臺北市立中正高中 114 學年度第 1 次專任教師甄選
一、填充題（每題 5 分）
解答:
$$假設\cases{B在平面z=5上\\ C在平面z=6上\\ D在平面z=7上\\P=\overline{BD}中點 \\ D在平面z=6的投影點D' \\ \triangle BCD與平面z=6的夾角\theta} \Rightarrow \cases{P也在平面z=6上,且\overline{PD}=5\sqrt 2/2\\ \overline{DD'}=1} \\\Rightarrow \sin \theta={\overline{DD'} \over \overline{PD}} ={2\over 5\sqrt 2} \Rightarrow \cos \theta={\sqrt{46}\over 5\sqrt 2}\\ 假設A在\triangle BCD的投影點G \Rightarrow G為\triangle BCD的重心 \Rightarrow G的z坐標={5+6+7\over 3}=6 \\\Rightarrow G也在在平面z=6上 \Rightarrow \overline{AG} ={\sqrt 6\over 3}\times 5\sqrt 2={10\sqrt 3\over 3}(正四面體的高) \\ \overline{AG} \bot \triangle BCD \Rightarrow \overline{AG}與平面z=6的夾角為{\pi\over 2}-\theta \Rightarrow A至平面z=6的距離= \overline{AG}\sin({\pi\over 2}-\theta) \\={10\sqrt 3\over 3} \cos \theta= {10\sqrt 3\over 3} \cdot {\sqrt{46}\over 5\sqrt 2} = {2\over 3}\sqrt{69} \Rightarrow A到地面距離=6-{2\over 3}\sqrt{69}= \bbox[red, 2pt]{18-2\sqrt{69}\over 3}$$
解答:$$\prod_{k=2}^{31}{\log_k \left(7^{k^2} \right) \over\log_{k+1} \left(7^{k^2-1} \right)} =\prod_{k=2}^{31}{k^2\log 7/\log k \over (k^2-1)\log 7/\log (k+1)} = \prod_{k=2}^{31}{k^2\log (k+1) \over (k^2-1) \log k} =\prod_{k=2}^{31}{k^2\log (k+1) \over (k-1)(k+1) \log k} \\={2^2\cdot 3^2\cdots31^2 \log 3\cdot \log 4\cdots\log 32\over (1\cdot 3)\cdot(2\cdot 4) \cdot(3\cdot 5)\cdots (30\cdot32)\log 2\cdot \log 3\cdots \log 31} ={2^2\cdot 3^2\cdots31^2 \log 32\over 30!\cdot 32!/2\cdot \log 2} \\={2^2\cdot 3^2\cdots31^2 \cdot 5\log 2\over 2^2\cdot 3^2\cdots30^2\cdot 31\cdot 32/2 \cdot \log 2} ={31\cdot 5\over 16} =\bbox[red, 2pt]{155\over 15}$$
解答:$$\cases{A(4\sin 45^\circ \cos 60^\circ, 4\sin 45^\circ \sin 60^\circ, 4\cos 45^\circ) = (\sqrt 2, \sqrt 6, 2\sqrt 2)\\ B(4\sin 135^\circ \cos(-30^\circ), 4\sin 135^\circ \sin(-30^\circ), 4 \cos 135^\circ) =(\sqrt 6, -\sqrt 2, -2\sqrt 2)} \\ \Rightarrow \cos \theta={-8\over 4\cdot 4}=-{1\over 2} \Rightarrow \theta={2\pi\over 3} \Rightarrow 弧長=4\cdot {2\pi \over 3}=\bbox[red, 2pt]{8\pi\over 3}$$
解答:$$\cases{P(X=0) =1/2^7 \approx 0.0078\\ P(X=1)= P(X=6)=C^7_1/2^7=7/2^7 \approx 0.0547\\ P(X=2) =P(X=5)=C^7_2/2^7 =21/2^7 \\ P(X=3)= P(X=4) =C^7_3/2^7 =35/2^7} \\ \alpha=0.05 \Rightarrow X=x \in 拒絕域 \Rightarrow P(X=x)< 0.025 \Rightarrow X=0,7 \Rightarrow 拒絕域\bbox[red, 2pt]{\{0,7\}}$$
解答:$$P\in L \Rightarrow P(2t+3,-2t,-t) \\\Rightarrow \overline{PA}-\overline{PB} =\sqrt{(2t+4)^2+(-2t+2)^2+ (-t-5)^2} -\sqrt{(2t+2)^2+ (-2t-5)^2+(-t-4)^2} \\=f(t)=\sqrt{9t^2+18t+45} -\sqrt{9t^2+36t+45} \\ f'(t)=0 \Rightarrow {t+1\over \sqrt{t^2+2t+5}} ={t+2\over \sqrt{t^2+4t+5}} \Rightarrow 3t^2+14t+15=(3t+5)(t+3) =0 \Rightarrow t=-3,-5/3 \\ \Rightarrow \cases{f(-3)=\sqrt{73}-\sqrt{18}=3\sqrt 2\\ f(-5/3)= \sqrt{40}-\sqrt{10} =\sqrt{10}} \Rightarrow t=-3 \Rightarrow P=(-3,6,3) \Rightarrow \overline{PA}-\overline{PB}有最大值3\sqrt 2 \\ \Rightarrow (a,b,c,d)= \bbox[red, 2pt]{(-3,6,3, 3\sqrt 2)}$$
解答:$$\text{taylor series: } \cases{\tan x=x +\displaystyle {x^3\over 3} +{2x^5\over 15}+ \cdots \\ \sin x=x-\displaystyle {x^3\over 6}+{x^5\over 120}+ \cdots} \Rightarrow \cases{\tan 2x=2x+ \displaystyle {8x^3\over 3} +{64x^5\over 15}+ \cdots \\ x^2\sin bx= bx^3-\displaystyle {b^3x^5\over 6}+{b^5 x^7\over 120}+ \cdots} \\ \Rightarrow \tan 2x +ax+x^2\sin bx =(2+a)x+({8\over 3}+b)x^3 +({64\over 15}-{b^3\over 6})x^5 +\cdots\\\lim_{x\to 0} \left( {\tan 2x\over x^3}+{a\over x^2}+ {\sin bx\over x}\right) =\lim_{x\to 0} \left( {\tan 2x +ax+x^2\sin bx\over x^3}\right) = 0 \Rightarrow \cases{2+a=0\\ 8/3+b=0} \\ \Rightarrow (a,b)= \bbox[red, 2pt]{\left(-2,-{8\over 3} \right)}$$
解答:$$L通過F(5,0)及M(2,-2) \Rightarrow L:2x-3y=10 \Rightarrow 斜率m={2\over 3} \\ 假設\cases{A(x_1,y_1) \\ B(x_2,y_2) \\ 橢圓E:{x^2\over \alpha}+{y^2\over \beta} =1} \Rightarrow \cases{M=(A+B)/2 \Rightarrow \cases{x_1+x_2=4\\ y_1+y_2=-4} \\ \cases{x_1^2/\alpha+y_1^2/\beta=1\\ x_2^2/\alpha+ y_2^2/\beta =1} \Rightarrow {1\over \alpha}(x_1^2-x_2^2)+{1\over \beta}(y_1^2-y_2^2)=0} \\ \Rightarrow {4\over \alpha}(x_1-x_2)-{4\over \beta}(y_1-y_2) =0 \Rightarrow m={2\over 3}={y_2-y_1\over x_2-x_1} ={\beta\over \alpha} \Rightarrow \alpha={3\over 2}\beta \\ \Rightarrow \alpha-\beta={1\over 2}\beta=5^2 \Rightarrow \beta=50\Rightarrow \alpha=75 \Rightarrow 橢圓E: \bbox[red, 2pt]{{x^2\over 75}+{y^2\over 50}=1}$$
解答:
$$假設\cases{\Gamma:y=|x^2-4x+3|\\ 直線L:y=x+a} \Rightarrow \Gamma與x軸交於\cases{A(1,0) \\B(3,0)} \Rightarrow 過A且與L平行的直線L_1:y=x-1\\ 若1\le x\le 3 \Rightarrow \Gamma:y=-x^2+4x-3 \Rightarrow y'=-2x+4=1 \Rightarrow x={3\over 2} \Rightarrow 切點P({3\over 2},{3\over 4}) \\ \Rightarrow 過P且與L平行的直線L_2: y=x-{3\over 4} \Rightarrow 在L_1與L_2之間(不含切點) 有4個交點\\ \Rightarrow \bbox[red, 2pt]{-1\lt a\lt -{3\over 4}}$$
解答:
$$作\overline{AF}\bot \overline{BF}，使得FBCD為一邊長12的正方形，並假設\cases{\overline{CE}=a\\ \overline{AF}=b\\ \angle EBC=\alpha\\ \angle ABF=\beta} \Rightarrow \cases{\tan \alpha=a/12\\ \tan \beta=b/12} \\ \Rightarrow \tan(\alpha+\beta) =\tan 45^\circ ={\tan \alpha+\tan \beta\over 1-\tan \alpha \tan \beta} \Rightarrow {{a+b\over 12} \over 1-{ab\over 144}} =1 \Rightarrow a+b=12-{ab\over 12} \cdots(1)\\ 直角\triangle ADE中，(12-a)^2+(12-b)^2=10^2 \Rightarrow a^2+b^2-24(a+b)+188=0 \\ 把(1)代入\Rightarrow a^2+b^2-24\left(12-{ab\over 12} \right)+188=(a+b)^2-100=0 \Rightarrow a+b=10 \\ \Rightarrow ab=24 \Rightarrow a=\bbox[red, 2pt]{4或 6}$$
解答:$$f(x)={1-x^7\over 1-x}=1+x+x^2+ \cdots+x^6 =(x-\omega)(x-\omega^2)\cdots (x-\omega^6) =\prod_{k=1}^{6}(x-\omega^k) \\ \prod_{k=0}^{6}(\omega^{2k}+ 2\omega^k+4) =7\prod_{k=1}^{6}(\omega^{2k}+ 2\omega^k+4) = 7\prod_{k=1}^{6}(( \omega^{k}+1)^2 +3) \\= 7\prod_{k=1}^{6}((\omega^{k}+1) +\sqrt 3i)(\omega^k+1-\sqrt 3i)) = 7\prod_{k=1}^{6}((-1-\sqrt 3i-\omega^{k})(-1+\sqrt 3i-\omega^k )) \\=7f(-1-\sqrt 3i)\cdot f(-1+\sqrt 3i) =7\cdot {1-(-1-\sqrt 3i)^7\over 2-\sqrt 3i} \cdot {1-(-1+\sqrt 3 i)^7 \over 2-\sqrt 3i} \\= \left[1-(-1-\sqrt 3i)^7 \right] \left[1-(-1+\sqrt 3i)^7 \right] =(65+64\sqrt 3 i)(65-64\sqrt 3i) =65^2+3\cdot 64^2 = \bbox[red, 2pt]{16513}$$
二、計算與教學題（除第 6 題 10 分外，其餘每題 8 分）
解答:$$\cases{\Gamma_1:y=f(x)=2^x\\\Gamma_2: y=g(x)=\log_2 x} \Rightarrow \cases{f=g^{-1} \\ g=f^{-1}} \Rightarrow 兩圖形\Gamma_1,\Gamma_2不相交且對稱於y=x \\ \Rightarrow \Gamma_1 恆在y=x上方$$
解答:$$假設\cases{f(x,y)=x^2+y^2 \\ g(x,y)=2xy(x^2-y^2)-x^2-y^2} \Rightarrow \cases{f_x=\lambda g_x\\ f_y= \lambda g_y\\ g=0} \Rightarrow \cases{2x=\lambda(6x^2y-2y^3-2x) \cdots(1)\\ 2y=\lambda(2x^3-6xy^2-2y) \cdots(2)\\ 2xy(x^2-y^2)=x^2+y^2 \cdots(3)} \\ \Rightarrow {(1) \over (2)} ={x\over y}={6x^2y-2y^3-2x\over 2x^3-6xy^2-2y}\Rightarrow x^4+y^4=6x^2y^2 \Rightarrow \cases{(x^2+y^2)^2 =8x^2y^2\\ (x^2-y^2)^2 = 4x^2y^2} \\ \Rightarrow {(x^2+y^2)^2 \over (x^2-y^2)^2} =2 代入(3)\Rightarrow {(x^2+y^2) \over (x^2-y^2)} =2xy=\pm \sqrt 2 \Rightarrow f(x,y)=\sqrt{8x^2y^2} = \bbox[red, 2pt]2$$
解答:$$\cases{\vec a=(\cos{3x\over 2},\sin{3x\over 2}) \\\vec b=(\cos{x\over 2}, -\sin{x\over 2})} \Rightarrow \cases{\vec a\cdot \vec b= \cos{3x\over 2}\cos {x\over 2}-\sin{3x\over 2}\sin {x\over 2} =\cos({3x\over 2}+{x \over 2}) =\cos 2x \\ |\vec a+\vec b|=\sqrt{(\cos{3x\over 2}+\cos{x\over 2})^2+ (\sin{3x\over 2}-\sin{x\over 2})^2} =\sqrt{2+2\cos 2x}= 2\cos x} \\ \Rightarrow f(x)=\cos 2x-4\lambda \cos x \Rightarrow f'(x)=-2\sin 2x+4\lambda \sin x \Rightarrow f''(x)=-4\cos 2x+4\lambda \cos x \\ 若f'(x)=0 \Rightarrow -4\sin x\cos x+4\lambda \sin x=4\sin x(\lambda-\cos x)=0 \\ \textbf{Case I } \sin x=0 \Rightarrow x=0 \Rightarrow f(0)= 1-4\lambda=-{11\over 2} \Rightarrow \lambda={13\over 8} \Rightarrow f''(0)=-4+4\lambda\gt 0 符合要求 \\\textbf{Case II } \cos x=\lambda \Rightarrow f(x)=2\lambda^2-1-4\lambda^2=-2\lambda^2-1=-{11\over 2} \Rightarrow \lambda^2= {9\over 4} \\\qquad \Rightarrow f''(x) =-4(2\lambda^2-1)+4\lambda^2 =-4\lambda^2+4 \lt 0 \Rightarrow -{11\over 2}為極大值, 不符要求\\ 因此\lambda= \bbox[red, 2pt]{13\over 8}$$
解答:$$\cases{a_1=1\\ S_{n+1}=4a_n+2} \Rightarrow a_2=S_2-a_1 =4a_1+2-a_1=5\\S_{n+1}=4a_n+2 \Rightarrow a_{n+1}=S_{n+1}-S_n = 4a_n+2-(4a_{n-1}+2) =4(a_n-a_{n-1}) \\ \Rightarrow a_{n+1}-4a_n+4a_{n-1}=0 \Rightarrow x^2-4x+4=0 \Rightarrow x=2 \Rightarrow a_n =(\alpha+\beta n) \cdot2^n \\ \Rightarrow \cases{a_1= 1=(\alpha+\beta)2\\ a_2=5= (\alpha+2\beta)2^2} \Rightarrow \cases{\alpha=-1/4\\ \beta=3/4} \Rightarrow a_n=(-{1\over 4}+{3n\over 4})\cdot 2^n \Rightarrow a_n= \bbox[red, 2pt]{(3n-1)\cdot 2^{n-2}}$$
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解題僅供參考，其它教甄試題及詳解
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