114年華江高中教甄-數學詳解

臺北市立華江高級中學 114 學年度正式教師甄選

一、 填充題: (每題 5 分，該題全對才給分，共 60 分)

解答:$$\cases{\tan B=3/4\\ \tan C=3/2} \Rightarrow \cases{\overline{AH}=3k\\ \overline{BH}=4k\\ \overline{CH}=2k} \Rightarrow \overline{MH}=5=k \Rightarrow \overline{AH} =15 \Rightarrow \overline{AM} =\sqrt{15^2+5^2} = \bbox[red, 2pt]{5\sqrt{10}}$$

解答:$$\lim_{x \to \infty} \left(n\sqrt{n-3\over n+3}-n \right) =\lim_{x \to \infty} \left({\sqrt{n-3\over n+3}-1 \over 1/n}\right) =\lim_{x \to \infty} \left({{3\over (n+3)\sqrt{n^2-9}} \over -1/n^2}\right) \\= \lim_{x \to \infty}{-3n^2\over (n+3)\sqrt{n^2-9}} =\bbox[red, 2pt]{-3}$$

解答:

$$\cases{小圓C_1:x^2+(y-\sqrt 6)^2=6\\ 大圓C_2:(x-3\sqrt 2)^2+y^2 =18} \Rightarrow \cases{小圓圓心O_1(0,\sqrt 6), 半徑r_1=\sqrt 6\\ 大圓圓心O_2(3\sqrt 2,0),半徑r_2=3\sqrt 2} \\\cases{A(0,0) \\ P(a,b)} \Rightarrow \cases{\overrightarrow{PA} =(-a,-b) \\ \overrightarrow{PB}=(-a,2\sqrt 6-b) \\ \overrightarrow{PD} =(6\sqrt 2-a,-b)}  \Rightarrow   \cases{\overrightarrow{PA} \cdot \overrightarrow{PB}=0\\ \overrightarrow{PA} \cdot \overrightarrow{PD}=0}\Rightarrow \cases{a^2+b^2-2\sqrt 6b=0\\ a^2+b^2-6\sqrt 2a=0} \Rightarrow a={b\over \sqrt 3} \\ \Rightarrow {b^2\over 3}+b^2-2\sqrt 6 b=0 \Rightarrow b={3\over 2}\sqrt 6 \Rightarrow a={3\over 2}\sqrt 2 \Rightarrow P({3\over 2}\sqrt 2, {3\over 2}\sqrt 6) \Rightarrow \overline{AP} = 3\sqrt 2  \\ \Rightarrow \cos \angle PO_1A=-{1\over 2} \Rightarrow \angle PO_1A=120^\circ, 又 \overrightarrow{PO_1}\cdot \overrightarrow{PO_2}=0  \Rightarrow P, A皆為圓切點 \\ \Rightarrow \angle PO_2A=60^\circ \Rightarrow \cases{S_A+ S_B+ S_C=\sqrt 6\cdot 3\sqrt 2=6\sqrt 3\\ S_A+S_B=6\pi/3=2\pi\\ S_B+S_C =18\pi/6=3\pi} \Rightarrow S_B=2\pi+3\pi-6\sqrt 3=\bbox[red, 2pt]{5\pi-6\sqrt 3}$$

解答:$$(1+2\cos \theta)^3=(2+\sin\theta)^3 \Rightarrow 1+2\cos \theta=2+\sin \theta \Rightarrow 2\cos \theta-\sin \theta =1 \\ \Rightarrow (2\cos \theta-\sin \theta)^2=1 \Rightarrow 3\cos^2\theta-4\sin \theta\cos \theta= 0\Rightarrow \cos\theta(3 \cos \theta-4\sin \theta)=0 \\ \Rightarrow 3\cos \theta =4\sin \theta \Rightarrow \tan\theta= {3\over 4} \Rightarrow \tan^2 \theta= \bbox[red, 2pt]{9\over 16}$$

解答:

$$假設圓心A，圓半徑r, 圓與x軸交於P、與y軸交於Q\\ 由於\angle POQ=90^\circ \Rightarrow \overline{PAQ}為直徑，又\stackrel{\Large\frown}{OQ} =2 \stackrel{ \Large\frown}{OP} \Rightarrow \angle OAQ=2\angle OAP \Rightarrow \cases{\angle OAP=60^\circ\\ \angle OAQ=120^\circ} \\ \stackrel{\Large\frown}{OP}=\pi=r\cdot {\pi\over 3} \Rightarrow r=\bbox[red, 2pt]3$$

解答:$$f(x)={ax^2+2x+b\over x^2+1} \Rightarrow \cases{f(x)=5 \Rightarrow (a-5)x^2+2x+b-5=0\\ f(x)=3 \Rightarrow (a-3)x^2 +2x+b-3=0}\;判別式皆為0 \\\Rightarrow \cases{(a-5)(b-5)=1\\ (a-3)(b-3)=1} \Rightarrow (a-5)(b-5)=(a-3)(b-3) \Rightarrow a+b=8 \Rightarrow (a-5)(3-a)=1\\ \Rightarrow a=4 \Rightarrow b=4\Rightarrow (a,b)= \bbox[red, 2pt]{(4,4)}$$

解答:$$B=\{5\} \Rightarrow A= \{1,2,3,4\} 所有非空子集合，共有2^4-1=15種\\ B=\{4\},\{4,5\}\Rightarrow A= \{1,2,3\} 所有非空子集合，共有2^3-1=7種，A,B合計7\cdot 2=14種\\ B=\{3\}, \{3,4\}, \{3,5\}, \{3,4,5\} \Rightarrow A=\{1,2\} 所有非空子集合，共有2^2-1=3種\\\qquad ，A,B合計3\cdot 4=12種 \\B=\{2\}, \{2,3\}, \{2,4\}, \{2,5\}, \{2,3,4\},\{2,3,5\}, \{2,4,5\},\{2,3,4,5\}, \Rightarrow A=\{1\},A,B合計8種\\ 總共有15+14+12+8=\bbox[red, 2pt]{49}種$$

解答:$$假設\cases{A=(\alpha,\log_a \alpha) \\B=(\beta, 2\log_a \beta) \\C=(\gamma,3\log_a\gamma)}, 由於\ \cases{\overline{AB}平行x軸 \\ \overline{AB}=6} \Rightarrow \cases{\log_a \alpha=2\log_a \beta\\ \alpha-\beta=6} \Rightarrow \cases{\alpha=9\\ \beta=3} \Rightarrow \cases{A(9,\log_a 9) \\B(3,2\log_a 3)} \\ 又\cases{\overline{BC}=6\\ \overline{BC}平行y 軸} \Rightarrow \cases{3\log_a \gamma-2\log_a \beta=6\\\gamma=\beta=3} \Rightarrow 3\log_a 3-2\log_a 3= \log_a 3=6 \\ \Rightarrow a^6=3 \Rightarrow a= \bbox[red, 2pt]{\sqrt[6]3}$$

解答:$$f(x)= \begin{cases} ax+b\lfloor x\rfloor, & x\ge 2\\ x^2,& x\lt 2\end{cases} \Rightarrow f(2)=2^2=2a+2b \Rightarrow a+b=2 \\ f'(x)= \begin{cases} a , & x\ge 2\\ 2x,& x\lt 2\end{cases} \Rightarrow  f'(2)=4=a \Rightarrow b=-2 \Rightarrow (a,b)= \bbox[red, 2pt]{(4,-2)}$$

解答:$$(\sqrt 2+\sqrt[4]3)^{100} =\sum_{n=0}^{100}C^{100}_n (\sqrt 2)^n(\sqrt[4]{3})^{100-n} \Rightarrow (\sqrt 2)^n(\sqrt[4]{3})^{100-n} =\begin{cases} 常數& n=0\\ 2^{1/2} \cdot 3^{3/4} & n=1\\ 3^{1/2}& n=2 \\ 2^{1/2}\cdot 3^{1/4} & n=3\\ 常數& n=4\\ 2^{1/2}\cdot 3^{3/4} & n=5 \\ \cdots& \cdots\end{cases} \\ \Rightarrow 循環數為4 \Rightarrow 共\bbox[red, 2pt]4項$$

解答:$$\cases{a_{n+1}=3a_n-2b_{n+1} \cdots(1)\\b_{n+1}=a_{n+1}-3b_n \cdots(2)} \Rightarrow a_{n+1} =3a_n-2(a_{n+1}-3b_n) \Rightarrow a_{n+1}=a_n+2b_n =a_n+ (3a_{n-1}-a_n) \\ \Rightarrow a_{n+1}=3a_{n-1} \Rightarrow 243 =a_9=3a_7=3^2a_5 =\cdots =3^4a_1 \Rightarrow a_1=3\\ 同理, 將(1)代入(2) \Rightarrow b_{n+1}=(3a_n-2b_{n+1})-3b_n \Rightarrow b_{n+1}=a_n-b_n =(b_n+3b_{n-1})-b_n \\ \Rightarrow b_{n+1}=3b_{n-1} \Rightarrow 324=b_{10} =3b_8=\cdots=3^4b_2 \Rightarrow b_2=4\\ 由(1)\Rightarrow a_2=3a_1-2b_2  =9-8=1 \Rightarrow a_2=1代入(2) \Rightarrow b_2=a_2-3b_1 \Rightarrow 4=1-3b_1 \Rightarrow b_1=-1\\ \Rightarrow (a_1,b_1) =\bbox[red, 2pt]{(3,-1)}$$

解答:$$|z|=\sqrt 3 \Rightarrow z \bar z=3 \Rightarrow \left| {1+2iz\over z-2i}+4i\right| = \left| {9+6iz\over z-2i} \right| =\left| {3z \bar z+6iz\over z-2i} \right| =\left| {3z( \bar z+2i) \over z-2i} \right| \\ =|3z|= \bbox[red, 2pt]{3\sqrt 3} \quad (\overline{z-2i} =\bar z+2i \Rightarrow \left| {  \bar z+2i  \over z-2i} \right|=1)$$

二、 計算題: (共 40 分， 請寫出詳細的計算過程，否則不予計分)

解答:$$y=f(x)=a(x+1)^3+ b(x+1)^2+c(x+1)+d \Rightarrow f'(x)=3a(x+1)^2+2b(x+1)+c \\ \Rightarrow f''(x)=6a(x+1)+2b\\ 廣域看y=f(x)圖形接近y=4x^3 \Rightarrow a=4 \\ 在x=-1附近圖形似近y=-2x-3 \Rightarrow\cases{f'(-1)=-2\\ f(-1) =-2\cdot (-1)-3=-1} \Rightarrow \cases{c=-2\\ d=-1} \\ 對稱中心在x=1處\Rightarrow f''(1)=0 \Rightarrow 24\cdot 2+2b=0 \Rightarrow b=-24\\ 因此(a,b,c,d)= \bbox[red, 2pt]{(4,-24,-2,-1)}$$

解答:

$$\textbf{(1) }五頂點皆在球上，且頂點E至四邊形頂點等距，因此底面箏形ABCD在一外接圓上\\\quad \overline{AC}為外接圓直徑\Rightarrow \angle ABC=\angle ADC=90^\circ  \Rightarrow 面積=\overline{AB}\times \overline{BC}= \bbox[red, 2pt]{80}$$

$$\textbf{(2) }P=\overline{AC}中點\Rightarrow \overline{AP}={1\over 2}\sqrt{10^2+8^2} =\sqrt{41} \Rightarrow \overline{PE}=\sqrt{(3\sqrt{10})^2-(\sqrt{41})^2} =7 \\\quad 假設球心O, 球半徑r, 在直角\triangle APO中, \overline{AO}^2=\overline{AP}^2+\overline{OP}^2 \Rightarrow r^2=41+(7-r)^2 \\\quad \Rightarrow r=\bbox[red, 2pt]{45\over 7} \\\textbf{(3) }角錐體積={1\over 3}\cdot 80\cdot 7 = \bbox[red, 2pt]{560\over 3}$$

解答:$$假設過P(1,2)直線L:y=m(x-1)+2與圓交於A(a,b)及B(c,d )兩點 \\ \Rightarrow P={1\over 3}(A+2B) \Rightarrow \cases{a+2c=3\\ b+2d=6} \Rightarrow \cases{c=(3-a)/2\\ d=(6-b)/2} \Rightarrow \cases{a^2+b^2=37\\ {(3-a)^2\over 4}+ {(6-b)^2\over 4}=37}\\ \Rightarrow \cases{a^2+b^2=37\\ a^2+b^2-6a-12b=103} \Rightarrow a+2b+11=0 \Rightarrow a=-2b-11 \Rightarrow (-2b-11)^2+b^2=37 \\ \Rightarrow 5b^2+44b+84=0 \Rightarrow (5b+14)(b+6) =0 \Rightarrow \cases{b= -14/5 \Rightarrow a=-27/5\\b=-6 \Rightarrow a=1} \Rightarrow L: \bbox[red, 2pt]{\cases{3x-4y+5=0\\x=1}}$$

解答:$$假設\overline{AC}=a \Rightarrow \cases{\triangle ABC周長=a+3\\ \triangle ACD周長=a+2} \Rightarrow \cases{S^2=-(a^2-9)(a^2-1)/8\\ T^2=-a^2(a^2-4)/8} \\ \Rightarrow S^2+T^2=f(a)=-{1\over 8}(2a^4-14a^2+9) \Rightarrow f'(a)=-{1\over 2}a(2a^2-7) =0 \Rightarrow a= \bbox[red, 2pt]{\sqrt{7\over 2}}\\ \Rightarrow f(\sqrt{7\over 2})= \bbox[red, 2pt]{31\over 16}$$

 

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