114年中崙高中教甄3-數學詳解

 臺北市立中崙高級中學114學年度第三次教師甄選

一、填充題(每格7分,共77分)

解答:$$\text{Shell Method: }圓: x^2+y^2=25 \Rightarrow x= \sqrt{25-y^2} \Rightarrow V=\int_3^4 2\pi\cdot y\cdot 2x\,dy \\ =\int_3^4 2\pi\cdot y\cdot 2\sqrt{25-y^2}\,dy  =\int_{16}^{9} -2\pi \sqrt u\,du = -2\pi \left. \left[ {2\over 3}u^{3/2} \right] \right|_{16}^9 = \bbox[red, 2pt]{{148\over 3}\pi}$$


解答:$$\cases{F(x)=\int_0^x f(t)\, dt \Rightarrow F'(x)=f(x) \cdots(1)\\ G(x)=\int_{-x}^{2x} f(t) \,dt \Rightarrow G'(x)=2f(2x)+f(-x) \cdots(2)} \Rightarrow \cases{a=2 \\b=1\\ c=0} \\又\int_0^x G'(t)\,dt =p\cdot F(2x)+q \cdot F(-x) \Rightarrow G'(x)=2p\cdot F'(2x)-qF'(-x)\cdots(3) \\由式(2) \Rightarrow G'(x)=2 f(2x) +f(-x) 再由式(1) 可得G'(x)=2 F'(2x)+F'(-x) \cdots(4) \\ (3) =(4) \Rightarrow \cases{p=1\\ \\q=-1} \Rightarrow 10p+q=9 \Rightarrow 10000a+1000b+100c+10p+q= \bbox[red, 2pt]{21009}$$

解答:$$f(x)=b \cdot a^x  \xrightarrow{向右平移2單位} f(x-2) =b\cdot a^{x-2} \xrightarrow{垂直伸縮3倍}3b\cdot a^{x-2} \\ \Rightarrow b\cdot a^x =3b\cdot a^{x-2} \Rightarrow a^2=2 \Rightarrow a=\sqrt 3 \\f(x)=b\cdot a^x \xrightarrow {水平伸縮3倍}b\cdot a^{x/3} \xrightarrow {水平右移2單位}b\cdot a^{(x-2)/3 } \Rightarrow g(x)=b\cdot a^{(x-2)/3 } \\ \Rightarrow f(x)= b\cdot a^x =g(x)^3 = \left( b\cdot a^{(x-2)/3 } \right)^3 =b^3\cdot a^{x-2} \Rightarrow b^2=a^2 \Rightarrow b=\sqrt 3 \\ \Rightarrow y=f(x)=\sqrt 3 \left( \sqrt 3 \right)^x =\left( \sqrt 3 \right)^{x+1}  \Rightarrow \log_{\sqrt 3} y=x+1 \Rightarrow x=-1+\log_{\sqrt 3}y \\ \Rightarrow f^{-1}(x)=-1+\log_{\sqrt 3}x=-1+{2\over \log 3} \log x =(r+s\log x) \Rightarrow (r,s )= \bbox[red, 2pt]{\left( -1,{2\over \log 3} \right)}$$

解答:

$$P_\theta(\cos \theta, \sin \theta), 0\le \theta\le \pi \Rightarrow P_\theta在單位圓的上半部, \left| \overrightarrow{OQ}\right| =\overline{OQ}\le 2 \Rightarrow Q(x,y)在半徑為2的圓內\\ 又\overrightarrow{OP_\theta} \cdot \overrightarrow{OQ} =x\cos \theta+ y\sin \theta=1為一直線L \Rightarrow \cases{\theta=0^\circ \Rightarrow L: x=1\\ \theta=90^\circ \Rightarrow L:y=1\\ \theta=180^\circ \Rightarrow L:x=-1} \\ \Rightarrow \Lambda區域即為上圖著色區域 \Rightarrow 面積=S(綠色)+2R(棕色) \\\Rightarrow \cases{S= \displaystyle {1\over 2}(2^2-1^2)\pi ={3\over 2}\pi \\ R=扇形OAB-\triangle OBC=\displaystyle {2\over 3}\pi-{\sqrt 3\over 2}} \Rightarrow S+2R={3\over 2}\pi+{4\over 3}\pi-\sqrt 3= \bbox[red, 2pt]{{17\over 6}\pi-\sqrt 3}$$

解答:$$假設\cases{L_1:{x-3\over 2} ={y-3\over 2} ={z-1\over 1} \\L_2:\cases{x=1\\ 4y-3z=4}} \Rightarrow \cases{L_1方向向量 \vec u_1=(2,2,1)\\ L_2方向向量\vec u_2=(1,0,0)\times (0,4,-3) =(0,3,4)}\\ \Rightarrow \cases{\vec e_1={ \vec u_1/||\vec u_1||} =(2/3,2/3,1/3) \\ \vec e_2=\vec u_2/||\vec u_2|| =(0,3/5,4/5)}  \Rightarrow \cases{\overrightarrow{AB} =\overline{AB}\cdot \vec e_1 =(4,4,2) \\ \overrightarrow{AC} = \overline{AC}\cdot \vec e_2 =(0,3,4)} \\ 底面積大小= || \overrightarrow{AB} \times \overrightarrow{AC}|| = ||(10,-16,12) ||= \sqrt{10^2+ 16^2+ 12^2} =10\sqrt 5\\底面法向量\vec n \parallel (\overrightarrow{AB} \times \overrightarrow{AC}) \Rightarrow \vec n=(5,-8,6) \Rightarrow A=L_1\cap L_2 =(1,1,0)  \\ \Rightarrow 底面通過A點且法向量為\vec n之平面\Rightarrow 5(x-1)-8(y-1)+6z=0 \Rightarrow 5x-8y+6z+3=0\\ 假設另一平面頂點P\in yz平面\Rightarrow P(0,y_0,z_0), 又\overline{OP}=1 \Rightarrow y_0^2+z_0^2=1 \\ \Rightarrow 六面積的高h=d(P,底面) = {|-8y_0+6z_0+3|\over \sqrt{5^2+8^2+6^2}} \Rightarrow 體積最大值取決於h的最大值\\ 柯西不等等: ((-8)^2+6^2)(y_0^2+z_0^2) \ge(-8y_0 +6z_0)^2 \Rightarrow -8y_0 +6z_0\le 10 \\ \Rightarrow h最大值={10+3\over \sqrt{125}} \Rightarrow 體積最大值=10\sqrt 5\times {13\over 5\sqrt 5} =\bbox[red, 2pt]{26}$$

解答:$$\cases{\Gamma: {x^2\over a^2}+{y^2\over b^2}=1 \\ P(x_0,y_0)\in \Gamma} \Rightarrow \cases{{x_0^2\over a^2} +{y_0^2\over b^2}=1 \\ Q (x_0,0) \\ R(0,y_0) } \Rightarrow \cases{\overline{PQ} =|y_0|\\ \overline{PR} =|x_0|\\ \overline{QR} =\sqrt{x_0^2+y_0^2}}\\ 已知\cases{面積=ab \pi=5\pi \\ \overline{QR} = \sqrt 6=\sqrt{x_0^2+y_0^2} \\ \overline{PQ}: \overline{PR} =|y_0|:|x_0|=b^2:a^2} \Rightarrow \cases{ab=5\\ x_0^2+y_0^2=6\\ y_0^2:x_0^2=b^4:a^4} \\ \Rightarrow y_0^2={b^4\over a^4}x_0^2 \Rightarrow {x_0^2\over a^2} +{b^2x_0^2\over a^4} =1 \Rightarrow x_0^2 ={a^4\over a^2+b^2} \Rightarrow y_0^2 ={b^4\over a^2+b^2} \\\Rightarrow x_0^2+y_0^2 =6={a^4+b^4\over a^2+b^2} ={(a^2+b^2)^2- 2a^2b^2\over a^2+b^2} ={(a^2+b^2)^2- 50\over a^2+b^2} \\ \Rightarrow (a^2+b^2)^2-6(a^2+b^2)-50=0 \Rightarrow a^2+b^2={ 6+\sqrt{236}\over 2} = \bbox[red, 2pt]{3+\sqrt{59}}$$

解答:$$\begin{bmatrix}1&2& 3\\ 2& 3& 1\\ 3& 2& 1 \end{bmatrix} \begin{bmatrix}{1\over x-2y} \\{1\over y-2z} \\{1\over z-2x} \end{bmatrix} = \begin{bmatrix}-2\\ -3\\ -4 \end{bmatrix} \Rightarrow \begin{bmatrix}{1\over x-2y} \\{1\over y-2z} \\{1\over z-2x} \end{bmatrix} =\begin{bmatrix}1&2& 3\\ 2& 3& 1\\ 3& 2& 1 \end{bmatrix}^{-1} \begin{bmatrix}-2\\ -3\\ -4 \end{bmatrix} = \begin{bmatrix} \frac{-1}{12} & \frac{-1}{3} & \frac{7}{12} \\\frac{-1}{12} & \frac{2}{3} & \frac{-5}{12} \\\frac{5}{12} & \frac{-1}{3} & \frac{1}{12} \end{bmatrix}  \begin{bmatrix}-2\\ -3\\ -4 \end{bmatrix} \\= \begin{bmatrix}\frac{-7}{6} \\\frac{-1}{6} \\\frac{-1}{6} \end{bmatrix} \Rightarrow \cases{x-2y=-6/7\\ y-2z=-6\\ z-2x=-6} \Rightarrow \begin{bmatrix}1&-2 &0 \\0& 1& -2\\ -2&0& 1 \end{bmatrix} \begin{bmatrix}x\\ y\\ z \end{bmatrix} = \begin{bmatrix}-6/7\\ -6\\ -6 \end{bmatrix} \\\Rightarrow \begin{bmatrix}x\\ y\\ z \end{bmatrix} =\begin{bmatrix}1&-2 &0 \\0& 1& -2\\ -2&0& 1 \end{bmatrix}^{-1} \begin{bmatrix}-6/7\\ -6\\ -6 \end{bmatrix} = \begin{bmatrix} \frac{-1}{7} & \frac{-2}{7} & \frac{-4}{7} \\\frac{-4}{7} & \frac{-1}{7} & \frac{-2}{7} \\ \frac{-2}{7} & \frac{-4}{7} & \frac{-1}{7} \end{bmatrix} \begin{bmatrix}-6/7\\ -6\\ -6 \end{bmatrix} = \begin{bmatrix} \frac{258}{49} \\\frac{150}{49} \\\frac{222}{49} \end{bmatrix} \\ \Rightarrow x+y+z={630\over 49} = \bbox[red, 2pt]{90\over 7}$$

解答:$$假設\cases{事件A:2黑球不相鄰 \\事件B:3紅球不相鄰}\\ \quad 2白、3紅球排列數:{5!\over 2!3!}=10，再將2黑球插入6個間隙，有{6\choose 2}=15種插法\\ \qquad \Rightarrow n(A)=10\times 15=150\\ 2黑、2白球排列數:{4!\over 2!2!}=6，再將3紅球插入5個間隙，有{5\choose 3}=10種插法\\ \Rightarrow n(B)=6\times 10=60\\接著計算n(A'\cap B), 2個相鄰黑球與2個白球排列數:{3!\over 2!}=3，再將3紅球插入4個間隙，\\有{4\choose 3}=4種插法\Rightarrow n(A'\cap B) =3\times 4=12 \Rightarrow n(A\cap B)=60-12=48 \\ \Rightarrow P(B\mid A)= {n(A\cap B) \over n(A)} ={48\over 150} = \bbox[red, 2pt]{8\over 25}$$

解答:$$\log_a \sqrt b+ \log_{a^2} \sqrt[3] b +\log_{a^3} \sqrt[4]b + \cdots + \log_{a^{10}} \sqrt[11]b \\ ={1\over 2}  \log_a b +{1\over 2\cdot 3}\log_a b +{1\over 3\cdot 4}\log_a b+ \cdots +{1\over 10\cdot 11} \log_a b \\ =\log_a b \left( 1-{1\over 2}+{1\over 2}-{1\over 3}+{1\over 3}-{1\over 4}+\cdots+{1\over 10}-{1\over 11} \right) =\log_a b \left( 1-{1\over 11} \right) ={10\over 11} \log_a b=1 \\ \Rightarrow \log_a b={11\over 10} \Rightarrow \log_{a^{n}} \sqrt[n+1]b ={1\over n\cdot (n+1)} \log_a b={1\over n(n+1)} \cdot {11\over 10} \lt 0.001 \\ \Rightarrow {1\over n(n+1)} \lt {0.01\over 11} ={1\over 1100} \Rightarrow n=\bbox[red, 2pt]{33}$$

解答:$$假設\cases{A(a,0) \\ \overleftrightarrow{AB} 在x軸上 \\ C(\alpha,\beta)}  \Rightarrow \cases{B(ta,0) \\ D(k\alpha, k\beta)} \Rightarrow \cases{\overrightarrow{AD} =(k\alpha-a,k\beta) \\ \overrightarrow{AC} =(\alpha-a,\beta) \\ \overrightarrow{AB} =((t-1)a,0)} \\ \Rightarrow (k\alpha-a,k\beta)=3 ((t-1)a,0)+2 (\alpha-a,\beta)  \Rightarrow \cases{k \alpha-a=(3t-5)a+2\alpha\\ k\beta=2\beta} \Rightarrow \cases{k=2 \\ t=4/3} \\ \Rightarrow \cases{A(a,0) \\B({4a\over 3},0) \\C(\alpha, \beta) \\D(2\alpha, 2\beta )} \Rightarrow  \begin{vmatrix} a&0\\ 4a/3& 0 \end{vmatrix} + \begin{vmatrix} 4a/3& 0 \\ \alpha& \beta\end{vmatrix} + \begin{vmatrix} \alpha & \beta\\ 2\alpha & 2\beta\end{vmatrix} +  \begin{vmatrix}  2\alpha & 2\beta \\ a& 0\end{vmatrix} =1 \\ \Rightarrow 0+ {4\over 3} a\beta+0-2a\beta =1 \Rightarrow a\beta=-{3\over 2} \Rightarrow \triangle ABC ={1\over 2} \begin{Vmatrix} \overrightarrow{AB} \\ \overrightarrow{AC} \end{Vmatrix} ={1\over 2}\begin{Vmatrix} a/3& 0 \\ \alpha-a& \beta\end{Vmatrix} \\={1\over 6} |a\beta|= \bbox[red, 2pt]{1\over 4}$$

解答:$$假設\cases{O為外心\\ I為內心} \Rightarrow d=\overline{OI} \Rightarrow d^2=R(R-2r)\ge 0 \Rightarrow R\ge 2r\Rightarrow R/r\ge 2 \Rightarrow 最小值\bbox[red, 2pt]2$$

二、計算證明題(23分)

解答:$$k=n^3-13n-9=(n-1)n(n+1)-3(4n+3)為3的倍數 \\ \Rightarrow k\bbox[red, 2pt]{是質數又是3的倍數}\Rightarrow k=3 \Rightarrow n^3-13n-9=3 \Rightarrow n^3-13n-12=0 \\ \Rightarrow (n+1)(n+3)(n-4)=0 \Rightarrow n=\bbox[red, 2pt]{-3或-1或4}$$

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