105年鳳山高中教甄-數學詳解

 國立鳳山高級中學105學年度第1次專任教師甄選

一、填充題

解答： $$\cases{A(0,0)\\ B(b,11) \\C(c,37)} \Rightarrow \cases{C=B逆時鐘旋轉60度\\ C=B順時鐘旋轉60度}\qquad \quad\Rightarrow \cases{\begin{bmatrix} 1/2 & -\sqrt 3/2\\ \sqrt 3/2 & 1/2\end{bmatrix}\begin{bmatrix} b\\ 11\end{bmatrix} =\begin{bmatrix} c\\ 37\end{bmatrix} \\\begin{bmatrix} 1/2 & \sqrt 3/2\\ -\sqrt 3/2 & 1/2\end{bmatrix}\begin{bmatrix} b\\ 11\end{bmatrix} =\begin{bmatrix} c\\ 37\end{bmatrix}} \\ \Rightarrow \cases{\cases{b-11\sqrt 3=2c\\ \sqrt 3b+11=74} \Rightarrow \cases{b= 21\sqrt 3\\c=5 \sqrt 3} \\\cases{b+11\sqrt 3=2c\\ -\sqrt 3b+11=74} \Rightarrow \cases{b= -21\sqrt 3\\c= -5\sqrt 3}} \Rightarrow bc=105\times 3= \bbox[red,2pt]{315}$$

解答：

$$\overline{AB}為直徑 \Rightarrow \cases{\overline{BC}=\sqrt{8^2-2^2} =2\sqrt{15} \\\overline{AD} =\sqrt{8^2-7^2} =\sqrt{15}} \Rightarrow \cases{\tan \theta_1=\tan \angle ABC=2/2\sqrt{15} =1/\sqrt{15}\\ \tan (\theta_1+\theta_2)=\tan \angle ABD = \sqrt{15}/7} \\ 由\tan(\theta_1+\theta_2) ={\tan\theta_1+ \tan \theta_2\over 1-\tan\theta_1 \tan \theta_2 } \Rightarrow {\sqrt {15}\over 7} ={{1\over \sqrt{15}}+ \tan \theta_2\over 1-{1\over \sqrt{15}} \tan \theta_2 } \Rightarrow \tan \theta_2={1\over \sqrt{15}}\\ \Rightarrow \cos\theta_2={\sqrt{15}\over 4} ={(2\sqrt{15})^2+7^2-\overline{CD}^2 \over 28\sqrt{15}} \Rightarrow \overline{CD}= \bbox[red,2pt]{2}$$

解答： $$令\cases{P_n=(x_n,y_n),1\le n\le 179\\ P_0=A=(1,0)\\ P_{180}=B=(-1,0)}，則\angle P_{n+1}OP_n=1^\circ,0\le n \le 179 \Rightarrow x_n=\cos n^\circ\\ 因此\sum_{n=1}^{179} x_n^2 =\sum_{n=1}^{179} \cos^2 n^\circ \\先求\cos^2 1^\circ +\cos^2 2^\circ+\cdots +\cos^2 90^\circ \\=\cos^2 1^\circ +\cos^2 2^\circ+\cdots +\cos^2 88^\circ+\cos^2 89^\circ +0 \\=\cos^2 1^\circ +\cos^2 2^\circ+\cdots +\cos^2 (90^\circ-2^\circ)+\cos^2 (90^\circ-1^\circ)   \\=\cos^2 1^\circ +\cos^2 2^\circ+\cdots +\sin^2 2^\circ +\sin^2 1^\circ \\ =(\cos^2 1^\circ+ \sin^2 1^\circ)+(\cos^2 2^\circ+ \sin^2 2^\circ) +\cdots +(\cos^2 44^\circ+ \sin^2 44^\circ)+\cos^2 45^\circ\\ =1+1+\cdots +1 +{1\over 2} =44+{1\over 2}\\ 同理\cos^2 91^\circ +\cos^2 92^\circ +\cdots +\cos^2 179^\circ =44+{1\over 2} \\因此\sum_{n=1}^{179} x_n^2 =2(44+{1\over 2} )=\bbox[red, 2pt]{89}$$

解答：

$$剩下的水體積相當於上圖著色區域繞x軸旋轉的體積，即\pi\int_{3\sqrt 2}^636-x^2\;dx \\ =\pi\left. \left[36x-{1\over 3}x^3 \right] \right|_{3\sqrt 2}^6 =\pi(144-90\sqrt 2)，而半球體積={2\over 3}\pi \cdot 6^3=144\pi\\ 因此溢出的水體積=144\pi-(144-90\sqrt 2)\pi = \bbox[red,2pt]{90\sqrt 2\pi}$$

解答： $$\lim_{n\to \infty}\cfrac{\left({1\over 2n} \right)^p+ \left({2\over 2n} \right)^p+ \cdots +\left({2n\over 2n} \right)^p}{\left({1\over 2}+{1\over 2n} \right)^p+ \left({1\over 2}+{2\over 2n} \right)^p+ \cdots +\left({1\over 2}+{n\over 2n} \right)^p} \\ =\lim_{n\to \infty}\cfrac{{1\over n}\left(\left({1\over 2n} \right)^p+ \left({2\over 2n} \right)^p+ \cdots +\left({2n\over 2n} \right)^p\right)}{{1\over n}\left(\left({1\over 2}+{1\over 2n} \right)^p+ \left({1\over 2}+{2\over 2n} \right)^p+ \cdots +\left({1\over 2}+{n\over 2n} \right)^p\right)} \\ =\cfrac{\lim_{n\to \infty}{1\over n}\left(\left({1\over 2n} \right)^p+ \left({2\over 2n} \right)^p+ \cdots +\left({2n\over 2n} \right)^p\right)}{\lim_{n\to \infty}{1\over n}\left(\left({1\over 2}+{1\over 2n} \right)^p+ \left({1\over 2}+{2\over 2n} \right)^p+ \cdots +\left({1\over 2}+{n\over 2n} \right)^p\right)} \\ =\cfrac{\lim_{n\to \infty}{1\over n}\sum_{k=1}^{2n} \left({k\over 2n} \right)^p}{\lim_{n\to \infty}{1\over n}\sum_{k=1}^{n} \left({1\over 2}+{k\over 2n} \right)^p} =\cfrac{\int_0^2 ({x\over 2})^pdx}{\int_0^1 ({1\over 2}+{x\over 2})^p\;dx} \\ =\cfrac{2^{p+1}/2^p(p+1)}{(2^{p+1}-1)/(p+1)2^p} =\bbox[red,2pt]{2^{p+1} \over 2^{p+1}-1}$$

解答： $$2a+b+{2\over a}+{18\over ab} ={a\over 2} +{a\over 2} +{a\over 2} +{a\over 2} + {b\over 3} + {b\over 3}+ {b\over 3}+ {2\over a}+{6\over ab}+ {6\over ab}+ {6\over ab}\\ \ge 11\sqrt[11]{{a^4\over 2^4}\cdot {b^3\over 3^3}\cdot {2\over a}\cdot {6^3\over a^3b^3}} = \bbox[red,2pt]{11}$$

解答： $$由題意可知，轉換矩陣M=\begin{bmatrix}0 & 1/3 & 1/3 & 1/3\\ 1/3 & 0 & 1/3 & 1/3\\ 1/3 & 1/3 & 0 & 1/3\\ 1/3 & 1/3 & 1/3 & 0\end{bmatrix} \Rightarrow M=PAP^{-1} \\ =\begin{bmatrix}1 & -1 & -1 & -1\\ 1 & 1 & 0 & 0\\ 1 & 0 & 1 & 0\\ 1 & 0 & 0 & 1\end{bmatrix} \begin{bmatrix}1 & 0 & 0 & 0\\ 0 & -1/3 & 0 & 0\\ 0 & 0 & -1/3 & 0\\ 0 & 0 & 0 & -1/3\end{bmatrix} \begin{bmatrix}1/4 & 1/4 & 1/4 & 1/4\\ -1/4 & 3/4 & -1/4 & -1/4\\ -1/4 & -1/4 & 3/4 & -1/4\\ -1/4 & -1/4 & -1/4 & -3/4 \end{bmatrix}\\ \Rightarrow M^{60}=PA^{60}P^{-1} =P\begin{bmatrix}1 & 0 & 0 & 0\\ 0 & 1/3^{60} & 0 & 0\\ 0 & 0 & 1/3^{60} & 0\\ 0 & 0 & 0 & 1/3^{60}\end{bmatrix}P^{-1} \\=\begin{bmatrix}1 & -1/3^{60} & -1/3^{60} & -1/3^{60}\\ 1 & 1/3^{60} & 0 & 0\\ 1 & 0 & 1/3^{60} & 0\\ 1 & 0 & 0 & 1/3^{60}\end{bmatrix}\begin{bmatrix}1/4 & 1/4 & 1/4 & 1/4\\ -1/4 & 3/4 & -1/4 & -1/4\\ -1/4 & -1/4 & 3/4 & -1/4\\ -1/4 & -1/4 & -1/4 & -3/4 \end{bmatrix}\\= \begin{bmatrix} {1\over 4}+{1\over 4}\cdot {1\over 3^{59}} & * & * & * \\ * & * & * & * \\ * & * & * & * \\ * & * & * & *\end{bmatrix} ，其中無需計算之值以*代表\\ 由於從頂點A出發，因此所求之值為PA^{60}P^{-1}\begin{bmatrix} 1\\ 0\\ 0\\0 \end{bmatrix}的第一個元素值，也就是\bbox[red, 2pt]{{1\over 4}+{1\over 4}\cdot {1\over 3^{59}}}$$

解答： $$a,b,c成等差，可取\cases{a=3\\b=4 \\c=5}，則\angle C=90^\circ；因此\cases{\tan {C\over 2} =\tan 45^\circ =1\\ \tan A={3\over 4}={2\tan(A/2)\over 1-\tan^2(A/2)} \Rightarrow \tan {A\over 2}={1\over 3}} \\ \Rightarrow \tan{A\over 2}\cdot \tan{C\over 2} ={1\over 3}\cdot 1=\bbox[red, 2pt]{1\over 3}$$

解答： $$令\cases{L_1:x+3y=4\\ L_2:5x+3y=-16\\ A(4,0)\\ B(4,-12)} \Rightarrow \cases{  L_1與L_2的交點P(-5,3) \\\overline{AB}中點Q(4, -6)} \Rightarrow \cases{\overleftrightarrow{PQ}:x+y=-2\\ \overline{PQ}的中點C(-{1\over 2},-{3\over 2})}\\  拋物線\;\Gamma\;的對稱軸與\overleftrightarrow{PQ}平行，因此將\Gamma 逆時鐘旋轉45^\circ 變成\Gamma'，\Gamma'對稱軸與x軸平行；\\(x,y)逆時鐘旋轉45^\circ變為(x',y')，即\cases{x'=x-y\\ y'=x+y} \Rightarrow \cases{A(4,0)\to A'(4,4)\\ B(4,-12)\to B'(16,-8)\\ C(-1/2,-3/2)\to C'(1,-2)\\ \Gamma':x'=ay'^2+by'+c} \\ A',B',C'皆在\Gamma'上\Rightarrow \cases{16a+4b+c=4\\ 64a-8b+c=16\\ 4a-2b+c=1} \Rightarrow \cases{a=1/4\\ b=c=0} \Rightarrow x'={1\over 4}y'^2 \Rightarrow (x-y)={1\over 4}(x+y)^2 \\ \Rightarrow \Gamma:\bbox[red,2pt]{x^2+2xy+y^2-4x+4y=0}$$

解答： $$\alpha,\beta均為質數，且為x^2+(k^2+ak)x+ k^2+ak+127=0之二根\\ \Rightarrow \cases{\alpha+\beta =-k^2-ak\\ \alpha\beta=k^2+ak+127} \Rightarrow \alpha\beta+\alpha+\beta=127 \Rightarrow (\alpha+1)(\beta+1)=128=32\times 4 \\ \Rightarrow \cases{\alpha=31\\ \beta=3} 代回原式\Rightarrow \cases{31^2+31(k^2+ak)+k^2+ak+127=0\\ 3^2+3(k^2+ak)+k^2+ak+127=0 } \Rightarrow k^2+ak+34=0 \\僅有一根，其判別式為0，即a^2=4\times 34 \Rightarrow a= \bbox[red, 2pt]{2\sqrt{34}} (-2\sqrt{34}不合，因為a為正實數)$$

解答： $$f(x)+f(1-{1\over x})=x+1+{1\over x-1} \Rightarrow f(x)+f({x-1\over x})=x+{x\over x-1} \cdots(1)\\ 將x={x-1\over x}代入(1) \Rightarrow f({x-1\over x})+f\left({{x-1\over x}-1 \over {x-1\over x}} \right) ={x-1\over x}+{{x-1\over x} \over {x-1\over x}-1}\\\quad\Rightarrow f({x-1\over x})+f({1\over 1-x})={x-1\over x}+1-x\cdots(2)\\ 將x={x-1\over x}代入(2) \Rightarrow f\left( {{x-1\over x}-1 \over {x-1\over x}}\right)+ f\left({1\over 1-{x-1\over x}} \right) ={{x-1\over x}-1\over {x-1\over x}}+1-{x-1\over x}\\\quad \Rightarrow f({1\over 1-x})+f(x)={1\over 1-x}+{1\over x} \cdots(3)\\ (1)+(2)+(3) \Rightarrow 2\left( f(x)+f({x-1\over x})+ f({1\over 1-x})\right)=3 \Rightarrow f(x)+f({x-1\over x})+ f({1\over 1-x})={3\over 2} \cdots(4)\\ (4)-(2) \Rightarrow f(x)={1\over 2}-{x-1\over x}+x =\bbox[red,2pt]{x+{1\over x}-{1\over 2}}$$

解答： $$因式分解公式:x^3+y^3+z^3-3xyz =(x+y+z)(x^2+y^2+z^2-(xy+yz+zx)) \\={1\over 2}(x+y+z)((x-y)^2+ (y-z)^2+ (z-x)^2 )\\ \Rightarrow x^3+(-y)^3+(-z)^3-3x(-y)(-z)={1\over 2}(x-y-z)((x+y)^2+(-y+z)^2+(x+z)^2=0\\ \Rightarrow x-y-z=0 \Rightarrow x= y+z\cdots(1)\\將(1)代入\;x^2=2(y+z) \Rightarrow (y+z)^2=2(y+z) \Rightarrow (y+z)(y+z-2)=0\\ \Rightarrow y+z=2 代回(1)\Rightarrow x=2 \Rightarrow a+b+c=x+y+z= 2+2= \bbox[red,2pt]{4}$$

解答：

$$z=a+bi \Rightarrow {z-2\over z+2} ={a-2+bi\over a+2+bi} ={(a-2+bi)(a+2-bi)\over (a+2+bi)(a+2-bi)} ={(a^2+b^2-4)+4bi\over (a+2)^2+b^2} 為一純虛數\\ \Rightarrow \cases{a^2+b^2=4\\ b\ne 0} \Rightarrow z 在圓C_1:x^2+y^2=4上(上圖黑色圓)\\ \Rightarrow \omega=z+i在圓C_2:x^2+(y-1)^2=4 (上圖紅色圓，即紅色圓往上移1單位)\\ M=|\omega+1|^2+|\omega-1|^2 = \overline{RP}^2 +\overline{RQ}^2，其中\cases{P(1,0)\\ Q(-1,0)\\ R在圓C_2上} \\ \Rightarrow R(0,3)，即\omega=3i時M有最大值\Rightarrow |\omega|=\bbox[red,2pt]{3}$$

解答：

$$橢圓\cases{x=m+2\cos \theta\\ y=\sqrt 3\sin \theta} \Rightarrow ({x-m\over 2})^2 +({y\over \sqrt 3})^2=1 \Rightarrow {(x-m)^2\over 4} +{y^2\over 3}=1\Rightarrow \cases{a'=2\\ b'=\sqrt 3} \\  \Rightarrow \cases{左頂點P(m-a',0)=(m-2,0) \\ 右頂點Q(m+a',0)=(m+2,0)}\\ 拋物線\cases{x=t^2+{3\over 2}\\ y=\sqrt 6\cdot t} \Rightarrow x=({y\over \sqrt 6})^2+{3\over 2} ={y^2\over 6}+{3\over 2}\Rightarrow y^2 =6x-9 =6(x-{3\over 2})\Rightarrow 頂點R({3\over 2},0)\\ 橢圓與拋物線均對稱x軸，因此兩圖形有交點代表R\in \overline{PQ}\\，也就是\cases{P=R \Rightarrow m-2=3/2 \Rightarrow m=7/2\\ Q=R \Rightarrow m+2=3/2 \Rightarrow m=-1/2}\\ \Rightarrow 當-{1\over 2}\le m\le {7\over 2}時，兩圖形有交點\Rightarrow a+b=-{1\over 2}+{7\over 2}= \bbox[red,2pt]{3}$$

解答： $$\log_2 x-4\log_{2^2}x+12\log_{2^3}x+ \cdots+n(-2)^{n-1}\log_{2^n}x\\ = \log_2 x-2\log_2 x+4\log_2 x+ \cdots +(-2)^{n-1}\log_2 x \\ =\log_2\left( x\cdot x^{-2}\cdot x^4\cdots x^{(-2)^{n-1}}\right) =(1-2+4-\cdots -2^{n-1})\log_2 x  ={1-(-2)^n \over 3}\log_2 x\\ 因此原式變為:{1-(-2)^n \over 3}\log_2 x \gt {1-(-2)^n \over 3}\log_2 (x^2-2)\\ \Rightarrow x \lt x^2-2 (\because 1-(-2)^n \lt 0) \Rightarrow x^2-x-2 \gt 0\Rightarrow (x-2)(x+1) \gt 0 \\ \Rightarrow \bbox[red,2pt]{x\gt 2}(x為真數，需為正數，因此x\not \lt -1)$$

解答：

$$將圓台展開後如上圖AA'B'B，其中\cases{\overset{\Large\frown}{AA'} =上底面圓周長=2\pi \\\overset{\Large\frown}{BB'} =下底面圓周長=10\pi} \qquad，則\overline{QR}即為所求；\\ 令\angle AOA'=\theta \Rightarrow 2\pi = r\theta = 3\theta \Rightarrow \theta ={2\over 3}\pi \Rightarrow \cos \theta ={\overline{OP}^2 +\overline{OB'}^2- \overline{PB'}^2 \over 2\times \overline{OP}\times \overline{OB'}} \\ \Rightarrow -{1\over 2}={81+225-\overline{PB'}^2 \over 270} \Rightarrow \overline{PB'}=21\\ \triangle OPB'面積={1\over 2}\cdot \overline{OQ}\cdot \overline{PB'} ={1\over 2}\cdot \overline{OP}\cdot \overline{OB'}\sin \angle POB' \Rightarrow 21 \cdot \overline{OQ} =135\cdot {\sqrt 3\over 2}\\ \Rightarrow \overline{OQ}={45\over 14}\sqrt 3 \Rightarrow \overline{QR}= \overline{OQ}-3 ={45\over 14}\sqrt 3-3 ={45\sqrt 3-42\over 14} ={a\sqrt 3+b\over 14} \Rightarrow \cases{a=45\\ b=-42}\\ \Rightarrow a+b= \bbox[red,2pt]{3}\\註:題目的{a\sqrt 3+b\over \color{blue}{12}}，分母的12應該是14$$

二、計算題

解答：

(1) $$f(x)=x^6+x^4+x^2+1=(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_6)\\ \Rightarrow f(-2i)=-64+16-4+1=(-2i-\alpha_1)(-2i-\alpha_2) \cdots (-2i-\alpha_6)\\ \Rightarrow (2i+\alpha_1)(2i+\alpha_2)\cdots (2i+\alpha_6) =\bbox[red,2pt]{-51}$$ (2)

$$x^6+x^4+x^2+1=0 \Rightarrow (x^2-1)(x^6+x^4+x^2+1) =0  \Rightarrow x^8=1\\ \Rightarrow P_1,P_2,\dots,P_6為單位圓上的正八邊形的八個頂點扣除(1,0),(-1,0)所剩的六個頂點，見上圖；\\六邊形面積=正方形P_1P_3P_4P_6+ \triangle P_1P_2P_3 + \triangle P_4P_5P_6 = (\sqrt 2)^2 + \sqrt 2\times (1-{\sqrt 2\over 2})\\ =2+\sqrt 2-1 =\bbox[red,2pt]{\sqrt 2+1}$$

解答： $$f(x)=\int_0^x(x-t)\cos^3t\;dt = \left. \left[ {1\over 36}(3(x-t)(9\sin t+\sin(3t))-27\cos t-\cos(3t))\right] \right|_0^x\\ ={1\over 36}(-27\cos x-\cos(3x))-{1\over 36}(-27-1) ={1\over 36}(28-27\cos x-\cos(3x)) \\f'(x)=0 \Rightarrow {1\over 36}(27\sin x+3\sin(3x))=0 \Rightarrow 9\sin x+\sin (3x)=0 \\ \Rightarrow 9\sin x+3\sin x-3\sin^2x -\sin^3 x=0 \Rightarrow \sin^3x-3\sin^2x+12\sin x=0\\ \Rightarrow \sin x(\sin^2x-3\sin x+12)=0 \Rightarrow \sin x=0 \Rightarrow x=0,\pi \Rightarrow \bbox[red,2pt]{\cases{f(0)=0為極小值\\ f(\pi) =14/9為極大值\quad}}$$
 

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