分，共 60 分，請將答案化到最簡)
解答:
$$\triangle ABC: \cos \angle B =\cos 120^\circ=-{1\over 2}={3^2+5^2-\overline{AC}^2 \over 2\cdot 3\cdot 5} \Rightarrow \overline{AC}=7 \\ \triangle ABC面積={1\over 2}\cdot 3\cdot 5\sin B={15\over 2}\cdot {\sqrt 3\over 2} ={15\sqrt 3\over 4}\\ 假設\cases{\vec a= \overrightarrow{AB} \\ \vec b = \overrightarrow{BC}} \Rightarrow \overrightarrow AC=\vec a+\vec b, \quad 又\cases{\overline{BC} \parallel \overline{AD} \Rightarrow \overrightarrow{AD}=k \vec b, k\gt 0 \\ \overline{AB} \parallel \overline{CE} \Rightarrow \overrightarrow{CE}=-m \vec a, m\gt 0} \\ \Rightarrow \overrightarrow{DE}= \overrightarrow{DA} +\overrightarrow{AC}+ \overrightarrow{CE} =-k \vec b+(\vec a+\vec b)-m\vec a =(1-m)\vec a+(1-k) \vec b \\ \overline{DE} \parallel \overline{AC} \Rightarrow ((1-m)\vec a+(1-k) \vec b ) \parallel (\vec a+ \vec b) \Rightarrow 1-m=1-k \Rightarrow m=k \Rightarrow \overrightarrow{DE}= (1-k)(\vec a+\vec b) \\ 又{\overline{DE} \over \overline{AC}} ={14\over 7}=2 \Rightarrow |1-k|=2 \Rightarrow k=3 \Rightarrow \overrightarrow{AD} =3\vec b\Rightarrow \overrightarrow{AE} =\overrightarrow{AC}+ \overrightarrow{CE} =-2\vec a+ \vec b\\ 令\triangle ABC面積=S={1\over 2}|\vec a\times \vec b|={15\sqrt 3\over 4} \Rightarrow \triangle ABD面積={1\over 2}|\overrightarrow{AB} \times \overrightarrow{AD}| ={1\over 2}|\vec a\times 3\vec b|=3S \\ \Rightarrow \triangle ADE面積={1\over 2}|\overrightarrow{AD} \times \overrightarrow{AE}| ={1\over 2}|3\vec b\times (-2\vec a+\vec b)|= 6S \\ \Rightarrow \triangle ABE面積= {1\over 2}| \overrightarrow{AB} \times \overrightarrow{AE}| ={1\over 2}|\vec a\times (-2\vec a+\vec b)|= S \\\Rightarrow \triangle BDE面積=四邊形ABDE面積-\triangle ABE面積=(3S+6S)-S= 8S =8\cdot {15\sqrt 3\over 4} =\bbox[red, 2pt]{30\sqrt 3}$$
解答:$$ \begin{cases} x+y+z=3 \\ xy+yz+zx=-9 \end{cases} \Rightarrow \cases{x+y=3-z\\ xy=z^2-3z-9} \Rightarrow x,y為t^2-(3-z)t+(z^2-3z-9)=0的兩實根 \\ \Rightarrow \Delta = [-(3-z)]^2 - 4(1)(z^2 - 3z - 9) \ge 0 \Rightarrow z^2-2z-15\le 0 \Rightarrow (z-5)(z+3)\le 0 \\ \Rightarrow -3\le z\le 5 \Rightarrow (M,m) = \bbox[red, 2pt]{(5,-3)}$$
解答:
$$折起前:\cases{A(0,3) \\B(0,0) \\C(4,0) \\D(4,3)}, 並假設D在\overline{AC}的垂足為M \Rightarrow \cases{\overline{AC} =5\\ \overleftrightarrow{AC}:3x+4y=12\\ \overleftrightarrow{MD}:4x-3y=7} \\ \Rightarrow \cases{\overline{MD} =(3\times 4)/5= 12/5\\ M= \overleftrightarrow{AC} \cap \overleftrightarrow{MD} = ({64\over 25},{ 27\over 25})} \\ \Rightarrow 折起後:\cases{A(0,3,0) \\B(0,0,0) \\C(4,0,0)\\ D({64\over 25},{27\over 25},{12\over 5})} \Rightarrow \overline{BD} = \sqrt{({64\over 25})^2+ ({27\over 25})^2+ ({12\over 5})^2} =\sqrt{337\over 25} =\bbox[red, 2pt]{\sqrt{337} \over 5}$$
解答:
$$假設\cases{山頂位於P\\ 山底位於O} \Rightarrow \cases{山高=\overline{OP} =h \\ \overline{OC}=a \\ \tan \theta= x} \Rightarrow h=(50+a)\tan 2\theta =(200+a)\tan \theta={a\over \tan \theta} \\ \Rightarrow (50+a)\cdot {2x\over 1-x^2}=(200+a)x ={a\over x} \Rightarrow \cases{(50+a)\cdot 2x^2= (1-x^2)a \\ (200+a)x^2=a} \Rightarrow x^2= {1\over 5} \Rightarrow a=50 \\ \Rightarrow h={50\over 1/\sqrt 5} = \bbox[red, 2pt]{50\sqrt 5}$$
解答:$$假設\cases{A 群 有 a 人 \\B 群 有 b 人 \\ C群有\cases{C_T人:奇數題說實話,偶數題說謊話\\C_L人: 偶數題說實話,奇數題說謊話}} \\ \Rightarrow \cases{問題一: a+b+C_L=21\\ 問題二: b+C_L=12\\ 問題三: C_L=8 } \Rightarrow \cases{a= \bbox[red, 2pt]9\\ b=4}$$
$$假設矩形ABCD的\cases{寬=\overline{AB}=W\\ 高=\overline{BC}=H \\ T的三頂點為P,Q,C} 及\cases{ \overline{AP} =a\\ \overline{AQ}=b} \Rightarrow \cases{\overline{PB}=W-a\\ \overline{QD}=H-b} \\ \Rightarrow \cases{ab=6\\ H(W-a)=8 \Rightarrow aH=WH-8\\ W(H-b)=10 \Rightarrow bW=WH-10} \Rightarrow aH\cdot bW=(WH-8)(WH-10)\\ \Rightarrow 6WH=(WH-8)(WH-10) \Rightarrow (WH)^2 - 24WH + 80 = 0 \Rightarrow (WH - 4)(WH - 20) = 0 \\ \Rightarrow 矩形面積=WH=20 (WH=4\lt \triangle CDQ=5, 不合) \Rightarrow T面積=20-3-4-5= \bbox[red, 2pt]8$$
解答:$$柯西不等式: \left( \frac{\sin^4 \theta}{4} + \frac{\cos^4 \theta}{9} \right)(4 + 9) \ge (\sin^2 \theta + \cos^2 \theta)^2=1 \Rightarrow \frac{\sin^4 \theta}{4} + \frac{\cos^4 \theta}{9} \ge \frac{1}{13} \\ 依題意等號成立，因此 \frac{\frac{\sin^2 \theta}{2}}{2} = \frac{\frac{\cos^2 \theta}{3}}{3} \implies \frac{\sin^2 \theta}{4} = \frac{\cos^2 \theta}{9} =k \Rightarrow \cases{\sin^2 \theta= 4k\\ \cos^2\theta=9k} \\\Rightarrow 4k+9k=13k=1 \Rightarrow k={1\over 13} \Rightarrow \cases{\sin^2\theta=4 /13\\ \cos^2\theta=9 /13} \\\Rightarrow \frac{\sin^{2026} \theta}{4^{1013}} + \frac{\cos^{2026} \theta}{9^{1013}} = \frac{(\sin^2 \theta)^{1013}}{4^{1013}} + \frac{(\cos^2 \theta)^{1013}}{9^{1013}}= \frac{\left(\frac{4}{13}\right)^{1013}}{4^{1013}} + \frac{\left(\frac{9}{13} \right)^{1013}}{9^{1013}} \\= \left( \frac{1}{13} \right)^{1013} + \left( \frac{1}{13} \right)^{1013} = \bbox[red, 2pt]{\frac{2}{13^{1013}} }$$
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解題僅供參考，其他教甄試題及詳解
解答:$$\cases{A(-3,-2,1) \\B(3,1,1) \\C(-1,0,2)} \Rightarrow \overleftrightarrow{AC}: {x+3\over 2}={y+2\over 2}={z-1} \equiv (2t-3,2t-2,t+1), t\in \mathbb R \\ \Rightarrow d(B,\overleftrightarrow{AC}) = \sqrt{(2t-6)^2+(2t-3)^2+t^2} =\sqrt{9(t-2)^2+9} \ge \bbox[red, 2pt]3$$
解答:$$ P = \int_{8}^{17} \log_2(32x) dx $ = \int_{8}^{17} (5 + \log_2 x) dx = \int_{8}^{17} 5 dx + \int_{8}^{17} \log_2 x dx \\ Q = \int_{1}^{10} \log_2(x+7) dx = \int_{8}^{17} \log_2(u) du = \int_{8}^{17} \log_2(x) dx \\ \Rightarrow P-Q= \int_{8}^{17} 5 dx =5\times(17-8)=\bbox[red,2pt]{45}$$
解答:$$ A(z) 以原點為中心逆時針旋轉 120^\circ 後會與點 B(w) 重合\Rightarrow w = z(\cos 120^\circ + i\sin 120^\circ) = z \cdot e^{i 120^\circ} \\旋轉後重合 \Rightarrow w=\bar z \Rightarrow \bar z=z\cdot e^{i120^\circ} \Rightarrow 4 e^{-i\theta} = 4 e^{i\theta} \cdot e^{i 120^\circ} \Rightarrow e^{-i\theta} = e^{i(\theta + 120^\circ)} \\ \Rightarrow -\theta \equiv \theta + 120^\circ \pmod{360^\circ} \Rightarrow 2\theta \equiv -120^\circ \pmod{360^\circ} \Rightarrow \theta = -60^\circ + 180^\circ k \quad k\in \mathbb Z \\在 0^\circ \le \theta < 360^\circ 的範圍內, \theta=120^\circ 或300^\circ \\ \theta=120^\circ \Rightarrow z = 4(\cos 120^\circ + i\sin 120^\circ) = 4(-\frac{1}{2} + \frac{\sqrt{3}}{2}i) = -2 + 2\sqrt{3}i \Rightarrow w=\bar z=-2-2\sqrt 3i\\ \qquad \Rightarrow \cases{兩根之和:z+w=-4 \Rightarrow p=4\\ 兩根之積:z\cdot w=16 \Rightarrow q=16} \Rightarrow p+q=20 \\ \theta=300^\circ \Rightarrow z = 4(\cos(-60^\circ) + i\sin(-60^\circ)) = 4(\frac{1}{2} - \frac{\sqrt{3}}{2}i) = 2 - 2\sqrt{3}i \Rightarrow w=2+2\sqrt 3i\\ \qquad \Rightarrow \cases{兩根之和:z+w= 4 \Rightarrow p=-4\\ 兩根之積:z\cdot w=16 \Rightarrow q=16} \Rightarrow p+q=12 \\ \Rightarrow p+q= \bbox[red, 2pt]{12或20}$$
二、計算題： (5 題，每題 8 分，共 40 分)
解答:
$$假設\cases{M=\overline{BD}中點 \Rightarrow M=(B+D)/2= (9/2,9/2,0)\\ 平面E與\overline{BC}交於P} \Rightarrow \frac{ \triangle MBP }{ \triangle BCD } = \frac{\overline{BM}}{\overline{BD}} \times \frac{\overline{BP}}{\overline{BC}} \\ \Rightarrow {1\over 3}={1\over 2} \times \frac{\overline{BP}}{\overline{BC}} \Rightarrow \frac{\overline{BP}}{ \overline{BC}}={2\over 3} \Rightarrow \overline{BP} : \overline{PC} = 2 : 1 \Rightarrow P={1\over 3}(B+2C) =(1,2,0)\\ 平面E上有三點\cases{A(0,0,2) \\M(9/2,9/2,0) \\ P(1,2,0)} \Rightarrow E的法向量\vec n\parallel (\overrightarrow{AP} \times \overrightarrow{AM}) \Rightarrow \vec n=(10,-14,-9) \\ \Rightarrow E:10x-14y-9(z-2)=0 \Rightarrow \bbox[red, 2pt]{10x-14y-9z+18=0}$$
解答:$$f(x)=ax^2+bx+c \Rightarrow f'(x)=2ax+b\\在 x =1附近的局部特徵近似於直線 y = 2x+1 \Rightarrow 直線經過(1,3)\Rightarrow \cases{f(1)=3\\ f'(1)=2} \\ \Rightarrow \cases{a+b+c=3\\ 2a+b=2} \Rightarrow 兩式相加:3a+2b+c= 5\\ g(x)=ax^3+bx^2+cx \Rightarrow g'(x)=3ax^2+2bx+c \Rightarrow \cases{g(1)=a+b+c=3\\ g'(1)=3a+2b+c=5} \\ \Rightarrow y=g(x)在x=1的近似直線: y=5(x-1)+3 \Rightarrow \bbox[red, 2pt]{y=5x-2}$$
解答:
$$\textbf{(1) }兩中線交點即為重心G \Rightarrow x軸與y軸的交點為原點\Rightarrow G=(0,0) \\ 假設\cases{B(x_B,0) \\ C(0,y_C)} \Rightarrow G= {1\over 3}(A+B+C) \Rightarrow \cases{-8+x_B=0\\ -6+y_C=0} \Rightarrow \cases{x_B=8\\ y_C=6} \Rightarrow \cases{B(8,0) \\C(0,6)} \\ \Rightarrow \triangle GBC ={1\over 2}\times 8\times 6=24 \Rightarrow \triangle ABC=24\times 3=\bbox[red, 2pt]{72} \\ \textbf{(2) }假設外心O(x,y) \Rightarrow \cases{\overline{OA}^2= (x+8)^2+(y+6)^2\\ \overline{OB}^2=(x-8)^2+ y^2 \\ \overline{OC}^2 =x^2+(y-6)^2} \Rightarrow \cases{\overline{OB}^2= \overline{OC}^2 \Rightarrow 4x-3y=7\\ \overline{OA}^2= \overline{OB}^2=8x+3y=-9} \\ \Rightarrow \cases{x=-1/6\\ y=-23/9} \Rightarrow 外心坐標 \bbox[red, 2pt]{\left( -{1\over 6},-{23\over 9} \right)}$$
解答:$$\textbf{(1) }\Gamma_1: y=x^2 \Rightarrow \Gamma_2: y = (x - 2)^2 + 4 = x^2 - 4x + 8 \Rightarrow \cases{對\Gamma_1微分: y'=2x\\ 對\Gamma_2微分: y'=2x-4}\\ 假設\cases{L與\Gamma_1 相切於P(a,a^2) \Rightarrow 斜率2a \Rightarrow L:y=2ax-a^2\\L與\Gamma_2 相切於Q(b,b^2-4b+8) \Rightarrow 斜率2b-4 \Rightarrow L:y= (2b-4)x-b^2+8} \\ \Rightarrow \cases{斜率相等: 2a= 2b-4 \Rightarrow b=a+2\\ y斜距相等:-a^2=-b^2 +8} \Rightarrow -a^2=-(a+2)^2+8 \Rightarrow 4a=4\Rightarrow a=1 \Rightarrow b=3 \\ \Rightarrow L: \bbox[red, 2pt]{y=2x-1} \\ \textbf{(2) }求\Gamma_1與 \Gamma_2交點: x^2=x^2-4x+8 \Rightarrow x=2 \Rightarrow y=4 \Rightarrow 交點為(2,4) \\ \Rightarrow 所圍面積A= \int_1^2 (\Gamma_1-L)\,dx + \int_2^3 (\Gamma_2-L)\,dx = \int_{1}^{2} (x^2 - 2x + 1) dx + \int_{2}^{3} (x^2 - 6x + 9) dx \\ = \int_{1}^{2} (x - 1)^2 dx + \int_{2}^{3} (x - 3)^2 dx = \left. \left[ {1\over 3}(x-1)^3 \right] \right|_1^2 + \left. \left[ {1\over 3}(x-3)^3 \right] \right|_2^3={1\over 3} +{1\over 3}= \bbox[red, 2pt]{2\over 3}$$
解題僅供參考，其他教甄試題及詳解
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